You are given a string s(only lowercase alphabets) with length n. Print the position of every character of the String it must acquire so that it will form a palindromic string.

Examples :

Input: c b b a aOutput: 3 1 5 2 4 To make string palindrome 'c' must be at position 3, 'b' at 1 and 5, 'a' at 2 and 4.Input: a b cOutput: Not Possible Any permutation of string cannot form palindrome .

The idea is to create an array of vectors (or dynamic size array) which stores all positions of every character. After storing positions, we check if count of odd characters is more than one. If yes, we return “Not Possible”. Otherwise, we first print first half positions from the array, then one position of odd character (if it is present) and finally second half positions.

## C++

`// CPP program to print original ` `// positions of characters in a ` `// string after rearranging and ` `// forming a palindrome ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Maximum number of characters ` `const` `int` `MAX = 256; ` ` ` `void` `printPalindromePos(string &str) ` `{ ` ` ` `// Insert all positions of every ` ` ` `// character in the given string. ` ` ` `vector<` `int` `> pos[MAX]; ` ` ` `int` `n = str.length(); ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `pos[str[i]].push_back(i+1); ` ` ` ` ` `/* find the number of odd elements. ` ` ` `Takes O(n) */` ` ` `int` `oddCount = 0; ` ` ` `char` `oddChar; ` ` ` `for` `(` `int` `i=0; i<MAX; i++) { ` ` ` `if` `(pos[i].size() % 2 != 0) { ` ` ` `oddCount++; ` ` ` `oddChar = i; ` ` ` `} ` ` ` `} ` ` ` ` ` `/* A palindrome cannot contain more than 1 ` ` ` `odd characters */` ` ` `if` `(oddCount > 1) ` ` ` `cout << ` `"NO PALINDROME"` `; ` ` ` ` ` `/* Print positions in first half ` ` ` `of palindrome */` ` ` `for` `(` `int` `i=0; i<MAX; i++) ` ` ` `{ ` ` ` `int` `mid = pos[i].size()/2; ` ` ` `for` `(` `int` `j=0; j<mid; j++) ` ` ` `cout << pos[i][j] << ` `" "` `; ` ` ` `} ` ` ` ` ` `// Consider one instance odd character ` ` ` `if` `(oddCount > 0) ` ` ` `{ ` ` ` `int` `last = pos[oddChar].size() - 1; ` ` ` `cout << pos[oddChar][last] << ` `" "` `; ` ` ` `pos[oddChar].pop_back(); ` ` ` `} ` ` ` ` ` `/* Print positions in second half ` ` ` `of palindrome */` ` ` `for` `(` `int` `i=MAX-1; i>=0; i--) ` ` ` `{ ` ` ` `int` `count = pos[i].size(); ` ` ` `for` `(` `int` `j=count/2; j<count; j++) ` ` ` `cout << pos[i][j] << ` `" "` `; ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `string s = ` `"geeksgk"` `; ` ` ` `printPalindromePos(s); ` ` ` `return` `0; ` `} ` |

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**Output:**

2 1 4 5 7 6 3

**Time Complexity** : O ( n )

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