#2977Hard~60 min

Minimum Cost to Convert String II

Time O(o * l + k * eloge + n * l) · Space O(t + k * v + l) · Official statement on LeetCode

dynamic programming dijkstras algorithm floyd warshall algorithm dynamic programming trie

Solutions

// Time:  O(o * l + k * eloge + n * c * l), e is the number of edges reachable from a given node u, o = len(original), l = max(len(x) for x in original), k = len(lookups), c = len({len(x) for x in original})
// Space: O(o * l + k * v + c + l), v is the number of nodes reachable from a given node u

// hash table, dijkstra's algorithm, dp, memoization
class Solution {
public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();

        unordered_map<int, unordered_map<string, int>> lookups;
        for (const auto& x : original) {
            const int l = size(x);
            auto& lookup = lookups[l];
            if (!lookup.count(x)) {
                lookup[x] = size(lookup);
            }
        }
        for (const auto& x : changed) {
            const int l = size(x);
            auto& lookup = lookups[l];
            if (!lookup.count(x)) {
                lookup[x] = size(lookup);
            }
        }
        unordered_map<int, vector<vector<int64_t>>> dists;
        for (const auto& [l, lookup] : lookups) {
            auto& dist = dists[l];
            dist.assign(size(lookup), vector<int64_t>(size(lookup), INF));
            for (int u = 0; u < size(dist); ++u) {
                dist[u][u] = 0;
            }
        }
        for (int i = 0; i < size(original); ++i) {
            const int l = size(original[i]);
            auto& lookup = lookups[l];
            auto& dist = dists[l];
            const int u = lookup[original[i]];
            const int v = lookup[changed[i]];
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        const auto& dijkstra = [&](const auto& dist, int start) {
            vector<int64_t> best(size(dist), INF);
            best[start] = 0;
            priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>> min_heap;
            min_heap.emplace(0, start);
            while (!empty(min_heap)) {
                const auto [curr, u] = min_heap.top(); min_heap.pop();
                if (curr > best[u]) {
                    continue;
                }
                for (int v = 0; v < size(dist[u]); ++v) {
                    const auto& w = dist[u][v];
                    if (best[v] - curr <= w) {
                        continue;
                    }
                    best[v] = curr + w;
                    min_heap.emplace(best[v], v);
                }
            }
            return best;
        };
        
        unordered_map<int, unordered_map<int, vector<int64_t>>> memo;
        const auto& memoization = [&](int l, const auto& dist, int u, int v) {   
            if (!memo[l].count(u)) {
                memo[l][u] = dijkstra(dist, u);
            }
            return memo[l][u][v];
        };

        unordered_set<int> candidates;
        int l = 0;
        for (const auto& x : original) {
            candidates.emplace(size(x));
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
            }
            for (const auto& l : candidates) {
                if (i + l > size(source)) {
                    continue;
                }
                auto& lookup = lookups[l];
                auto& dist = dists[l];
                const auto& u = source.substr(i, l);
                const auto& v = target.substr(i, l);
                if (lookup.count(u) && lookup.count(v) && memoization(l, dist, lookup[u], lookup[v]) != INF) {
                    dp[(i + l) % size(dp)] = min(dp[(i + l) % size(dp)], dp[i % size(dp)] + memoization(l, dist, lookup[u], lookup[v]));
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

// Time:  O(o * l + k^3 + n * c * l), o = len(original), l = max(len(x) for x in original), k = len(lookup), c = len({len(x) for x in original})
// Space: O(o * l + k^2 + c + l)
// hash table, Floyd-Warshall algorithm, dp
class Solution2 {
public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();
        
        const auto& floydWarshall = [](auto& dist) {
            for (int k = 0; k < size(dist); ++k) {
                for (int i = 0; i < size(dist); ++i) {
                    if (dist[i][k] == INF) {
                        continue;
                    }
                    for (int j = 0; j < size(dist[0]); ++j) {
                        if (dist[k][j] == INF) {
                            continue;
                        }
                        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        };

