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Minimize swaps to make remainder equal when an element and its index is divided by K

  • Difficulty Level : Medium
  • Last Updated : 16 Dec, 2021

Given an array arr[] of positive integers and a positive number K, the task is to find the minimum swaps of elements required such that for every element at index i, the following condition holds true:

 arr[i] % K =  i % K 

Example:

Input: arr = {4, 3, 5, 2, 9, 7}, K=3
Output: 3
Explanation: Index % 3 = 0 1 2 0 1 2 and arr[i] % 3 = 1 0 2 2 0 1
swap index 0 with index 1 => 0 1 2 2 0 1
swap index 3 with index 4 => 0 1 2 0 2 1
swap index 4 with index 5 => 1 0 2 0 1 2

Input: arr = {0, 1, 2, 3, 4, 5}, K=1
Output: 0

 

Approach: The given problem can be solved using a greedy approach. The idea is to traverse the array and make swaps such that two elements are brought to their accurate positions if possible, or else the correct element is brought at the current position. Below steps can be followed to solve the problem:

  • Check if it is possible to complete the task by comparing frequencies of index % k with element % k
    • If the frequencies do not match then return -1
  • Iterate the array and at every index i:
    • If arr[i] % 3 == i % 3 then continue to the next index
    • Else if arr[i] % 3 != i % 3 then find the index j from i+1 to N-1, where i % 3 = arr[j] % 3 and j % 3 = arr[i] % 3 that will bring two elements to their accurate positions
    • Else find the index k, by iterating from i+1 to N-1, where i % 3 = arr[k] % 3, and swap the elements

Below is the implementation of the above approach:

C++




// C++ implementation for the above approach
 
#include <iostream>
using namespace std;
 
// Function to swap the values
void swapping(int arr[], int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}
 
// Function to find the minimum swaps
// required such that arr[i] % k = i % k
int CountMinSwaps(int arr[], int N, int K)
{
 
    // initialize matrix with 0 values
    int mat[K][2] = { 0 };
 
    int i, j, count = 0;
 
    for (i = 0; i < N; i++) {
 
        // Count the frequency of
        // index % k
        mat[i % K][0] += 1;
 
        // Count the frequency of
        // index % k
        mat[arr[i] % K][1] += 1;
    }
 
    // If the count of indexes % k and
    // array elements % K are not same
    // then the task is not possible
    // therefore return -1
    for (i = 0; i < K; i++) {
 
        if (mat[i][0] != mat[i][1])
            return -1;
    }
 
    // Count the swaps
    for (i = 0; i < N; i++) {
 
        // If condition is already true
        // move to the next index
        if (i % K == arr[i] % K)
            continue;
 
        // Current index remainder
        int ind = i % K;
 
        // Current element remainder
        int ele = arr[i] % K;
 
        // Boolean variable to indicate
        // if the swap was made with the
        // element such that both the swapped
        // elements would be at correct place
        bool swapped = false;
 
        // Search for the element from
        // i + 1 till end of the array
        for (j = i + 1; j < N; j++) {
 
            // Expected index remainder
            int ind_exp = j % K;
 
            // Expected element remainder
            int ele_exp = arr[j] % K;
 
            if (ind == ele_exp
                && ele == ind_exp) {
 
                // Swap the element if found
                swapping(arr, i, j);
 
                // Update the boolean
                // variable swap to true
                swapped = true;
 
                // Increment count of swaps
                count++;
 
                break;
            }
        }
 
        // If the swap didnt take place
        if (swapped == false) {
 
            // Iterate from i+1 till end and
            // find the accurate element for
            // the current index
            for (j = i + 1; j < N; j++) {
 
                // Expected element remainder
                int ele_exp = arr[j] % K;
 
                if (ind == ele_exp) {
 
                    // Swap after finding
                    // the element
                    swapping(arr, i, j);
 
                    // Increment the count
                    count++;
 
                    break;
                }
            }
        }
    }
 
    // Return the result
    return count;
}
 
// Driver code
int main()
{
 
    int arr[6] = { 0, 1, 2, 3, 4, 5 };
    int K = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Call the function
    int swaps = CountMinSwaps(arr, N, K);
 
    // Print the answer
    cout << swaps << endl;
}

Java




// Java implementation for the above approach
public class GFG {
     
    // Function to swap the values
    static void swapping(int arr[], int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
     
    // Function to find the minimum swaps
    // required such that arr[i] % k = i % k
    static int CountMinSwaps(int arr[], int N, int K)
    {
     
        // initialize matrix with 0 values
        int mat[][] = new int[K][2] ;
        int i, j, count = 0;
     
        for (i = 0; i < N; i++) {
     
            // Count the frequency of
            // index % k
            mat[i % K][0] += 1;
     
            // Count the frequency of
            // index % k
            mat[arr[i] % K][1] += 1;
        }
     
        // If the count of indexes % k and
        // array elements % K are not same
        // then the task is not possible
        // therefore return -1
        for (i = 0; i < K; i++) {
     
            if (mat[i][0] != mat[i][1])
                return -1;
        }
     
        // Count the swaps
        for (i = 0; i < N; i++) {
     
            // If condition is already true
            // move to the next index
            if (i % K == arr[i] % K)
                continue;
     
