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Minimize swaps of pairs of characters required such that no two adjacent characters in the string are same

  • Difficulty Level : Hard
  • Last Updated : 13 Aug, 2021

Given a string S consisting of N characters, the task is to find the minimum number of pairs of characters that are required to be swapped such that no two adjacent characters are the same. If it is not possible to do so, then print “-1”.

Examples:

Input: S = “ABAACD”
Output: 1
Explanation: Swapping S[3] and S[4] modifies the given string S to “ABACAD”. Since no two adjacent characters are the same, the minimum number of operations required is 1.

Input: S = “AABA”
Output: -1

Approach: The given problem can be solved by using Backtracking. The idea is to generate all possible combinations of swapping of a pair of indices and then if the string is generated having no adjacent characters same with minimum swap, then print that minimum number of swaps operations performed. Follow the steps below to solve the problem:



  • Define a function minimumSwaps(string &str, int &minimumSwaps, int swaps=0, int idx) and perform the following operations:
    • If the adjacent characters of the string str are different then update the value of minimumSwaps to the minimum of minimumSwaps and swaps.
    • Iterate over the range [idx, N] using the variable i and performing the following operations:
      • Iterate over the range [i + 1, N] using the variable j and performing the following operations:
        • Swap the characters at positions i and j in string S.
        • Call for the function minimumSwaps(str, minimumSwaps, swaps+1, i + 1) to find other possible pairs of swapping to generate the resultant string.
        • Swap the characters in positions i and j in the string S.
  • Initialize the variable, say ansSwaps as INT_MAX to store the count of minimum swaps required.
  • Call the function minimumSwaps(str, ansSwaps) to find the minimum number of swaps required to make all the adjacent characters different.
  • After completing the above steps, if the value of ansSwaps is INT_MAX, then print -1. Otherwise, print the value of ansSwaps as the resultant minimum swaps.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if S
// contains any pair of
// adjacent characters that are same
bool check(string& S)
{
    // Traverse the string S
    for (int i = 1; i < S.length(); i++) {
 
        // If current pair of adjacent
        // characters are the same
        if (S[i - 1] == S[i]) {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Utility function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
void minimumSwaps(string& S, int& ansSwaps,
                  int swaps = 0, int idx = 0)
{
    // Check if the required string
    // is formed already
    if (check(S)) {
        ansSwaps = min(ansSwaps, swaps);
    }
 
    // Traverse the string S
    for (int i = idx;
         i < S.length(); i++) {
 
        for (int j = i + 1;
             j < S.length(); j++) {
 
            // Swap the characters at i
            // and j position
            swap(S[i], S[j]);
            minimumSwaps(S, ansSwaps,
                         swaps + 1, i + 1);
 
            // Swap for Backtracking
            // Step
            swap(S[i], S[j]);
        }
    }
}
 
// Function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
void findMinimumSwaps(string& S)
{
 
    // Stores the resultant minimum
    // number of swaps required
    int ansSwaps = INT_MAX;
 
    // Function call to find the
    // minimum swaps required
    minimumSwaps(S, ansSwaps);
 
    // Print the result
    if (ansSwaps == INT_MAX)
        cout << "-1";
    else
        cout << ansSwaps;
}
 
// Driver Code
int main()
{
    string S = "ABAACD";
    findMinimumSwaps(S);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG
{
    static int ansSwaps ;
   
// Function to check if S
// contains any pair of
// adjacent characters that are same
static boolean check(char[] S)
{
   
    // Traverse the String S
    for (int i = 1; i < S.length; i++) {
 
        // If current pair of adjacent
        // characters are the same
        if (S[i - 1] == S[i]) {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Utility function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
static void minimumSwaps(char[] S,
                  int swaps, int idx)
{
   
    // Check if the required String
    // is formed already
    if (check(S)) {
        ansSwaps = Math.min(ansSwaps, swaps);
    }
 
    // Traverse the String S
    for (int i = idx;
         i < S.length; i++) {
 
        for (int j = i + 1;
             j < S.length; j++) {
 
            // Swap the characters at i
            // and j position
            swap(S,i,j);
            minimumSwaps(S,
                         swaps + 1, i + 1);
 
            // Swap for Backtracking
            // Step
            S= swap(S,i,j);
        }
    }
}
static char[] swap(char []arr, int i, int j){
    char temp= arr[i];
    arr[i]=arr[j];
    arr[j]=temp;
    return arr;
}
   