        unordered_map<int, unordered_map<string, int>> lookups;
        for (const auto& x : original) {
            const int l = size(x);
            auto& lookup = lookups[l];
            if (!lookup.count(x)) {
                lookup[x] = size(lookup);
            }
        }
        for (const auto& x : changed) {
            const int l = size(x);
            auto& lookup = lookups[l];
            if (!lookup.count(x)) {
                lookup[x] = size(lookup);
            }
        }
        unordered_map<int, vector<vector<int64_t>>> dists;
        for (const auto& [l, lookup] : lookups) {
            auto& dist = dists[l];
            dist.assign(size(lookup), vector<int64_t>(size(lookup), INF));
            for (int u = 0; u < size(dist); ++u) {
                dist[u][u] = 0;
            }
        }
        for (int i = 0; i < size(original); ++i) {
            const int l = size(original[i]);
            auto& lookup = lookups[l];
            auto& dist = dists[l];
            const int u = lookup[original[i]];
            const int v = lookup[changed[i]];
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        for (auto& [_, dist] : dists) {
            floydWarshall(dist);
        }
        unordered_set<int> candidates;
        int l = 0;
        for (const auto& x : original) {
            candidates.emplace(size(x));
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
            }
            for (const auto& l : candidates) {
                if (i + l > size(source)) {
                    continue;
                }
                auto& lookup = lookups[l];
                auto& dist = dists[l];
                const auto& u = source.substr(i, l);
                const auto& v = target.substr(i, l);
                if (lookup.count(u) && lookup.count(v) && dist[lookup[u]][lookup[v]] != INF) {
                    dp[(i + l) % size(dp)] = min(dp[(i + l) % size(dp)], dp[i % size(dp)] + dist[lookup[u]][lookup[v]]);
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

// Time:  O(o * l + k * eloge + n * c * l), e is the number of edges reachable from a given node u, o = len(original), l = max(len(x) for x in original), k = len(lookups), c = len({len(x) for x in original})
// Space: O(t + k * v + c + l), v is the number of nodes reachable from a given node u
// hash table, dijkstra's algorithm, dp, memoization
class Solution3 {
public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();

        unordered_map<string, int> lookup;
        for (const auto& x : original) {
            if (lookup.count(x)) {
                continue;
            }
            lookup[x] = size(lookup);
        }
        for (const auto& x : changed) {
            if (lookup.count(x)) {
                continue;
            }
            lookup[x] = size(lookup);
        }
        unordered_map<int, unordered_map<int, int64_t>> dist;
        for (int i = 0; i < size(original); ++i) {
            const int u = lookup[original[i]];
            const int v = lookup[changed[i]];
            if (!dist[u].count(v)) {
                dist[u][v] = INF;
            }
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        const auto& dijkstra = [&](int start) {
            unordered_map<int, int64_t> best = {{start, 0}};
            priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>> min_heap;
            min_heap.emplace(0, start);
            while (!empty(min_heap)) {
                const auto [curr, u] = min_heap.top(); min_heap.pop();
                if (curr > best[u]) {
                    continue;
                }
                for (auto [v, w] : dist[u]) {
                    if (best.count(v) && best[v] - curr <= w) {
                        continue;
                    }
                    best[v] = curr + w;
                    min_heap.emplace(best[v], v);
                }
            }
            return best;
        };
        
        unordered_map<int, unordered_map<int, int64_t>> memo;
        const auto& memoization = [&](int u, int v) {            
            if (!memo.count(u)) {
                memo[u] = dijkstra(u);
            }
            return memo[u].count(v) ? memo[u][v] : INF;
        };
        
        unordered_set<int> candidates;
        int l = 0;
        for (const auto& x : original) {
            candidates.emplace(size(x));
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
            }
            for (const auto& l : candidates) {
                if (i + l > size(source)) {
                    continue;
                }
                const auto& u = source.substr(i, l);
                const auto& v = target.substr(i, l);
                if (lookup.count(u) && lookup.count(v) && memoization(lookup[u], lookup[v]) != INF) {
                    dp[(i + l) % size(dp)] = min(dp[(i + l) % size(dp)], dp[i % size(dp)] + memoization(lookup[u], lookup[v]));
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

