#3197Hard~60 min

Find the Minimum Area to Cover All Ones II

Time O(max(n, m)^2) · Space O(max(n, m)^2) · Official statement on LeetCode

dynamic programming prefix sum binary search dynamic programming

Solutions

// Time:  O(max(n, m)^2)
// Space: O(max(n, m)^2)

// dp
class Solution {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& cmp = [](int x, int y) {
            return x < y ? -1 : (x > y ? 1 : 0);
        };

        const auto& count = [&](int start1, int end1, int start2, int end2) {
            vector<vector<int>> dp(size(grid), vector<int>(size(grid[0])));
            vector<int> up(size(grid[0]), size(grid));
            vector<int> down(size(grid[0]), -1);
            for (int i = start1, d1 = cmp(end1, start1); i != end1; i += d1) {
                int l = size(grid[0]);
                int r = -1;
                int u = size(grid);
                int d = -1;
                for (int j = start2, d2 = cmp(end2, start2); j != end2; j += d2) {
                    if (grid[i][j]) {
                        up[j] = min(up[j], i);
                        down[j] = max(down[j], i);
                    }
                    u = min(u, up[j]);
                    d = max(d, down[j]);
                    if (down[j] >= 0) {
                        l = min(l, j);
                        r = max(r, j);
                    }
                    dp[i][j] = r >= 0 ? (r - l + 1) * (d - u + 1) : 0;
                }
            }
            return dp;
        };
        
        const auto& count2 = [&](bool is_vertical) {
            const auto& get_n = [&]() {
                return !is_vertical ? size(grid) : size(grid[0]);
            };

            const auto& get_m = [&]() {
                return !is_vertical ? size(grid[0]) : size(grid);
            };

            const auto& get = [&](int i, int j) {
                return !is_vertical ? grid[i][j] : grid[j][i];
            };

            vector<int> left(get_n(), get_m());
            vector<int> right(get_n(), -1);
            for (int i = 0; i < get_n(); ++i) {
                for (int j = 0; j < get_m(); ++j) {
                    if (get(i, j) == 0) {
                        continue;
                    }
                    left[i] = min(left[i], j);
                    right[i] = max(right[i], j);
                }
            }
            vector<vector<int>> dp(get_n(), vector<int>(get_n()));
            for (int i = 0; i < size(dp); ++i) {
                int l = get_m();
                int r = -1;
                int u = get_n();
                int d = -1;
                for (int j = i; j < size(dp[0]); ++j) {
                    if (right[j] != -1) {
                        l = min(l, left[j]);
                        r = max(r, right[j]);
                        u = min(u, j);
                        d = max(d, j);
                    }
                    dp[i][j] = r >= 0 ? (r - l + 1) * (d - u + 1) : 0;
                }
            }
            return dp;
        };

        const auto& up_left = count(0, size(grid), 0, size(grid[0]));
        const auto& up_right = count(0, size(grid), size(grid[0]) - 1, -1);
        const auto& down_left = count(size(grid) - 1, -1, 0, size(grid[0]));
        const auto& down_right = count(size(grid) - 1, -1, size(grid[0]) - 1, -1);
        const auto& horizon = count2(false);
        const auto& vertical = count2(true);
        int result = numeric_limits<int>::max();
        for (int i = 0; i + 1 < size(grid); ++i) {
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                result = min({result,
                    up_left[i][j] + up_right[i][j + 1] + horizon[i + 1][size(grid) - 1],
                    horizon[0][i] + down_left[i + 1][j] + down_right[i + 1][j + 1],
                    up_left[i][j] + down_left[i + 1][j] + vertical[j + 1][size(grid[0]) - 1],
                    vertical[0][j] + up_right[i][j + 1] + down_right[i + 1][j + 1]
                });
            }
        }
        for (int i = 0; i + 2 < size(grid); ++i) {
            for (int j = i + 1; j + 1 < size(grid); ++j) {
                result = min(result, horizon[0][i] + horizon[i + 1][j] + horizon[j + 1][size(grid) - 1]);
            }
        }
        for (int i = 0; i + 2 < size(grid[0]); ++i) {
            for (int j = i + 1; j + 1 < size(grid[0]); ++j) {
                result = min(result, vertical[0][i] + vertical[i + 1][j] + vertical[j + 1][size(grid[0]) - 1]);
            }
        }
        return result;
    }
};

