CSES Solutions Palindrome Reorder (original) (raw)

Last Updated : 23 Jul, 2025

Given a string **S, your task is to reorder its letters in such a way that it becomes a palindrome (i.e., it reads the same forwards and backwards).

**Examples:

**Input: S = "AAAACACBA"
**Output: AAACBCAAA
**Explanation: We can reorder "AAAACACBA" to "AAACBCAAA" to get a palindrome string.

**Input: S = "AAABBB"
**Output: NO SOLUTION
**Explanation: We cannot reorder "AAABBB" to get a palindrome string.

**Approach: To solve the problem, follow the below idea:

We can construct the palindrome by recording the frequency of every character in the string **S. If the character occurs **even number of times in the string, then we can construct the palindromic string by placing equal number of characters at same distance from the middle of the string. Otherwise, if the character occurs **odd number of times, then we can only have one occurrence in the middle and then distribute the remaining even number of characters equally at same distance from the middle of the string. Also, we can have **at most one character that occurs odd number of times as that character will be placed in the middle.

**Step-by-step algorithm:

Count the frequency of all characters in the string **S.
If more than one character has odd frequency, then it is not possible to reorder the string to make it palindrome.
Use two pointers, say **left and **right to keep track of the positions where we have to keep the next pair of same characters.
If exactly one character has odd frequency, then place one such character at the middle of the string and now we will be left will all characters having even frequency which we can distribute equally with same distance from the middle of the string.
If no character has odd frequency, then also we can distribute equally with same distance from the middle of the string.
Finally, return the constructed palindromic string.

Below is the implementation of the algorithm:

C++ `

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

// function to construct a palindromic string string solve(string S) { int N = S.length(); string ans(N, ' ');

// frequency array to count the occurrence of each
// character
int freq[26] = {};
for (int i = 0; i < N; i++) {
    freq[S[i] - 'A'] += 1;
}

// Count the number of character having odd frequency
int cnt = 0;
for (int i = 0; i < 26; i++) {
    if (freq[i] % 2 != 0) {
        cnt += 1;
    }
}

// If more than one characters have odd frequency, then
// no solution exists
if (cnt > 1)
    return "NO SOLUTION";

int left = 0, right = N - 1;
for (int i = 0; i < N; i++) {
    if (freq[S[i] - 'A'] % 2 == 1) {
        ans[N / 2] = S[i];
        freq[S[i] - 'A'] -= 1;
    }
    while (freq[S[i] - 'A'] > 0) {
        ans[left++] = ans[right--] = S[i];
        freq[S[i] - 'A'] -= 2;
    }
}
return ans;

}

int main() { // Sample Input string S = "AAAACACBA"; cout << solve(S) << endl; return 0; }

Java

import java.util.Arrays;

public class PalindromicString {

// function to construct a palindromic string
static String solve(String S) {
    int N = S.length();
    char[] ans = new char[N];

    // frequency array to count the occurrence of each character
    int[] freq = new int[26];
    for (int i = 0; i < N; i++) {
        freq[S.charAt(i) - 'A'] += 1;
    }

    // Count the number of characters having odd frequency
    int cnt = 0;
    for (int i = 0; i < 26; i++) {
        if (freq[i] % 2 != 0) {
            cnt += 1;
        }
    }

    // If more than one character has odd frequency, then no solution exists
    if (cnt > 1)
        return "NO SOLUTION";

    int left = 0, right = N - 1;
    for (int i = 0; i < 26; i++) {
        while (freq[i] > 1) {
            ans[left++] = ans[right--] = (char) ('A' + i);
            freq[i] -= 2;
        }
    }

    for (int i = 0; i < 26; i++) {
        if (freq[i] == 1) {
            ans[N / 2] = (char) ('A' + i);
            break;
        }
    }

    return new String(ans);
}

public static void main(String[] args) {
    // Sample Input
    String S = "AAAACACBA";
    System.out.println(solve(S));
}

}

// This code is contributed by shivamgupta0987654321

JavaScript

// Function to construct a palindromic string function solve(S) { const N = S.length; const ans = new Array(N);

// Frequency array to count the occurrence of each character
const freq = new Array(26).fill(0);
for (let i = 0; i < N; i++) {
    freq[S.charCodeAt(i) - 65] += 1;
}

// Count the number of characters having odd frequency
let cnt = 0;
for (let i = 0; i < 26; i++) {
    if (freq[i] % 2 !== 0) {
        cnt += 1;
    }
}

// If more than one character has odd frequency, then no solution exists
if (cnt > 1)
    return "NO SOLUTION";

let left = 0, right = N - 1;
for (let i = 0; i < 26; i++) {
    while (freq[i] > 1) {
        ans[left++] = ans[right--] = String.fromCharCode(65 + i);
        freq[i] -= 2;
    }
}

for (let i = 0; i < 26; i++) {
    if (freq[i] === 1) {
        ans[Math.floor(N / 2)] = String.fromCharCode(65 + i);
        break;
    }
}

return ans.join('');

}

// Sample Input const S = "AAAACACBA"; console.log(solve(S));

Python3

function to construct a palindromic string

def solve(S): N = len(S) ans = [' '] * N

# frequency array to count the occurrence of each character
freq = [0] * 26
for i in range(N):
    freq[ord(S[i]) - ord('A')] += 1

# Count the number of characters having odd frequency
cnt = 0
for i in range(26):
    if freq[i] % 2 != 0:
        cnt += 1

# If more than one character has odd frequency, then no solution exists
if cnt > 1:
    return "NO SOLUTION"

left, right = 0, N - 1
for i in range(N):
    if freq[ord(S[i]) - ord('A')] % 2 == 1:
        ans[N // 2] = S[i]
        freq[ord(S[i]) - ord('A')] -= 1
    while freq[ord(S[i]) - ord('A')] > 0:
        ans[left] = ans[right] = S[i]
        left += 1
        right -= 1
        freq[ord(S[i]) - ord('A')] -= 2

return ''.join(ans)

Sample Input

S = "AAAACACBA" print(solve(S))

**Time Complexity: O(N), where **N is the length of input string **S.
**Auxiliary Space: O(1)