Program to Find Out the Cost after Finding k Unique Subsequences From a Given String in C++

Suppose, we have a string s and another value k. We have to select some subsequences of s, so that we can get k unique subsequences. Here, the cost of selecting a subsequence equals the length of (s) - length of (subsequence). So, we have to find the lowest total cost possible after selecting k unique subsequences. If we are unable to find out this set, we will return -1. We will consider the empty string as a valid subsequence.

So, if the input is like s = "pqrs", k = 4, then the output will be 3.

To solve this, we will follow these steps −

• n := size of s

• Define one 2D array dp of size (n + 1) x (n + 1) and initialize it with 0

• Define one map last

• dp[0, 0] := 1

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • dp[i + 1, 0] := 1

  • for initialize j := (i + 1), when j >= 1, update (decrease j by 1), do −

    • dp[i + 1, j] := dp[i, j] + dp[i, j - 1]

  • if s[i] is not the end element of last, then −

    • for initialize j := 0, when j <= last[s[i]], update (increase j by 1), do −

      • dp[i + 1, j + 1] - = dp[last[s[i]], j]

  • last[s[i]] := i

• cost := 0

• for initialize i := n, when i >= 0, update (decrease i by 1), do −

  • val := minimum of k and dp[n, i]

  • cost := cost + (val * (n - i))

  • k := k - dp[n, i]

  • if k <= 0, then −

    • Come out from the loop

• if k <= 0, then −

  • return cost

• return -1

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int solve(string s, int k) {
   int n = s.size();
   vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
   unordered_map<char, int> last;
   dp[0][0] = 1;
   for (int i = 0; i < n; i++) {
      dp[i + 1][0] = 1;
      for (int j = (i + 1); j >= 1; j--) {
         dp[i + 1][j] = dp[i][j] + dp[i][j - 1];
      }
      if (last.find(s[i]) != last.end()) {
         for (int j = 0; j <= last[s[i]]; j++) {
            dp[i + 1][j + 1] -= dp[last[s[i]]][j];
         }
      }
      last[s[i]] = i;
   }
   int cost = 0;
   for (int i = n; i >= 0; i--) {
      int val = min(k, dp[n][i]);
      cost += (val * (n - i));
      k -= dp[n][i];
      if (k <= 0) {
         break;
      }
   }
   if (k <= 0) {
      return cost;
   }
   return -1;
}
int main(){
   cout << solve("pqrs",4) << endl;
   return 0;
}

Input:

"pqrs", 4

Output

3