Check if a substring can be Palindromic by replacing K characters for Q queries

Last Updated : 12 Jul, 2025

Given a string str and Q queries in form of [L, R, K], the task is to find whether characters from the string from [L, R] with at most K changes are allowed can be rearranged to make string palindromic or not. For each query, print "YES" if it can become a palindromic string else print "NO".
Examples:

Input: str = "GeeksforGeeks", Q = { { 1, 5, 3 }, { 5, 7, 0 }, { 8, 11, 3 }, {3, 10, 5 }, { 0, 9, 5 } }
Output:
YES
NO
YES
YES
YES
Explanation:
queries[0] : substring = "eeksf", could be changed to "eekee" which is palindrome.
queries[1] : substring = "for", is not palindrome and can't be made palindromic after replacing atmost 0 character..
queries[2] : substring = "Geek", could be changed to "GeeG" which is palindrome.
queries[3] : substring = "ksforGee", could be changed to "ksfoofsk" which is palindrome.
queries[4] : substring = "GeeksforGe", could be changed to "GeeksskeeG" which is palindrome.
Input: str = "abczwerte", Q = { { 3, 7, 4 }, { 1, 8, 10 }, { 0, 3, 1 } }
Output:
YES
YES
NO

Approach: This problem can be solved using Dynamic Programming.

Create a 2D matrix (say dp[i][j]) where dp[i][j] denotes the count of ith character in the substring str[0...j].
Below is the recurrence relation for the above approach:
- If str[i] is equals to str[j], then dp[i][j] = 1 + dp[i][j-1].
- If str[i] is not equals to str[j], then dp[i][j] = dp[i][j-1].
- if j equals 0, then dp[i][j] would be one of the first characters which are equal to ith characters.
For each query, find out the count of each character in the substring str[L...R] by the simple relation:

count =  dp[i][right] - dp[i][left] + (str[left] == i + 'a').

Get the count of unmatched pairs.
Now we need to convert the half unmatched characters to the remaining characters. If the count of half unmatched characters is less than or equals to K then, print "YES" else print "NO".

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find whether string can
// be made palindromic or not for each queries
void canMakePaliQueries(
    string str,
    vector<vector<int> >& Q)
{
    int n = str.length();

    // To store the count of ith character
    // of substring str[0...i]
    vector<vector<int> > dp(
        26,
        vector<int>(n, 0));

    for (int i = 0; i < 26; i++) {

        // Current character
        char currentChar = i + 'a';
        for (int j = 0; j < n; j++) {

            // Update dp[][] on the basis
            // recurrence relation
            if (j == 0) {
                dp[i][j]
                    = (str[j] == currentChar);
            }
            else {
                dp[i][j]
                    = dp[i][j - 1]
                      + (str[j] == currentChar);
            }
        }
    }

    // For each queries
    for (auto query : Q) {
        int left = query[0];
        int right = query[1];
        int k = query[2];

        // To store the count of
        // distinct character
        int unMatchedCount = 0;
        for (int i = 0; i < 26; i++) {

            // Find occurrence of i + 'a'
            int occurrence
                = dp[i][right]
                  - dp[i][left]
                  + (str[left] == (i + 'a'));

            if (occurrence & 1)
                unMatchedCount++;
        }

        // Half the distinct Count
        int ans = unMatchedCount / 2;

        // If half the distinct count is
        // less than equals to K then
        // palindromic string can be made
        if (ans <= k) {
            cout << "YES\n";
        }
        else {
            cout << "NO\n";
        }
    }
}

// Driver Code
int main()
{
    // Given string str
    string str = "GeeksforGeeks";

    // Given Queries
    vector<vector<int> > Q;
    Q = { { 1, 5, 3 }, { 5, 7, 0 },
 { 8, 11, 3 }, { 3, 10, 5 }, 
{ 0, 9, 5 } };

    // Function call
    canMakePaliQueries(str, Q);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find whether String can be 
// made palindromic or not for each queries
static void canMakePaliQueries(String str, 
                               int [][]Q)
{
    int n = str.length();

