Count minimum swap to make string palindrome

• Difficulty Level : Medium
• Last Updated : 04 Oct, 2021

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

Examples:

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

Output: -1

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.
4. If the same character found then swap the right pointer’s character towards the right until it is not placed at its correct position in a string.
5. Increase left pointer and repeat step 2.

Below is the implementation of the above approach:

C++

 // C++ program to Count// minimum swap to make// string palindrome#include using namespace std;  // Function to Count minimum swapint 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 codeint main(){    string s = "geeksfgeeks";      // Function calling    int ans1 = countSwap(s);      reverse(s.begin(), s.end());    int ans2 = countSwap(s);      cout << max(ans1, ans2);      return 0;}

Python3

 ''' 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:        return (count)    else:        return -1    # Driver Codes = 'geeksfgeeks'  # Length of stringn = len(s)  # Function callingans1 = CountSwap(s, n)ans2 = CountSwap(s[::-1], n)print(max(ans1, ans2))

C#

 // C# program to Count// minimum swap to make// string palindromeusing 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 ans1 = countSwap(s);          char[] charArray = s.ToCharArray();        Array.Reverse(charArray);        s = new string(charArray);          int ans2 = countSwap(s);          if (ans1 > ans2)            Console.WriteLine(ans1);        else            Console.WriteLine(ans2);    }}  // This code is contributed by sapnasingh4991
Output
9

Complexity Analysis
Time Complexity: Since we are running two nested loops on the length of string, the time complexity is O(n2)
Auxiliary Space: Since we aren’t using any extra space, Therefore Auxiliary space used is O(1)

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