Given an array A[], consisting of N elements, the task is to find the minimum number of swaps required such that array elements swapped to replace a higher element, in the original array, are maximized.
Examples:
Input: A[] = {4, 3, 3, 2, 5}
Output: 3
Explanation:
Swap 1: {4, 3, 3, 2, 5} -> {5, 3, 3, 2, 4}
Swap 2: {5, 3, 3, 2, 4} -> {3, 3, 5, 2, 4}
Swap 3: {3, 3, 5, 2, 4} -> {3, 3, 2, 5, 4}
Therefore, elements {4, 3, 2} occupies original position of a higher element after swapping
Input:. A[] = {6, 5, 4, 3, 2, 1}
Output: 5
Naive Approach: The simplest approach to solve the problem can be implemented as follows:
- Sort the array in ascending order.
- Initialize two variables, result, and index, to store the count and the index up to which it has been considered in the original array, respectively.
- Iterate over the array elements. For any element A[i], go to a value in the array which is greater than ai and increment the index variable accordingly.
- After finding an element greater than A[i], increment result, and index.
- If the index has reached the end of the array, no elements are left to be swapped with previously checked elements.
- Therefore, print count.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum// number of swaps requiredint countSwaps(int A[], int n){ // Sort the array in ascending order sort(A, A + n); int ind = 1, res = 0; for (int i = 0; i < n; i++) { // Iterate until a greater element // is found while (ind < n and A[ind] == A[i]) // Keep incrementing ind ind++; // If a greater element is found if (ind < n and A[ind] > A[i]) { // Increase count of swap res++; // Increment ind ind++; } // If end of array is reached if (ind >= n) break; } // Return the answer return res;}// Driver Codeint main(){ int A[] = { 4, 3, 3, 2, 5 }; cout << countSwaps(A, 5); return 0;} |
Java
// Java Program to implement// the above approachimport java.util.*;class GFG{// Function to find the minimum// number of swaps requiredstatic int countSwaps(int A[], int n){ // Sort the array in ascending order Arrays.sort(A); int ind = 1, res = 0; for (int i = 0; i < n; i++) { // Iterate until a greater element // is found while (ind < n && A[ind] == A[i]) // Keep incrementing ind ind++; // If a greater element is found if (ind < n && A[ind] > A[i]) { // Increase count of swap res++; // Increment ind ind++; } // If end of array is reached if (ind >= n) break; } // Return the answer return res;}// Driver Codepublic static void main(String[] args){ int A[] = { 4, 3, 3, 2, 5 }; System.out.print(countSwaps(A, 5));}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program to implement# the above approach# Function to find the minimum# number of swaps requireddef countSwaps(A, n): # Sort the array in ascending order A.sort() ind, res = 1, 0 for i in range(n): # Iterate until a greater element # is found while (ind < n and A[ind] == A[i]): # Keep incrementing ind ind += 1 # If a greater element is found if (ind < n and A[ind] > A[i]): # Increase count of swap res += 1 # Increment ind ind += 1 # If end of array is reached if (ind >= n): break # Return the answer return res# Driver CodeA = [ 4, 3, 3, 2, 5 ]print (countSwaps(A, 5))# This code is contributed by chitranayal |
C#
// C# Program to implement// the above approachusing System;class GFG{// Function to find the minimum// number of swaps requiredstatic int countSwaps(int []A, int n){ // Sort the array in ascending order Array.Sort(A); int ind = 1, res = 0; for (int i = 0; i < n; i++) { // Iterate until a greater element // is found while (ind < n && A[ind] == A[i]) // Keep incrementing ind ind++; // If a greater element is found if (ind < n && A[ind] > A[i]) { // Increase count of swap res++; // Increment ind ind++; } // If end of array is reached if (ind >= n) break; } // Return the answer return res;}// Driver Codepublic static void Main(String[] args){ int []A = { 4, 3, 3, 2, 5 }; Console.Write(countSwaps(A, 5));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program for the above approach// Function to find the minimum// number of swaps requiredfunction countSwaps(A, n){ // Sort the array in ascending order A.sort(); let ind = 1, res = 0; for (let i = 0; i < n; i++) { // Iterate until a greater element // is found while (ind < n && A[ind] == A[i]) // Keep incrementing ind ind++; // If a greater element is found if (ind < n && A[ind] > A[i]) { // Increase count of swap res++; // Increment ind ind++; } // If end of array is reached if (ind >= n) break; } // Return the answer return res;} // Driver Code let A = [ 4, 3, 3, 2, 5 ]; document.write(countSwaps(A, 5)); </script> |
Output:
3
Time Complexity: O(N * log N)
Auxiliary Space: O(1)
Efficient Approach:
Since any swap between two unequal elements leads to an element replacing a higher element, it can be observed that the minimum number of swaps required is N – (the maximum frequency of an array element). Therefore, find the most frequent element in the array using HashMap, and print the result.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum// number of swaps requiredint countSwaps(int A[], int n){ // Stores the frequency of the // array elements map<int, int> mp; // Stores maximum frequency int max_frequency = 0; // Find the max frequency for (int i = 0; i < n; i++) { // Update frequency mp[A[i]]++; // Update maximum frequency max_frequency = max(max_frequency, mp[A[i]]); } return n - max_frequency;}// Driver Codeint main(){ int A[] = { 6, 5, 4, 3, 2, 1 }; // function call cout << countSwaps(A, 6); return 0;} |