    // To store the count of ith character
    // of subString str[0...i]
    int [][]dp = new int[26][n];

    for(int i = 0; i < 26; i++)
    {
       
       // Current character
       char currentChar = (char)(i + 'a');
       for(int j = 0; j < n; j++)
       {
       
          // Update dp[][] on the basis
          // recurrence relation
          if (j == 0)
          {
              dp[i][j] = (str.charAt(j) == 
                          currentChar) ? 1 : 0;
          }
          else
          {
              dp[i][j] = dp[i][j - 1] +
                         ((str.charAt(j) == 
                           currentChar) ? 1 : 0);
          }
       }
    }

    // For each queries
    for(int []query : Q)
    {
       int left = query[0];
       int right = query[1];
       int k = query[2];
       
       // To store the count of
       // distinct character
       int unMatchedCount = 0;
       for(int i = 0; i < 26; i++)
       {
           
          // Find occurrence of i + 'a'
          int occurrence = dp[i][right] - 
                           dp[i][left] + 
                           (str.charAt(left) == 
                           (i + 'a') ? 1 : 0);
          
          if (occurrence % 2 == 1)
              unMatchedCount++;
       }
       
       // Half the distinct Count
       int ans = unMatchedCount / 2;
       
       // If half the distinct count is
       // less than equals to K then
       // palindromic String can be made
       if (ans <= k)
       {
           System.out.print("YES\n");
       }
       else
       {
           System.out.print("NO\n");
       }
    }
}

// Driver Code
public static void main(String[] args)
{
    
    // Given a String str
    String str = "GeeksforGeeks";

    // Given Queries
    int [][]Q = { { 1, 5, 3 }, 
                  { 5, 7, 0 },
                  { 8, 11, 3 },
                  { 3, 10, 5 }, 
                  { 0, 9, 5 } };
                  
    // Function call
    canMakePaliQueries(str, Q);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program for 
# the above approach

# Function to find whether 
# string can be made palindromic 
# or not for each queries 
def canMakePaliQueries(str, Q):
    n = len(str)

    # To store the count of 
    # ith character of substring 
    # str[0...i] 
    dp = [[0 for i in range(n)] 
             for j in range(26)]\
    
    for i in range(26):

        # Current character
        currentChar = chr(i + ord('a'))
        
        for j in range(n):

            # Update dp[][] on the basis 
            # recurrence relation
            if(j == 0):
                dp[i][j] = (str[j] == 
                            currentChar)
            else:
                dp[i][j] = dp[i][j - 1] + 
                           (str[j] == currentChar)

    # For each queries
    for query in Q:
        left = query[0]
        right = query[1]
        k = query[2]

        # To store the count of 
        # distinct character
        unMatchedCount = 0

        for i in range(26):

            # Find occurrence of 
            # i + 'a'
            occurrence = dp[i][right] - 
                         dp[i][left] + 
                         (str[left] == 
                          chr(i + ord('a')))
                
            if(occurrence & 1):
                unMatchedCount += 1

         # Half the distinct Count
        ans = int(unMatchedCount / 2)

        # If half the distinct count is 
        # less than equals to K then 
        # palindromic string can be made
        if(ans <= k):
            print("YES")
        else:
            print("NO")

# Driver Code

# Given string str
str = "GeeksforGeeks"

# Given Queries 
Q = [[1, 5, 3], 
     [5, 7, 0],
     [8, 11, 3], 
     [3, 10, 5],
     [0, 9, 5]]

# Function call 
canMakePaliQueries(str, Q)

# This code is contributed by avanitrachhadiya2155

// C# program for the above approach
using System;
class GFG{

// Function to find whether String can be 
// made palindromic or not for each queries
static void canMakePaliQueries(String str, 
                               int [,]Q)
{
    int n = str.Length;

    // To store the count of ith character
    // of subString str[0...i]
    int [,]dp = new int[26, n];

    for(int i = 0; i < 26; i++)
    {
       
       // Current character
       char currentChar = (char)(i + 'a');
       for(int j = 0; j < n; j++)
       {
       
          // Update [,]dp on the basis
          // recurrence relation
          if (j == 0)
          {
              dp[i,j] = (str[j] == 
                         currentChar) ? 1 : 0;
          }
          else
          {
              dp[i,j] = dp[i, j - 1] +
                         ((str[j] == 
                           currentChar) ? 1 : 0);
          }
       }
    }

