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Problem solution · C++

Find the Shortest Superstring

Find the Shortest Superstring: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
65 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Find the Shortest Superstring, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 65 lines of C++ from the credited upstream file 943-2.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 8 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Shortest Superstring · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string shortestSuperstring(vector<string>& words) {    const int n = words.size();    // cost[i][j] := the cost to append words[j] after words[i]    vector<vector<int>> cost(n, vector<int>(n));    // dp[s][j] := the minimum cost to visit nodes of s ending in j, s is a    // binary Value, e.g. dp[6][2] means the minimum cost to visit {1, 2} ending    // in 2 (6 = 2^1 + 2^2)    vector<vector<int>> dp(1 << n, vector<int>(n, INT_MAX / 2));    // parent[s][j] := the parent of "nodes of s ending in j"    vector<vector<int>> parent(1 << n, vector<int>(n, -1));     // Returns the cost to append b after a.    auto getCost = [](const string& a, const string& b) {      int cost = b.length();      const int minLength = min(a.length(), b.length());      for (int k = 1; k <= minLength; ++k)        if (a.substr(a.length() - k) == b.substr(0, k))          cost = b.length() - k;      return cost;    };     // Pre-calculate the `cost` array to save time.    for (int i = 0; i < n; ++i)      for (int j = i + 1; j < n; ++j) {        cost[i][j] = getCost(words[i], words[j]);        cost[j][i] = getCost(words[j], words[i]);      }     for (int i = 0; i < n; ++i)      dp[1 << i][i] = words[i].length();     // Enumerate all the states ending in different nodes.    for (int s = 1; s < (1 << n); ++s)      for (int i = 0; i < n; ++i) {        if ((s & (1 << i)) == 0)          continue;        for (int j = 0; j < n; ++j)          if (dp[s - (1 << i)][j] + cost[j][i] < dp[s][i]) {            dp[s][i] = dp[s - (1 << i)][j] + cost[j][i];            parent[s][i] = j;          }      }     string ans;    const vector<int>& dpBack = dp.back();    int j = distance(dpBack.begin(), ranges::min_element(dpBack));    int s = (1 << n) - 1;  // 2^0 + 2^1 + ... + 2^(n - 1)     // Traverse back to build the string.    while (s > 0) {      const int i = parent[s][j];      if (i == -1)        ans = words[j] + ans;      else        ans = words[j].substr(words[j].length() - cost[i][j]) + ans;      s -= 1 << j;      j = i;    }     return ans;  }};

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