            // Current index remainder
            int ind = i % K;
     
            // Current element remainder
            int ele = arr[i] % K;
     
            // Boolean variable to indicate
            // if the swap was made with the
            // element such that both the swapped
            // elements would be at correct place
            boolean swapped = false;
     
            // Search for the element from
            // i + 1 till end of the array
            for (j = i + 1; j < N; j++) {
     
                // Expected index remainder
                int ind_exp = j % K;
     
                // Expected element remainder
                int ele_exp = arr[j] % K;
     
                if (ind == ele_exp
                    && ele == ind_exp) {
     
                    // Swap the element if found
                    swapping(arr, i, j);
     
                    // Update the boolean
                    // variable swap to true
                    swapped = true;
     
                    // Increment count of swaps
                    count++;
     
                    break;
                }
            }
     
            // If the swap didnt take place
            if (swapped == false) {
     
                // Iterate from i+1 till end and
                // find the accurate element for
                // the current index
                for (j = i + 1; j < N; j++) {
     
                    // Expected element remainder
                    int ele_exp = arr[j] % K;
     
                    if (ind == ele_exp) {
     
                        // Swap after finding
                        // the element
                        swapping(arr, i, j);
     
                        // Increment the count
                        count++;
     
                        break;
                    }
                }
            }
        }
     
        // Return the result
        return count;
    }
     
    // Driver code
    public static void main (String[] args)
    {
     
        int arr[] = { 0, 1, 2, 3, 4, 5 };
        int K = 1;
        int N = arr.length;
     
        // Call the function
        int swaps = CountMinSwaps(arr, N, K);
     
        // Print the answer
        System.out.println(swaps);
    }
}
 
// This code is contributed by AnkThon

Python3




# python implementation for the above approach
 
# Function to swap the values
def swapping(arr, i, j):
 
    temp = arr[i]
    arr[i] = arr[j]
    arr[j] = temp
 
# Function to find the minimum swaps
# required such that arr[i] % k = i % k
def CountMinSwaps(arr, N, K):
 
    # initialize matrix with 0 values
    mat = [[0 for _ in range(2)] for _ in range(K)]
 
    count = 0
 
    for i in range(0, N):
 
        # Count the frequency of
        # index % k
        mat[i % K][0] += 1
 
        # Count the frequency of
        # index % k
        mat[arr[i] % K][1] += 1
 
    # If the count of indexes % k and
    # array elements % K are not same
    # then the task is not possible
    # therefore return -1
    for i in range(0, K):
 
        if (mat[i][0] != mat[i][1]):
            return -1
 
    # Count the swaps
    for i in range(0, N):
 
        # If condition is already true
        # move to the next index
        if (i % K == arr[i] % K):
            continue
 
        # Current index remainder
        ind = i % K
 
        # Current element remainder
        ele = arr[i] % K
 
        # Boolean variable to indicate
        # if the swap was made with the
        # element such that both the swapped
        # elements would be at correct place
        swapped = False
 
        # Search for the element from
        # i + 1 till end of the array
        for j in range(i+1, N):
 
            # Expected index remainder
            ind_exp = j % K
 
            # Expected element remainder
            ele_exp = arr[j] % K
 
            if (ind == ele_exp and ele == ind_exp):
 
                # Swap the element if found
                swapping(arr, i, j)
 
                # Update the boolean
                # variable swap to true
                swapped = True
 
                # Increment count of swaps
                count += 1
 
                break
 
        # If the swap didnt take place
        if (swapped == False):
 
            # Iterate from i+1 till end and
            # find the accurate element for
            # the current index
            for j in range(i+1, N):
 
                # Expected element remainder
                ele_exp = arr[j] % K
 
                if (ind == ele_exp):
 
                    # Swap after finding
                    # the element
                    swapping(arr, i, j)
 
                    # Increment the count
                    count += 1
 
                    break
 
    # Return the result
    return count
 
# Driver code
if __name__ == "__main__":
 
    arr = [0, 1, 2, 3, 4, 5]
    K = 1
    N = len(arr)
 
    # Call the function
    swaps = CountMinSwaps(arr, N, K)
 
    # Print the answer
    print(swaps)
 
# This code is contributed by rakeshsahni

C#




// C# implementation for the above approach
 
using System;
 
public class GFG
{
 
    // Function to swap the values
    static void swapping(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
 
    // Function to find the minimum swaps
    // required such that arr[i] % k = i % k
    static int CountMinSwaps(int[] arr, int N, int K)
    {
 
        // initialize matrix with 0 values
        int[,] mat = new int[K, 2];
        int i, j, count = 0;
 
        for (i = 0; i < N; i++)
        {
 
            // Count the frequency of
            // index % k
            mat[i % K, 0] += 1;
 
            // Count the frequency of
            // index % k
            mat[arr[i] % K, 1] += 1;
        }
 