// Time:  O(o * l + k^3 + n * c * l), o = len(original), l = max(len(x) for x in original), k = len(lookup), c = len({len(x) for x in original})
// Space: O(o * l + k^2 + c + l)
// hash table, Floyd-Warshall algorithm, dp
class Solution4 {
public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();
        
        const auto& floydWarshall = [](auto& dist) {
            for (const auto& [k, _] : dist) {
                for (const auto& [i, _] : dist) {
                    if (dist[i][k] == INF) {
                        continue;
                    }
                    for (const auto& [j, _] : dist) {
                        if (dist[k][j] == INF) {
                            continue;
                        }
                        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        };

        unordered_map<string, int> lookup;
        unordered_map<int, vector<int>> buckets;
        for (const auto& x : original) {
            if (lookup.count(x)) {
                continue;
            }
            lookup[x] = size(lookup);
            buckets[size(x)].emplace_back(lookup[x]);
        }
        for (const auto& x : changed) {
            if (lookup.count(x)) {
                continue;
            }
            lookup[x] = size(lookup);
            buckets[size(x)].emplace_back(lookup[x]);
        }
        unordered_map<int, unordered_map<int, unordered_map<int, int64_t>>> dists;
        for (const auto& [l, lookup] : buckets) {
            auto& dist = dists[l];
            for (const auto& u : lookup) {
                for (const auto& v : lookup) {
                    dist[u][v] = u == v ? 0 : INF;
                }
            }
        }
        for (int i = 0; i < size(original); ++i) {
            const int l = size(original[i]);
            auto& dist = dists[l];
            const int u = lookup[original[i]];
            const int v = lookup[changed[i]];
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        for (auto& [_, dist] : dists) {
            floydWarshall(dist);
        }
        unordered_set<int> candidates;
        int l = 0;
        for (const auto& x : original) {
            candidates.emplace(size(x));
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
            }
            for (const auto& l : candidates) {
                if (i + l > size(source)) {
                    continue;
                }
                auto& dist = dists[l];
                const auto& u = source.substr(i, l);
                const auto& v = target.substr(i, l);
                if (lookup.count(u) && lookup.count(v) && dist[lookup[u]][lookup[v]] != INF) {
                    dp[(i + l) % size(dp)] = min(dp[(i + l) % size(dp)], dp[i % size(dp)] + dist[lookup[u]][lookup[v]]);
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

// Time:  O(o * l + k * eloge + n * l), e is the number of edges reachable from a given node u, o = len(original), l = max(len(x) for x in original), k = trie.k
// Space: O(t + k * v + l), v is the number of nodes reachable from a given node u
// trie, dijkstra's algorithm, dp, memoization
class Solution5 {
private:
    class Trie {
    public:
        Trie() {
             new_node();
         }

        pair<bool, int> add(const string& s) {
            int curr = 0;
            for (const auto& c : s) {
                const int x = c - 'a';
                if (nodes_[curr][x] == -1) {
                    nodes_[curr][x] = new_node();
                }
                curr = nodes_[curr][x];
            }
            if (idxs_[curr] == -1) {
                idxs_[curr] = k++;
                return {true, idxs_[curr]};
            }
            return {false, idxs_[curr]};
        }

        int query(const string& s) {
            int curr = 0;
            for (const auto& c : s) {
                curr = nodes_[curr][c - 'a'];
            }
            return idxs_[curr];
        }

        int next(int curr, int c) {
            return nodes_[curr][c - 'a'];
        }

        int id(int curr) {
            return idxs_[curr];
        }

    public:
        int k = 0;

    private:
        int new_node() {
            nodes_.emplace_back(26, -1);
            idxs_.emplace_back(-1);
            return size(nodes_) - 1;
        }

        vector<vector<int>> nodes_;
        vector<int> idxs_;
    };

public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();