// Time:  O(n * m * log(max(n, m)) + max(n, m)^2)
// Space: O(n * m * log(max(n, m)))
// sparse table
class Solution2 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        vector<SparseTable> st_min_i, st_max_i, st_min_j, st_max_j;
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            const int min_r = min(st_min_i[(size(grid) - 1) - min_i].query(min_j, max_j), max_i + 1);
            const int max_r = max(st_max_i[max_i].query(min_j, max_j), min_i - 1);
            const int min_c = min(st_min_j[(size(grid[0]) - 1) - min_j].query(min_i, max_i), max_j + 1);
            const int max_c = max(st_max_j[max_j].query(min_i, max_i), min_j - 1);
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };

        vector<int> curr;
        curr.assign(size(grid[0]), size(grid));
        for (int i = size(grid) - 1; i >= 0; --i) {
            for (int j = 0; j < size(grid[0]); ++j) {
                if (grid[i][j]) {
                    curr[j] = i;
                }
            }
            st_min_i.emplace_back(SparseTable(curr, [&](int i, int j) { return min(i, j); }));
        }
        curr.assign(size(grid[0]), -1);
        for (int i = 0; i < size(grid); ++i) {
            for (int j = 0; j < size(grid[0]); ++j) {
                if (grid[i][j]) {
                    curr[j] = i;
                }
            }
            st_max_i.emplace_back(SparseTable(curr, [&](int i, int j) { return max(i, j); }));
        }
        curr.assign(size(grid), size(grid[0]));
        for (int j = size(grid[0]) - 1; j >= 0; --j) {
            for (int i = 0; i < size(grid); ++i) {
                if (grid[i][j]) {
                    curr[i] = j;
                }
            }
            st_min_j.emplace_back(SparseTable(curr, [&](int i, int j) { return min(i, j); }));
        }
        curr.assign(size(grid), -1);
        for (int j = 0; j < size(grid[0]); ++j) {
            for (int i = 0; i < size(grid); ++i) {
                if (grid[i][j]) {
                    curr[i] = j;
                }
            }
            st_max_j.emplace_back(SparseTable(curr, [&](int i, int j) { return max(i, j); }));
        }
        int result = numeric_limits<int>::max();
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(i + 1, size(grid) - 1, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, 0, j);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, 0, j);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = i + 1; j + 1 < size(grid); ++j) {
                const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid[0]); ++i) {
            int a = minimumArea(0, size(grid) - 1, 0, i);
            for (int j = i + 1; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, size(grid) - 1, i + 1, j);
                const int c = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        return result;
    }

private:
    // Reference: https://cp-algorithms.com/data_structures/sparse-table.html
    class SparseTable {
    public:
        SparseTable(const vector<int>& arr, function<int (int, int)> fn)
         :  fn(fn) {  // Time: O(nlogn) * O(fn) = O(nlogn), Space: O(nlogn)
            const int n = size(arr);
            const int k = __lg(n);
            st.assign(k + 1, vector<int>(n));
            st[0].assign(cbegin(arr), cend(arr));
            for (int i = 1; i <= k; ++i) {
                for (int j = 0; j + (1 << i) <= n; ++j) {
                    st[i][j] = fn(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
                }
            }
         }

        int query(int L, int R) const {
            const int i = __lg(R - L + 1);
            return fn(st[i][L], st[i][R - (1 << i) + 1]);  // Time: O(fn) = O(1)
        }
    
    private:
        vector<vector<int>> st;
        const function<int (int, int)>& fn;
    };
};

// Time:  O(n * m * log(max(n, m)) + max(n, m)^2)
// Space: O(n * m * log(max(n, m)))
// sparse table
class Solution3 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& rotate = [](const auto& grid) {
            vector<vector<int>> result(size(grid[0]), vector<int>(size(grid)));
            for (int i = 0; i < size(grid); ++i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    result[j][(size(grid) - 1) - i] = grid[i][j];
                }
            }
            return result;
        };

        vector<SparseTable> st_min_i, st_max_i, st_min_j, st_max_j;
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            const int min_r = min(st_min_i[(size(grid) - 1) - min_i].query(min_j, max_j), max_i + 1);
            const int max_r = max(st_max_i[max_i].query(min_j, max_j), min_i - 1);
            const int min_c = min(st_min_j[(size(grid[0]) - 1) - min_j].query(min_i, max_i), max_j + 1);
            const int max_c = max(st_max_j[max_j].query(min_i, max_i), min_j - 1);
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };
    