Java
// Java Program to implement// the above approachimport java.util.*;class GFG{// Function to find the minimum// number of swaps requiredstatic int countSwaps(int arr[], int n){ // Stores the frequency of the // array elements HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Stores maximum frequency int max_frequency = 0; // Find the max frequency for (int i = 0; i < n; i++) { // Update frequency if(mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i]) + 1); } else { mp.put(arr[i], 1); } // Update maximum frequency max_frequency = Math.max(max_frequency, mp.get(arr[i])); } return n - max_frequency;}// Driver Codepublic static void main(String[] args){ int A[] = { 6, 5, 4, 3, 2, 1 }; // function call System.out.print(countSwaps(A, 6));}}// This code is contributed by Rohit_ranjan |
Python3
# Python3 Program to implement # the above approach # Function to find the minimum # number of swaps required def countSwaps(A, n): # Stores the frequency of the # array elements mp = {} # Stores maximum frequency max_frequency = 0 # Find the max frequency for i in range(n): # Update frequency if A[i] in mp: mp[A[i]] += 1 else: mp[A[i]] = 1 # Update maximum frequency max_frequency = max(max_frequency, mp[A[i]]) return n - max_frequency # Driver codeif __name__ == "__main__": A = [6, 5, 4, 3, 2, 1] # function call print(countSwaps(A, 6))# This code is contributed by divyeshrabadiya07 |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimum// number of swaps requiredstatic int countSwaps(int []arr, int n){ // Stores the frequency of the // array elements Dictionary<int, int> mp = new Dictionary<int, int>(); // Stores maximum frequency int max_frequency = 0; // Find the max frequency for(int i = 0; i < n; i++) { // Update frequency if(mp.ContainsKey(arr[i])) { mp[arr[i]] = mp[arr[i]] + 1; } else { mp.Add(arr[i], 1); } // Update maximum frequency max_frequency = Math.Max(max_frequency, mp[arr[i]]); } return n - max_frequency;}// Driver Codepublic static void Main(String[] args){ int []A = { 6, 5, 4, 3, 2, 1 }; // Function call Console.Write(countSwaps(A, 6));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript Program to implement// the above approach// Function to find the minimum// number of swaps requiredfunction countSwaps(A, n){ // Stores the frequency of the // array elements var mp = new Map(); // Stores maximum frequency var max_frequency = 0; // Find the max frequency for (var i = 0; i < n; i++) { // Update frequency if(mp.has(A[i])) mp.set(A[i], mp.get(A[i])+1) else mp.set(A[i], 1); // Update maximum frequency max_frequency = Math.max(max_frequency, mp.get(A[i])); } return n - max_frequency;}// Driver Codevar A = [6, 5, 4, 3, 2, 1 ];// function calldocument.write( countSwaps(A, 6));</script> |
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(N)
Using Using Hash Map in python:
Approach:
In this approach, a hash map can be used to store the original position of each element. The array can be sorted and the number of swaps required can be counted. While swapping, the original position of the swapped elements can be updated in the hash map. Finally, the number of elements that occupy the original position of a higher element after swapping can be counted using the hash map.
Create a dictionary with the value as the key and index as the value.
Sort the array.
Loop through the array, and for each element that is not in the correct position, swap it with the element that should be in its place, and update the dictionary accordingly. Keep track of the number of swaps performed.
Count the number of elements that replace a greater element after sorting the array.
Python3
def count_swaps(arr): n = len(arr) position = {arr[i]: i for i in range(n)} # create a dictionary with the value as key and index as value arr.sort() # sort the array swaps = 0 for i in range(n): if position[arr[i]] != i: # if the current element is not in the correct position position[arr[i]], position[arr[position[arr[i]]]] = position[arr[position[arr[i]]]], position[arr[i]] # swap the elements swaps += 1 count = 0 for i in range(n-1): if arr[i] < arr[i+1]: # count the number of elements which replace a greater element after sorting count += 1 return count, swaps# Example usage:arr1 = [4, 3, 3, 2, 5]arr2 = [6, 5, 4, 3, 2, 1]count1, swaps1 = count_swaps(arr1)count2, swaps2 = count_swaps(arr2)print("Input: A[] = {}".format(arr1))print("Output: {}".format(count1))print("Number of swaps required: {}".format(swaps1))print("Input: A[] = {}".format(arr2))print("Output: {}".format(count2))print("Number of swaps required: {}".format(swaps2)) |
Output
Input: A[] = [2, 3, 3, 4, 5] Output: 3 Number of swaps required: 3 Input: A[] = [1, 2, 3, 4, 5, 6] Output: 5 Number of swaps required: 6
This approach has a time complexity of O(nlogn)
And an auxiliary space of O(n).
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