        Trie trie;
        for (const auto& x : original) {
            trie.add(x);
        }
        for (const auto& x : changed) {
            trie.add(x);
        }
        unordered_map<int, unordered_map<int, int64_t>> dist;
        for (int i = 0; i < size(original); ++i) {
            const int u = trie.query(original[i]);
            const int v = trie.query(changed[i]);
            if (!dist[u].count(v)) {
                dist[u][v] = INF;
            }
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        const auto& dijkstra = [&](int start) {
            unordered_map<int, int64_t> best = {{start, 0}};
            priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>> min_heap;
            min_heap.emplace(0, start);
            while (!empty(min_heap)) {
                const auto [curr, u] = min_heap.top(); min_heap.pop();
                if (curr > best[u]) {
                    continue;
                }
                for (auto [v, w] : dist[u]) {
                    if (best.count(v) && best[v] - curr <= w) {
                        continue;
                    }
                    best[v] = curr + w;
                    min_heap.emplace(best[v], v);
                }
            }
            return best;
        };
        
        unordered_map<int, unordered_map<int, int64_t>> memo;
        const auto& memoization = [&](int u, int v) {            
            if (!memo.count(u)) {
                memo[u] = dijkstra(u);
            }
            return memo[u].count(v) ? memo[u][v] : INF;
        };
        
        int l = 0;
        for (const auto& x : original) {
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
                
            }
            int u = 0, v = 0;
            for (int j = i; j < size(source); ++j) {
                u = trie.next(u, source[j]);
                v = trie.next(v, target[j]);
                if (u == -1 || v == -1) {
                    break;
                }
                if (trie.id(u) != -1 && trie.id(v) != -1 && memoization(trie.id(u), trie.id(v)) != INF) {
                    dp[(j + 1) % size(dp)] = min(dp[(j + 1) % size(dp)], dp[i % size(dp)] + memoization(trie.id(u), trie.id(v)));
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

// Time:  O(o * l + k^3 + n * l), o = len(original), l = max(len(x) for x in original), k = trie.k
// Space: O(t + k^2 + l)
// trie, Floyd-Warshall algorithm, dp
class Solution6 {
private:
    class Trie {
    public:
        Trie() {
             new_node();
         }

        pair<bool, int> add(const string& s) {
            int curr = 0;
            for (const auto& c : s) {
                const int x = c - 'a';
                if (nodes_[curr][x] == -1) {
                    nodes_[curr][x] = new_node();
                }
                curr = nodes_[curr][x];
            }
            if (idxs_[curr] == -1) {
                idxs_[curr] = k++;
                return {true, idxs_[curr]};
            }
            return {false, idxs_[curr]};
        }

        int query(const string& s) {
            int curr = 0;
            for (const auto& c : s) {
                curr = nodes_[curr][c - 'a'];
            }
            return idxs_[curr];
        }

        int next(int curr, int c) {
            return nodes_[curr][c - 'a'];
        }

        int id(int curr) {
            return idxs_[curr];
        }

    public:
        int k = 0;

    private:
        int new_node() {
            nodes_.emplace_back(26, -1);
            idxs_.emplace_back(-1);
            return size(nodes_) - 1;
        }

        vector<vector<int>> nodes_;
        vector<int> idxs_;
    };

public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        static const auto INF = numeric_limits<int64_t>::max();
        
        const auto& floydWarshall = [](auto& dist) {
            for (const auto& [k, _] : dist) {
                for (const auto& [i, _] : dist) {
                    if (dist[i][k] == INF) {
                        continue;
                    }
                    for (const auto& [j, _] : dist) {
                        if (dist[k][j] == INF) {
                            continue;
                        }
                        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        };