        vector<int> curr;
        int result = numeric_limits<int>::max();
        for (int _ = 0; _ < 4; ++_) {
            st_min_i.clear();
            curr.assign(size(grid[0]), size(grid));
            for (int i = size(grid) - 1; i >= 0; --i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    if (grid[i][j]) {
                        curr[j] = i;
                    }
                }
                st_min_i.emplace_back(SparseTable(curr, [&](int i, int j) { return min(i, j); }));
            }
            st_max_i.clear();
            curr.assign(size(grid[0]), -1);
            for (int i = 0; i < size(grid); ++i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    if (grid[i][j]) {
                        curr[j] = i;
                    }
                }
                st_max_i.emplace_back(SparseTable(curr, [&](int i, int j) { return max(i, j); }));
            }
            st_min_j.clear();
            curr.assign(size(grid), size(grid[0]));
            for (int j = size(grid[0]) - 1; j >= 0; --j) {
                for (int i = 0; i < size(grid); ++i) {
                    if (grid[i][j]) {
                        curr[i] = j;
                    }
                }
                st_min_j.emplace_back(SparseTable(curr, [&](int i, int j) { return min(i, j); }));
            }
            st_max_j.clear();
            curr.assign(size(grid), -1);
            for (int j = 0; j < size(grid[0]); ++j) {
                for (int i = 0; i < size(grid); ++i) {
                    if (grid[i][j]) {
                        curr[i] = j;
                    }
                }
                st_max_j.emplace_back(SparseTable(curr, [&](int i, int j) { return max(i, j); }));
            }
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = 0; j + 1 < size(grid[0]); ++j) {
                    const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                    const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            for (int i = 0; i + 2 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = i + 1; j + 1 < size(grid); ++j) {
                    const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                    const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            grid = rotate(grid);
        }
        return result;
    }

private:
    // Reference: https://cp-algorithms.com/data_structures/sparse-table.html
    class SparseTable {
    public:
        SparseTable(const vector<int>& arr, function<int (int, int)> fn)
         :  fn(fn) {  // Time: O(nlogn) * O(fn) = O(nlogn), Space: O(nlogn)
            const int n = size(arr);
            const int k = __lg(n);
            st.assign(k + 1, vector<int>(n));
            st[0].assign(cbegin(arr), cend(arr));
            for (int i = 1; i <= k; ++i) {
                for (int j = 0; j + (1 << i) <= n; ++j) {
                    st[i][j] = fn(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
                }
            }
         }

        int query(int L, int R) const {
            const int i = __lg(R - L + 1);
            return fn(st[i][L], st[i][R - (1 << i) + 1]);  // Time: O(fn) = O(1)
        }
    
    private:
        vector<vector<int>> st;
        const function<int (int, int)>& fn;
    };
};

// Time:  O(max(n, m)^2 * log(max(n, m)))
// Space: O(1)
// prefix sum, binary search
class Solution4 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& binary_search = [](auto left, auto right, const auto& check) {
            while (left <= right) {
                const auto mid = left + (right - left) / 2;
                if (check(mid)) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            return left;
        };

        const auto& binary_search_right = [](auto left, auto right, const auto& check) {
            while (left <= right) {
                const auto mid = left + (right - left) / 2;
                if (!check(mid)) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            return right;
        };
        
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            const auto& count = [&](int x1, int y1, int x2, int y2) {
                int cnt = grid[x2][y2];
                if (x1 - 1 >= 0) {
                    cnt -= grid[x1 - 1][y2];
                }
                if (y1 - 1 >= 0) {
                    cnt -= grid[x2][y1 - 1];
                }
                if (x1 - 1 >= 0 && y1 - 1 >= 0) {
                    cnt += grid[x1 - 1][y1 - 1];
                }
                return cnt;
            };

            const int min_r = binary_search(min_i, max_i, [&](int i) { return count(min_i, min_j, i, max_j); });
            const int max_r = binary_search_right(min_i, max_i, [&](int i) { return count(i, min_j, max_i, max_j); });
            const int min_c = binary_search(min_j, max_j, [&](int j) { return count(min_i, min_j, max_i, j); });
            const int max_c = binary_search_right(min_j, max_j, [&](int j) { return count(min_i, j, max_i, max_j); });
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };

        for (int i = 0; i < size(grid); ++i) {
            for (int j = 0; j < size(grid[0]); ++j) {
                if (i - 1 >= 0) {
                    grid[i][j] += grid[i - 1][j];
                }
                if (j - 1 >= 0) {
                    grid[i][j] += grid[i][j - 1];
                }
                if (i - 1 >= 0 && j - 1 >= 0) {
                    grid[i][j] -= grid[i - 1][j - 1];
                }
            }
        }
        int result = numeric_limits<int>::max();
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(i + 1, size(grid) - 1, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, 0, j);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, 0, j);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = i + 1; j + 1 < size(grid); ++j) {
                const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid[0]); ++i) {
            int a = minimumArea(0, size(grid) - 1, 0, i);
            for (int j = i + 1; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, size(grid) - 1, i + 1, j);
                const int c = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        return result;
    }
};

