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Print the arranged positions of characters to make palindrome
  • Difficulty Level : Easy
  • Last Updated : 31 Jan, 2020

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 a
Output : 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 c
Output : 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;
}

Java




// JAVA program to print original
// positions of characters in a
// String after rearranging and
// forming a palindrome
import java.util.*;
  
class GFG
{
  
// Maximum number of characters
static int MAX = 256;
  
static void printPalindromePos(String str)
{
    // Insert all positions of every
    // character in the given String.
    Vector<Integer> []pos = new Vector[MAX];
    for (int i = 0; i < MAX; i++)
        pos[i] = new Vector<Integer>();
    int n = str.length();
    for (int i = 0; i < n; i++)
        pos[str.charAt(i)].add(i + 1);
  
    /* find the number of odd elements.
    Takes O(n) */
    int oddCount = 0;
    char oddChar = 0;
    for (int i = 0; i < MAX; i++)
    {
        if (pos[i].size() % 2 != 0
        {
            oddCount++;
            oddChar = (char) i;
        }
    }
  
    /* A palindrome cannot contain more than 1
    odd characters */
    if (oddCount > 1)
        System.out.print("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++)
            System.out.print(pos[i].get(j) + " ");
    }
  
    // Consider one instance odd character
    if (oddCount > 0)
    {
        int last = pos[oddChar].size() - 1;
        System.out.print(pos[oddChar].get(last) + " ");
        pos[oddChar].remove(pos[oddChar].size() - 1);
    }
  
    /* 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++)
            System.out.print(pos[i].get(j) + " ");
    }
}
  
// Driver code
public static void main(String[] args)
{
    String s = "geeksgk";
    printPalindromePos(s);
}
}
  
// This code is contributed by 29AjayKumar

C#




// C# program to print original
// positions of characters in a
// String after rearranging and
// forming a palindrome
using System;
using System.Collections.Generic;
  
class GFG
{
  
// Maximum number of characters
static int MAX = 256;
  
static void printPalindromePos(String str)
{
    // Insert all positions of every
    // character in the given String.
    List<int> []pos = new List<int>[MAX];
    for (int i = 0; i < MAX; i++)
        pos[i] = new List<int>();
    int n = str.Length;
    for (int i = 0; i < n; i++)
        pos[str[i]].Add(i + 1);
  
    /* find the number of odd elements.
    Takes O(n) */
    int oddCount = 0;
    char oddChar = (char)0;
    for (int i = 0; i < MAX; i++)
    {
        if (pos[i].Count % 2 != 0) 
        {
            oddCount++;
            oddChar = (char) i;
        }
    }
  
    /* A palindrome cannot contain more than 1
    odd characters */
    if (oddCount > 1)
        Console.Write("NO PALINDROME");
  
    /* Print positions in first half
    of palindrome */
    for (int i = 0; i < MAX; i++)
    {
        int mid = pos[i].Count / 2;
        for (int j = 0; j < mid; j++)
            Console.Write(pos[i][j] + " ");
    }
  
    // Consider one instance odd character
    if (oddCount > 0)
    {
        int last = pos[oddChar].Count - 1;
        Console.Write(pos[oddChar][last] + " ");
        pos[oddChar].RemoveAt(pos[oddChar].Count - 1);
    }
  
    /* Print positions in second half
    of palindrome */
    for (int i = MAX - 1; i >= 0; i--)
    {
        int count = pos[i].Count;
        for (int j = count / 2; j < count; j++)
            Console.Write(pos[i][j] + " ");
    }
}
  
// Driver code
public static void Main(String[] args)
{
    String s = "geeksgk";
    printPalindromePos(s);
}
}
  
// This code is contributed by 29AjayKumar
Output:
2 1 4 5 7 6 3 

Time Complexity : O ( n )




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