// Function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
static void findMinimumSwaps(char[] S)
{
 
    // Stores the resultant minimum
    // number of swaps required
    ansSwaps = Integer.MAX_VALUE;
 
    // Function call to find the
    // minimum swaps required
    minimumSwaps(S, 0,0);
 
    // Print the result
    if (ansSwaps == Integer.MAX_VALUE)
        System.out.print("-1");
    else
        System.out.print(ansSwaps);
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "ABAACD";
    findMinimumSwaps(S.toCharArray());
 
}
}
 
// This code is contributed by 29AjayKumar

C#




// C# program for the above approach
using System;
 
public class GFG
{
    static int ansSwaps ;
   
// Function to check if S
// contains any pair of
// adjacent characters that are same
static bool check(char[] S)
{
   
    // Traverse the String S
    for (int i = 1; i < S.Length; i++) {
 
        // If current pair of adjacent
        // characters are the same
        if (S[i - 1] == S[i]) {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Utility function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
static void minimumSwaps(char[] S,
                  int swaps, int idx)
{
   
    // Check if the required String
    // is formed already
    if (check(S)) {
        ansSwaps = Math.Min(ansSwaps, swaps);
    }
 
    // Traverse the String S
    for (int i = idx;
         i < S.Length; i++) {
 
        for (int j = i + 1;
             j < S.Length; j++) {
 
            // Swap the characters at i
            // and j position
            swap(S,i,j);
            minimumSwaps(S,
                         swaps + 1, i + 1);
 
            // Swap for Backtracking
            // Step
            S= swap(S,i,j);
        }
    }
}
static char[] swap(char []arr, int i, int j){
    char temp= arr[i];
    arr[i]=arr[j];
    arr[j]=temp;
    return arr;
}
   
// Function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
static void findMinimumSwaps(char[] S)
{
 
    // Stores the resultant minimum
    // number of swaps required
    ansSwaps = int.MaxValue;
 
    // Function call to find the
    // minimum swaps required
    minimumSwaps(S, 0,0);
 
    // Print the result
    if (ansSwaps == int.MaxValue)
        Console.Write("-1");
    else
        Console.Write(ansSwaps);
}
 
// Driver Code
public static void Main(String[] args)
{
    String S = "ABAACD";
    findMinimumSwaps(S.ToCharArray());
 
}
}
 
// This code is contributed by 29AjayKumar.

Javascript




<script>
 
// javascript program for the above approach
var ansSwaps ;
   
// Function to check if S
// contains any pair of
// adjacent characters that are same
function check(S)
{
   
    // Traverse the String S
    for (var i = 1; i < S.length; i++) {
 
        // If current pair of adjacent
        // characters are the same
        if (S[i - 1] == S[i]) {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Utility function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
function minimumSwaps(S, swaps , idx)
{
   
    // Check if the required String
    // is formed already
    if (check(S)) {
        ansSwaps = Math.min(ansSwaps, swaps);
    }
 
    // Traverse the String S
    for (var i = idx;
         i < S.length; i++) {
 
        for (var j = i + 1;
             j < S.length; j++) {
 
            // Swap the characters at i
            // and j position
            swap(S,i,j);
            minimumSwaps(S,
                         swaps + 1, i + 1);
 
            // Swap for Backtracking
            // Step
            S= swap(S,i,j);
        }
    }
}
function swap(arr , i , j){
    var temp= arr[i];
    arr[i]=arr[j];
    arr[j]=temp;
    return arr;
}
   
// Function to find the minimum number
// of swaps of pair of characters required
// to make all pairs of adjacent characters different
function findMinimumSwaps(S)
{
 
    // Stores the resultant minimum
    // number of swaps required
    ansSwaps = Number.MAX_VALUE;
 
    // Function call to find the
    // minimum swaps required
    minimumSwaps(S, 0,0);
 
    // Prvar the result
    if (ansSwaps == Number.MAX_VALUE)
        document.write("-1");
    else
        document.write(ansSwaps);
}
 
// Driver Code
 
    var S = "ABAACD";
    findMinimumSwaps(S.split(''));
 
// This code is contributed by 29AjayKumar
</script>
Output: 
1

 

Time Complexity: O(N3*2N)
Auxiliary Space: O(1)

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