// Time:  O(max(n, m)^2 * log(max(n, m)))
// Space: O(n * m)
// prefix sum, binary search
class Solution5 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& rotate = [](const auto& grid) {
            vector<vector<int>> result(size(grid[0]), vector<int>(size(grid)));
            for (int i = 0; i < size(grid); ++i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    result[j][(size(grid) - 1) - i] = grid[i][j];
                }
            }
            return result;
        };
    
        const auto& binary_search = [](auto left, auto right, const auto& check) {
            while (left <= right) {
                const auto mid = left + (right - left) / 2;
                if (check(mid)) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            return left;
        };

        const auto& binary_search_right = [](auto left, auto right, const auto& check) {
            while (left <= right) {
                const auto mid = left + (right - left) / 2;
                if (!check(mid)) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            return right;
        };

        vector<vector<int>> prefix;
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            const auto& count = [&](int x1, int y1, int x2, int y2) {
                int cnt = prefix[x2][y2];
                if (x1 - 1 >= 0) {
                    cnt -= prefix[x1 - 1][y2];
                }
                if (y1 - 1 >= 0) {
                    cnt -= prefix[x2][y1 - 1];
                }
                if (x1 - 1 >= 0 && y1 - 1 >= 0) {
                    cnt += prefix[x1 - 1][y1 - 1];
                }
                return cnt;
            };

            const int min_r = binary_search(min_i, max_i, [&](int i) { return count(min_i, min_j, i, max_j); });
            const int max_r = binary_search_right(min_i, max_i, [&](int i) { return count(i, min_j, max_i, max_j); });
            const int min_c = binary_search(min_j, max_j, [&](int j) { return count(min_i, min_j, max_i, j); });
            const int max_c = binary_search_right(min_j, max_j, [&](int j) { return count(min_i, j, max_i, max_j); });
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };

        int result = numeric_limits<int>::max();
        for (int _ = 0; _ < 4; ++_) {
            prefix.assign(size(grid), vector<int>(size(grid[0])));
            for (int i = 0; i < size(grid); ++i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    prefix[i][j] = grid[i][j];
                    if (i - 1 >= 0) {
                        prefix[i][j] += prefix[i - 1][j];
                    }
                    if (j - 1 >= 0) {
                        prefix[i][j] += prefix[i][j - 1];
                    }
                    if (i - 1 >= 0 && j - 1 >= 0) {
                        prefix[i][j] -= prefix[i - 1][j - 1];
                    }
                }
            }
             for (int i = 0; i + 1 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = 0; j + 1 < size(grid[0]); ++j) {
                    const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                    const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            for (int i = 0; i + 2 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = i + 1; j + 1 < size(grid); ++j) {
                    const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                    const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            grid = rotate(grid);
        }
        return result;
    }
};

// Time:  O((n^2 + m^2 + 4 * n * m) * n * m) = O(max(n, m)^3 * min(n, m))
// Space: O(1)
// brute force
class Solution6 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            int min_r = max_i + 1;
            int max_r = min_i - 1;
            int min_c = max_j + 1;
            int max_c = min_j - 1;
            for (int i = min_i; i <= max_i; ++i) {
                for (int j = min_j; j <= max_j; ++j) {
                    if (grid[i][j] == 0) {
                        continue;
                    }
                    min_r = min(min_r, i);
                    max_r = max(max_r, i);
                    min_c = min(min_c, j);
                    max_c = max(max_c, j);
                }
            }
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };

        int result = numeric_limits<int>::max();
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(i + 1, size(grid) - 1, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 1 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = 0; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, 0, j);
                const int c = minimumArea(i + 1, size(grid) - 1, 0, j);
                result = min(result, a + b + c);
            }
        }
        for (int j = 0; j + 1 < size(grid[0]); ++j) {
            int a = minimumArea(0, size(grid) - 1, 0, j);
            for (int i = 0; i + 1 < size(grid); ++i) {
                const int b = minimumArea(0, i, j + 1, size(grid[0]) - 1);
                const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid); ++i) {
            int a = minimumArea(0, i, 0, size(grid[0]) - 1);
            for (int j = i + 1; j + 1 < size(grid); ++j) {
                const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        for (int i = 0; i + 2 < size(grid[0]); ++i) {
            int a = minimumArea(0, size(grid) - 1, 0, i);
            for (int j = i + 1; j + 1 < size(grid[0]); ++j) {
                const int b = minimumArea(0, size(grid) - 1, i + 1, j);
                const int c = minimumArea(0, size(grid) - 1, j + 1, size(grid[0]) - 1);
                result = min(result, a + b + c);
            }
        }
        return result;
    }
};

