Count minimum swap to make string palindrome

Given a string s, the task is to find out the minimum no of swaps required to make string s palindrome. If it is not possible, then return -1.

Examples:

Input : aabcb
Output : 3
Explanation :
After 1st swap: abacb
After 2nd swap: abcab
After 3rd swap: abcba

Input : adbcdbad
Output : -1



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

Approach

The following are detailed steps to solve this problem.

  1. Take two-pointer where the first pointer track from the left side of a string and second pointer keep track from the right side of a string.
  2. Till the time we find the same character, keep moving the right pointer to one step left.
  3. If the same character not found then return -1.
  4. If the same character found then swap the right pointer’s character towards right until it is not placed at it’s correct position in a string.
  5. Increase left pointer and repeat step 2.

Below is the implementation of the above approach:

C++

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// C++ program to Count
// minimum swap to make
// string palindrome
#include <bits/stdc++.h>
using namespace std;
  
// Function to Count minimum swap
int countSwap(string s)
{
    // calculate length of string as n
    int n = s.length();
      
    // counter to count minimum swap
    int count = 0;
  
    // A loop which run till mid of
    // string 
    for (int i = 0; i < n / 2; i++) {
        // Left pointer
        int left = i;
  
        // Right pointer
        int right = n - left - 1;
  
        // A loop which run from right
        // pointer towards left pointer
        while (left < right) {
            // if both char same then
            // break the loop.
            // If not, then we have to
            // move right pointer to one
            // position left
            if (s[left] == s[right]) {
                break;
            }
            else {
                right--;
            }
        }
  
        // If both pointers are at same
        // position, it denotes that we
        // don't have sufficient characters
        // to make palindrome string
        if (left == right) {
            return -1;
        }
          
        // else swap and increase the count
        for (int j = right; j < n - left - 1;
                j++) {
            swap(s[j], s[j + 1]);
            count++;
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    string s = "geeksfgeeks";
  
    // Function calling
    cout << countSwap(s);
  
    return 0;
}

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Java

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// Java program to Count
// minimum swap to make
// string palindrome
import java.util.*;
  
class GFG {
  
    // Function to Count minimum swap
    static int countSwap(String str)
    {
      
        // Legth of string
        int n = str.length();
          
        // it will convert string to
        // char array
        char s[] = str.toCharArray();
  
        // Counter to count minimum
        // swap
        int count = 0;
  
        // A loop which run in half
        // string from starting
        for (int i = 0; i < n / 2;
                i++) {
  
            // Left pointer
            int left = i;
  
            // Right pointer
            int right = n - left - 1;
  
            // A loop which run from
            // right pointer to left
            // pointer
            while (left < right) {
  
                // if both char same
                // then break the loop
                // if not same then we
                // have to move right
                // pointer to one step
                // left
                if (s[left] == s[right]) {
                    break;
                }
                else {
                    right--;
                }
            }
  
            // it denotes both pointer at
            // same position and we don't
            // have sufficient char to make
            // palindrome string
            if (left == right) {
                return -1;
            }
            else {
                for (int j = right;
                    j < n - left - 1; j++) {
                    char t = s[j];
                    s[j] = s[j + 1];
                    s[j + 1] = t;
                    count++;
                }
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String s = "geeksfgeeks";
  
        // Function calling
        int count = countSwap(s);
          
        System.out.println(count);
    }
}

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Python3

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''' Python3 program to Count 
    minimum swap to make
    string palindrome'''
  
# Function to Count minimum swap
def CountSwap(s, n):
    s = list(s)
  
    # Counter to count minimum swap
    count = 0
    ans = True
  
    # A loop which run in half string
    # from starting
    for i in range(n // 2):
  
        # Left pointer
        left = i
  
        # Right pointer
        right = n - left - 1
  
        # A loop which run from right pointer
        # to left pointer
        while left < right:
  
            # if both char same then
            # break the loop if not
            # same then we have to move
            # right pointer to one step left
            if s[left] == s[right]:
                break
            else:
                right -= 1
  
        # it denotes both pointer at
        # same position and we don't
        # have sufficient char to make
        # palindrome string
        if left == right:
            ans = False
            break
        else:
            for j in range(right, n - left - 1):
                (s[j], s[j + 1]) = (s[j + 1], s[j])
                count += 1
    if ans:
        print (count)
    else:
        print ('-1')
  
  
# Driver Code
s = 'geeksfgeeks'
  
# Legth of string
n = len(s)
  
# Function calling
CountSwap(s, n)

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C#

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// C# program to Count
// minimum swap to make
// string palindrome
using System;
  
class GFG {
   
    // Function to Count minimum swap
    static int countSwap(String str)
    {
       
        // Length of string
        int n = str.Length;
           
        // it will convert string to
        // char array
        char []s = str.ToCharArray();
   
        // Counter to count minimum
        // swap
        int count = 0;
   
        // A loop which run in half
        // string from starting
        for (int i = 0; i < n / 2;
                i++) {
   
            // Left pointer
            int left = i;
   
            // Right pointer
            int right = n - left - 1;
   
            // A loop which run from
            // right pointer to left
            // pointer
            while (left < right) {
   
                // if both char same
                // then break the loop
                // if not same then we
                // have to move right
                // pointer to one step
                // left
                if (s[left] == s[right]) {
                    break;
                }
                else {
                    right--;
                }
            }
   
            // it denotes both pointer at
            // same position and we don't
            // have sufficient char to make
            // palindrome string
            if (left == right) {
                return -1;
            }
            else {
                for (int j = right;
                    j < n - left - 1; j++) {
                    char t = s[j];
                    s[j] = s[j + 1];
                    s[j + 1] = t;
                    count++;
                }
            }
        }
   
        return count;
    }
   
    // Driver Code
    public static void Main(String[] args)
    {
        String s = "geeksfgeeks";
   
        // Function calling
        int count = countSwap(s);
           
        Console.WriteLine(count);
    }
}
  
// This code is contributed by sapnasingh4991

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Output:

9

Complexity Analysis
Time Complexity: Since we are running two nested loops on the length of string, the time complexity is O(n)
Auxiliary Space: Since we aren’t using any extra space, so auxiliary space is O(1)

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