Palindrome Partitioning III in C++

Suppose we have a string s that is containing lowercase letters and an integer k. We have to maintain some properties. These are −

First, we have to change some characters (if needed) of s to other lowercase English letters.
Then divide the string s into k substrings such that each substring is a palindrome.

We have to find the minimal number of characters that we need to change to divide the string.

So if the string is like “ababbc” and k = 2, then the answer will be 1, as we have to change one character to split this into two palindrome strings. So if we change c to b, or second last b to c, then we can get one palindrome like either “bbb” or “cbc”, and another palindrome is “aba”.

To solve this, we will follow these steps −

Define an array memo of size: 105 x 105.
Define a function solve(), this will take s, idx, k, one 2D array > dp,
if idx is same as size of s, then −
- return when k is 0 then 0 otherwise 1000
if memo[idx][k] is not equal to -1, then −
- return memo[idx][k]
if k <= 0, then −
- return inf
ans := inf
for initialize i := idx, when i < size of s, update (increase i by 1), do −
- ans := minimum of ans and dp[idx][i] + call the function solve(s, i + 1, k - 1, dp)
return memo[idx][k] = ans
From the main method, do the following
n := size of s
fill memo with -1
Define one 2D array dp of size n x n
for initialize l := 2, when l <= n, update (increase l by 1), do −
- for initialize i := 0, j := l - 1, when j < n, update (increase j by 1), (increase i by 1), do −
  - if l is same as 2, then −
  - dp[i][j] := true when s[i] is not same as s[j]
  - Otherwise
    - dp[i][j] := dp[i + 1][j - 1] + 0 when s[i] is same as s[j]
return solve(s, 0, k, dp)

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
   int memo[105][105];
   lli solve(string s, int idx, int k, vector < vector <int> > &dp){
      if(idx == s.size()) {
         return k == 0? 0 : 1000;
      }
      if(memo[idx][k] != -1) return memo[idx][k];
      if(k <= 0)return INT_MAX;
      lli ans = INT_MAX;
      for(int i = idx; i < s.size(); i++){
         ans = min(ans, dp[idx][i] + solve(s, i + 1, k - 1, dp));
      }
      return memo[idx][k] = ans;
   }
   int palindromePartition(string s, int k) {
      int n = s.size();
      memset(memo, -1, sizeof(memo));
      vector < vector <int> > dp(n, vector <int>(n));
      for(int l =2; l <= n; l++){
         for(int i = 0, j = l - 1; j <n; j++, i++){
            if(l==2){
               dp[i][j] = !(s[i] == s[j]);
            }else{
               dp[i][j] = dp[i+1][j-1] + !(s[i] == s[j]);
            }
         }
      }
      return solve(s, 0, k, dp);
   }
};
main(){
   Solution ob;
   cout << (ob.palindromePartition("ababbc", 2));
}

Input

“ababbc”

Output

Arnab Chakraborty

Published on 02-Jun-2020 11:29:14

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