// Time:  O((n^2 + m^2 + 4 * n * m) * n * m) = O(max(n, m)^3 * min(n, m))
// Space: O(n * m)
// brute force
class Solution7 {
public:
    int minimumSum(vector<vector<int>>& grid) {
        const auto& rotate = [](const auto& grid) {
            vector<vector<int>> result(size(grid[0]), vector<int>(size(grid)));
            for (int i = 0; i < size(grid); ++i) {
                for (int j = 0; j < size(grid[0]); ++j) {
                    result[j][(size(grid) - 1) - i] = grid[i][j];
                }
            }
            return result;
        };
    
        const auto& minimumArea = [&](int min_i, int max_i, int min_j, int max_j) {
            int min_r = max_i + 1;
            int max_r = min_i - 1;
            int min_c = max_j + 1;
            int max_c = min_j - 1;
            for (int i = min_i; i <= max_i; ++i) {
                for (int j = min_j; j <= max_j; ++j) {
                    if (grid[i][j] == 0) {
                        continue;
                    }
                    min_r = min(min_r, i);
                    max_r = max(max_r, i);
                    min_c = min(min_c, j);
                    max_c = max(max_c, j);
                }
            }
            return min_r <= max_i ? (max_r - min_r + 1) * (max_c - min_c + 1) : 0;
        };

        int result = numeric_limits<int>::max();
        for (int _ = 0; _ < 4; ++_) {
             for (int i = 0; i + 1 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = 0; j + 1 < size(grid[0]); ++j) {
                    const int b = minimumArea(i + 1, size(grid) - 1, 0, j);
                    const int c = minimumArea(i + 1, size(grid) - 1, j + 1, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            for (int i = 0; i + 2 < size(grid); ++i) {
                const int a = minimumArea(0, i, 0, size(grid[0]) - 1);
                for (int j = i + 1; j + 1 < size(grid); ++j) {
                    const int b = minimumArea(i + 1, j, 0, size(grid[0]) - 1);
                    const int c = minimumArea(j + 1, size(grid) - 1, 0, size(grid[0]) - 1);
                    result = min(result, a + b + c);
                }
            }
            grid = rotate(grid);
        }
        return result;
    }
};

Beginner Explanation

What is Find the Minimum Area to Cover All Ones II?

Find the Minimum Area to Cover All Ones II (LeetCode #3197) is a Hard problem that primarily trains dynamic programming.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with prefix sum, binary search, and dynamic programming.
Only then translate the idea into code.

Why this problem matters

Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Array, Brute Force, Prefix Sum, Binary Search.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Find the Minimum Area to Cover All Ones II

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to prefix sum, binary search, and dynamic programming.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(max(n, m)^2)) and space (O(max(n, m)^2)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(max(n, m)^2) time and O(max(n, m)^2) space.

Pattern focus: prefix sum, binary search, and dynamic programming

Use the pattern as a checklist:

prefix sum — confirm the invariant holds after each step
binary search — confirm the invariant holds after each step
dynamic programming — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	O(max(n, m)^2)
Space	O(max(n, m)^2)

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Find the Minimum Area to Cover All Ones II

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for prefix sum, binary search, and dynamic programming — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to prefix sum, binary search, and dynamic programming:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: dynamic programming.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Find the Minimum Area to Cover All Ones II in a second language (cpp, python).
Drill 3–5 more problems tagged dynamic programming.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Find the Minimum Area to Cover All Ones II: show input structure (dynamic programming), highlight the moving parts of the prefix sum, binary search, and dynamic programming approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the prefix sum, binary search, and dynamic programming approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Find the Minimum Area to Cover All Ones II (#3197) — Hard. Pattern: prefix sum, binary search, and dynamic programming. Complexity: O(max(n, m)^2) time / O(max(n, m)^2) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Find the Minimum Area to Cover All Ones II?+

The reference solutions aim for O(max(n, m)^2) time and O(max(n, m)^2) space. Always re-derive complexity from the code you write in the interview.

What pattern does Find the Minimum Area to Cover All Ones II use?+

It primarily maps to prefix sum, binary search, and dynamic programming, within the broader topic of dynamic programming.

Is Find the Minimum Area to Cover All Ones II good for interviews?+

Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/find-the-minimum-area-to-cover-all-ones-ii/

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