    // For each queries
    for(int l = 0; l < Q.GetLength(0);l++)
    {
        int []query = GetRow(Q,l);
       int left = query[0];
       int right = query[1];
       int k = query[2];
       
       // To store the count of
       // distinct character
       int unMatchedCount = 0;
       for(int i = 0; i < 26; i++)
       {
           
          // Find occurrence of i + 'a'
          int occurrence = dp[i, right] - 
                           dp[i, left] + 
                           (str[left] == 
                           (i + 'a') ? 1 : 0);
          
          if (occurrence % 2 == 1)
              unMatchedCount++;
       }
       
       // Half the distinct Count
       int ans = unMatchedCount / 2;
       
       // If half the distinct count is
       // less than equals to K then
       // palindromic String can be made
       if (ans <= k)
       {
           Console.Write("YES\n");
       }
       else
       {
           Console.Write("NO\n");
       }
    }
}
 public static int[] GetRow(int[,] matrix, int row)
  {
    var rowLength = matrix.GetLength(1);
    var rowVector = new int[rowLength];

    for (var i = 0; i < rowLength; i++)
      rowVector[i] = matrix[row, i];

    return rowVector;
  }
// Driver Code
public static void Main(String[] args)
{
    
    // Given a String str
    String str = "GeeksforGeeks";

    // Given Queries
    int [,]Q = { { 1, 5, 3 }, 
                 { 5, 7, 0 },
                 { 8, 11, 3 },
                 { 3, 10, 5 }, 
                 { 0, 9, 5 } };
                  
    // Function call
    canMakePaliQueries(str, Q);
}
}

// This code is contributed by Princi Singh

JavaScript

<script>

// JavaScript program for the above approach

// Function to find whether String can be
// made palindromic or not for each queries
function canMakePaliQueries(str,Q)
{
    let n = str.length;
 
    // To store the count of ith character
    // of subString str[0...i]
    let dp = new Array(26);
    for(let i=0;i<26;i++)
    {
        dp[i]=new Array(n);
        for(let j=0;j<n;j++)
            dp[i][j]=0;
    }
 
    for(let i = 0; i < 26; i++)
    {
        
       // Current character
       let currentChar = String.fromCharCode(i + 'a'.charCodeAt(0));
       for(let j = 0; j < n; j++)
       {
        
          // Update dp[][] on the basis
          // recurrence relation
          if (j == 0)
          {
              dp[i][j] = (str[j] ==
                          currentChar) ? 1 : 0;
          }
          else
          {
              dp[i][j] = dp[i][j - 1] +
                         ((str[j] ==
                           currentChar) ? 1 : 0);
          }
       }
    }
 
    // For each queries
    for(let query of Q.values())
    {
       let left = query[0];
       let right = query[1];
       let k = query[2];
        
       // To store the count of
       // distinct character
       let unMatchedCount = 0;
       for(let i = 0; i < 26; i++)
       {
            
          // Find occurrence of i + 'a'
          let occurrence = dp[i][right] -
                           dp[i][left] +
                           (str[left] ==
                           (i + 'a'.charCodeAt(0)) ? 1 : 0);
           
          if (occurrence % 2 == 1)
              unMatchedCount++;
       }
        
       // Half the distinct Count
       let ans = unMatchedCount / 2;
        
       // If half the distinct count is
       // less than equals to K then
       // palindromic String can be made
       if (ans <= k)
       {
           document.write("YES<br>");
       }
       else
       {
           document.write("NO<br>");
       }
    }
}

// Driver Code


// Given a String str
let str = "GeeksforGeeks";

// Given Queries
let Q=[[ 1, 5, 3 ],
                  [ 5, 7, 0 ],
                  [ 8, 11, 3 ],
                  [ 3, 10, 5 ],
                  [ 0, 9, 5 ]];
// Function call
canMakePaliQueries(str, Q);


// This code is contributed by unknown2108

</script>

Output

YES
NO
YES
YES
YES

Time Complexity: O(26*N), where N is the length of the string.
Auxiliary Space: O(26*N), where N is the length of the string.

Analysis of Algorithms

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