Print the arranged positions of characters to make palindrome in C Program.

You are provided with a string str with some length n. Print the position of every element of the string so it can form a palindrome, else print a message "No palindrome" on screen.

What is palindrome?

Palindrome is a word, sequence of characters which reads same from the reverse or backward as from the forward manner, like MADAM, racecar.

To find a sequence or a word is palindrome we generally store the reverse of a word in a separate string and compare both if they are same then the given word or sequence is palindrome. But in this question we have to print the arrangement to make a word or a sequence in palindrome.

Like, there is a string str = "tinni" then it can be intni or nitin so we have to return any one of the sequence of arrangement as index starting from 1 and result can be either 2 3 1 4 5 or 3 2 1 5 4 of the two.

Syntax

void printPalindromePos(char str[], int n);

Algorithm

The algorithm follows these key steps −

• Store positions of each character in the string
• Count characters with odd frequency (at most 1 allowed for palindrome)
• Print first half positions, then odd character (if any), then second half in reverse

Example

Here's a complete C program to print palindrome arrangement positions −

#include <stdio.h>
#include <string.h>

#define MAX 256

void printPalindromePos(char str[]) {
    int pos[MAX][100];  // Store positions for each character
    int count[MAX] = {0};  // Count of each character
    int n = strlen(str);
    
    // Store positions of each character
    for (int i = 0; i < n; i++) {
        pos[str[i]][count[str[i]]] = i + 1;
        count[str[i]]++;
    }
    
    // Count characters with odd frequency
    int oddCount = 0;
    char oddChar = 0;
    for (int i = 0; i < MAX; i++) {
        if (count[i] % 2 != 0) {
            oddCount++;
            oddChar = i;
        }
    }
    
    // Palindrome can't have more than 1 odd character
    if (oddCount > 1) {
        printf("NO PALINDROME<br>");
        return;
    }
    
    // Print first half positions
    for (int i = 0; i < MAX; i++) {
        int mid = count[i] / 2;
        for (int j = 0; j < mid; j++) {
            printf("%d ", pos[i][j]);
        }
    }
    
    // Print odd character position (middle)
    if (oddCount > 0) {
        printf("%d ", pos[oddChar][count[oddChar] - 1]);
    }
    
    // Print second half positions in reverse
    for (int i = MAX - 1; i >= 0; i--) {
        int total = count[i];
        for (int j = total / 2; j < total - (oddCount > 0 && i == oddChar ? 1 : 0); j++) {
            printf("%d ", pos[i][j]);
        }
    }
    printf("<br>");
}

int main() {
    char str1[] = "tinni";
    char str2[] = "baa";
    char str3[] = "abc";
    
    printf("Input: %s\nOutput: ", str1);
    printPalindromePos(str1);
    
    printf("Input: %s\nOutput: ", str2);
    printPalindromePos(str2);
    
    printf("Input: %s\nOutput: ", str3);
    printPalindromePos(str3);
    
    return 0;
}

Input: tinni
Output: 2 3 1 4 5 
Input: baa
Output: 2 1 3 
Input: abc
Output: NO PALINDROME

How It Works

The algorithm works by storing character positions and analyzing frequency patterns. For a valid palindrome arrangement −

• At most one character can have odd frequency (goes in middle)
• All other characters must have even frequency (split equally on both sides)
• Positions are printed as: first half + middle (if any) + second half reversed

Conclusion

This approach efficiently determines if a string can form a palindrome and prints the required character positions. The algorithm has O(n) time complexity where n is the string length.