        // If the count of indexes % k and
        // array elements % K are not same
        // then the task is not possible
        // therefore return -1
        for (i = 0; i < K; i++)
        {
 
            if (mat[i, 0] != mat[i, 1])
                return -1;
        }
 
        // Count the swaps
        for (i = 0; i < N; i++)
        {
 
            // If condition is already true
            // move to the next index
            if (i % K == arr[i] % K)
                continue;
 
            // Current index remainder
            int ind = i % K;
 
            // Current element remainder
            int ele = arr[i] % K;
 
            // Boolean variable to indicate
            // if the swap was made with the
            // element such that both the swapped
            // elements would be at correct place
            bool swapped = false;
 
            // Search for the element from
            // i + 1 till end of the array
            for (j = i + 1; j < N; j++)
            {
 
                // Expected index remainder
                int ind_exp = j % K;
 
                // Expected element remainder
                int ele_exp = arr[j] % K;
 
                if (ind == ele_exp
                    && ele == ind_exp)
                {
 
                    // Swap the element if found
                    swapping(arr, i, j);
 
                    // Update the boolean
                    // variable swap to true
                    swapped = true;
 
                    // Increment count of swaps
                    count++;
 
                    break;
                }
            }
 
            // If the swap didnt take place
            if (swapped == false)
            {
 
                // Iterate from i+1 till end and
                // find the accurate element for
                // the current index
                for (j = i + 1; j < N; j++)
                {
 
                    // Expected element remainder
                    int ele_exp = arr[j] % K;
 
                    if (ind == ele_exp)
                    {
 
                        // Swap after finding
                        // the element
                        swapping(arr, i, j);
 
                        // Increment the count
                        count++;
 
                        break;
                    }
                }
            }
        }
 
        // Return the result
        return count;
    }
 
    // Driver code
    public static void Main()
    {
 
        int[] arr = { 0, 1, 2, 3, 4, 5 };
        int K = 1;
        int N = arr.Length;
 
        // Call the function
        int swaps = CountMinSwaps(arr, N, K);
 
        // Print the answer
        Console.Write(swaps);
    }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript




<script>
 
        // JavaScript Program to implement
        // the above approach
 
        // Function to swap the values
        function swapping(arr, i, j) {
            let temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
 
        // Function to find the minimum swaps
        // required such that arr[i] % k = i % k
        function CountMinSwaps(arr, N, K) {
 
            // initialize matrix with 0 values
            let mat = new Array(K);
 
            for (let i = 0; i < mat.length; i++) {
                mat[i] = new Array(2).fill(0);
            }
 
            let i, j, count = 0;
 
            for (i = 0; i < N; i++) {
 
                // Count the frequency of
                // index % k
                mat[i % K][0] += 1;
 
                // Count the frequency of
                // index % k
                mat[arr[i] % K][1] += 1;
            }
 
            // If the count of indexes % k and
            // array elements % K are not same
            // then the task is not possible
            // therefore return -1
            for (i = 0; i < K; i++) {
 
                if (mat[i][0] != mat[i][1])
                    return -1;
            }
 
            // Count the swaps
            for (i = 0; i < N; i++) {
 
                // If condition is already true
                // move to the next index
                if (i % K == arr[i] % K)
                    continue;
 
                // Current index remainder
                let ind = i % K;
 
                // Current element remainder
                let ele = arr[i] % K;
 
                // Boolean variable to indicate
                // if the swap was made with the
                // element such that both the swapped
                // elements would be at correct place
                let swapped = false;
 
                // Search for the element from
                // i + 1 till end of the array
                for (j = i + 1; j < N; j++) {
 
                    // Expected index remainder
                    let ind_exp = j % K;
 
                    // Expected element remainder
                    let ele_exp = arr[j] % K;
 
                    if (ind == ele_exp
                        && ele == ind_exp) {
 
                        // Swap the element if found
                        swapping(arr, i, j);
 
                        // Update the boolean
                        // variable swap to true
                        swapped = true;
 
                        // Increment count of swaps
                        count++;
 
                        break;
                    }
                }
 
                // If the swap didnt take place
                if (swapped == false) {
 
                    // Iterate from i+1 till end and
                    // find the accurate element for
                    // the current index
                    for (j = i + 1; j < N; j++) {
 
                        // Expected element remainder
                        let ele_exp = arr[j] % K;
 
                        if (ind == ele_exp) {
 
                            // Swap after finding
                            // the element
                            swapping(arr, i, j);
 
                            // Increment the count
                            count++;
 
                            break;
                        }
                    }
                }
            }
 
            // Return the result
            return count;
        }
 
        // Driver code
 
 
        let arr = [0, 1, 2, 3, 4, 5];
        let K = 1;
        let N = arr.length;
 
        // Call the function
        let swaps = CountMinSwaps(arr, N, K);
 
        // Print the answer
        document.write(swaps + '<br>');
 
    // This code is contributed by Potta Lokesh
    </script>
Output
0

Time Complexity: O(N2)
Auxiliary Space: O(2 * K)


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