Medium
Minimum Number of Operations to Satisfy Conditions — C++
Full explanation · Time O(n * (m + 10)) · Space O(10)
// Time: O(n * (m + 10))
// Space: O(10)
// dp
class Solution {
public:
int minimumOperations(vector<vector<int>>& grid) {
static const int INF = numeric_limits<int>::max();
static const int MAX_VALUE = 9;
const auto& top = [](const auto& dp, int exclude) {
int j = -1;
for (int i = 0; i <= MAX_VALUE; ++i) {
if (i == exclude) {
continue;
}
if (j == -1 || dp[i] < dp[j]) {
j = i;
}
}
return j;
};
vector<int> dp(MAX_VALUE + 1);
for (int j = 0; j < size(grid[0]); ++j) {
vector<int> cnt(MAX_VALUE + 1);
for (int i = 0; i < size(grid); ++i) {
++cnt[grid[i][j]];
}
vector<int> new_dp(MAX_VALUE + 1, INF);
for (int i = 0; i <= MAX_VALUE; ++i) {
const int k1 = top(dp, -1);
const int k2 = top(dp, k1);
new_dp[i] = min(new_dp[i], (i != k1 ? dp[k1] : dp[k2]) + (static_cast<int>(size(grid)) - cnt[i]));
}
dp = move(new_dp);
}
return ranges::min(dp);
}
};
// Time: O(n * (m + 100))
// Space: O(10)
// dp
class Solution2 {
public:
int minimumOperations(vector<vector<int>>& grid) {
static const int INF = numeric_limits<int>::max();
static const int MAX_VALUE = 9;
vector<int> dp(MAX_VALUE + 1);
for (int j = 0; j < size(grid[0]); ++j) {
vector<int> cnt(MAX_VALUE + 1);
for (int i = 0; i < size(grid); ++i) {
++cnt[grid[i][j]];
}
vector<int> new_dp(MAX_VALUE + 1, INF);
for (int i = 0; i <= MAX_VALUE; ++i) {
int mn = INF;
for (int k = 0; k <= MAX_VALUE; ++k) {
if (k != i) {
mn = min(mn, dp[k]);
}
}
new_dp[i] = min(new_dp[i], mn + (static_cast<int>(size(grid)) - cnt[i]));
}
dp = move(new_dp);
}
return ranges::min(dp);
}
};