        Trie trie;
        unordered_map<int, vector<int>> buckets;
        for (const auto& x : original) {
            const auto& [not_duplicated, i] = trie.add(x);
            if (not_duplicated) {
                buckets[size(x)].emplace_back(i);
            }
        }
        for (const auto& x : changed) {
            const auto& [not_duplicated, i] = trie.add(x);
            if (not_duplicated) {
                buckets[size(x)].emplace_back(i);
            }
        }
        unordered_map<int, unordered_map<int, unordered_map<int, int64_t>>> dists;
        for (const auto& [l, lookup] : buckets) {
            auto& dist = dists[l];
            for (const auto& u : lookup) {
                for (const auto& v : lookup) {
                    dist[u][v] = u == v ? 0 : INF;
                }
            }
        }
        for (int i = 0; i < size(original); ++i) {
            const int l = size(original[i]);
            auto& dist = dists[l];
            const int u = trie.query(original[i]);
            const int v = trie.query(changed[i]);
            dist[u][v] = min(dist[u][v], static_cast<int64_t>(cost[i]));
        }
        for (auto& [_, dist] : dists) {
            floydWarshall(dist);
        }
        int l = 0;
        for (const auto& x : original) {
            l = max(l, static_cast<int>(size(x)));
        }
        vector<int64_t> dp(l + 1, INF);
        dp[0] = 0;
        for (int i = 0; i < size(source); ++i) {
            if (dp[i % size(dp)] == INF) {
                continue;
            }
            if (source[i] == target[i]) {
                dp[(i + 1) % size(dp)] = min(dp[(i + 1) % size(dp)], dp[i % size(dp)]);
                
            }
            int u = 0, v = 0;
            for (int j = i; j < size(source); ++j) {
                u = trie.next(u, source[j]);
                v = trie.next(v, target[j]);
                if (u == -1 || v == -1) {
                    break;
                }
                auto& dist = dists[j - i + 1];
                if (trie.id(u) != -1 && trie.id(v) != -1 && dist[trie.id(u)][trie.id(v)] != INF) {
                    dp[(j + 1) % size(dp)] = min(dp[(j + 1) % size(dp)], dp[i % size(dp)] + dist[trie.id(u)][trie.id(v)]);
                }
            }
            dp[i % size(dp)] = INF;
        }
        return dp[size(source) % size(dp)] != INF ? dp[size(source) % size(dp)] : -1;
    }
};

Beginner Explanation

What is Minimum Cost to Convert String II?

Minimum Cost to Convert String II (LeetCode #2977) 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 dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie.
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: Dijkstra's Algorithm, Floyd-Warshall Algorithm, DP, Memoization.

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

Interview Walkthrough

Interview approach for Minimum Cost to Convert String 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 dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie.

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(o * l + k * eloge + n * l)) and space (O(t + k * v + l)) 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(o * l + k * eloge + n * l) time and O(t + k * v + l) space.

Pattern focus: dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie

Use the pattern as a checklist:

dijkstras algorithm — confirm the invariant holds after each step
floyd warshall algorithm — confirm the invariant holds after each step
dynamic programming — confirm the invariant holds after each step
trie — 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(o l + k * eloge + n * l)**
Space	*O(t + k v + l)**

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 Minimum Cost to Convert String II

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie — 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 dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie:

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 Minimum Cost to Convert String 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 Minimum Cost to Convert String II: show input structure (dynamic programming), highlight the moving parts of the dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie 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 dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie 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

Minimum Cost to Convert String II (#2977) — Hard. Pattern: dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie. Complexity: O(o * l + k * eloge + n * l) time / O(t + k * v + l) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Minimum Cost to Convert String II?+

The reference solutions aim for O(o * l + k * eloge + n * l) time and O(t + k * v + l) space. Always re-derive complexity from the code you write in the interview.

What pattern does Minimum Cost to Convert String II use?+

It primarily maps to dijkstras algorithm, floyd warshall algorithm, dynamic programming, and trie, within the broader topic of dynamic programming.

Is Minimum Cost to Convert String 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/minimum-cost-to-convert-string-ii/

#	Title	Difficulty	Topics
2976	Minimum Cost to Convert String I	Medium	dynamic programming
1803	Count Pairs With XOR in a Range	Hard	dynamic programming
3213	Construct String with Minimum Cost	Hard	dynamic programming
2707	Extra Characters in a String	Medium	dynamic programming
3253	Construct String with Minimum Cost (Easy)	Medium	dynamic programming
10	Regular Expression Matching	Hard	dynamic programming
72	Edit Distance	Hard	dynamic programming
87	Scramble String	Hard	dynamic programming