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

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 : substring = “eeksf”, could be changed to “eekee” which is palindrome.
queries : substring = “for”, is not palindrome and can’t be made palindromic after replacing atmost 0 character..
queries : substring = “Gee”, could be changed to “GeG” which is palindrome.
queries : substring = “ksforGee”, could be changed to “ksfoofsk” which is palindrome.
queries : 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using Dynamic Programming.

1. Create a 2D matrix (say dp[i][j]) where dp[i][j] denotes the count of ith character in the substring str[0…j].
2. 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 is equals to 0, then dp[i][j] would be one of the first characters which is equals to ith characters.
3. For each query, find out the count of the each character in the substring str[L…R] by the simple relation:
```count =  dp[i][right] - dp[i][left] + (str[left] == i + 'a').
```
4. Get the count of unmatched pairs.
5. 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 ` `using` `namespace` `std; ` ` `  `// Function to find whether string can ` `// be made palindromic or not for each queries ` `void` `canMakePaliQueries( ` `    ``string str, ` `    ``vector >& Q) ` `{ ` `    ``int` `n = str.length(); ` ` `  `    ``// To store the count of ith character ` `    ``// of substring str[0...i] ` `    ``vector > 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; ` `        ``int` `right = query; ` `        ``int` `k = query; ` ` `  `        ``// 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 > 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 `

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.

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