Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Minimize swaps required to make the first and last elements the largest and smallest elements in the array respectively

  • Last Updated : 14 May, 2021

Given an array arr[] consisting of N integers, the task is to find the minimum number of adjacent swaps required to make the first and the last elements in the array the largest and smallest element present in the array respectively.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {1, 3, 2}
Output: 2
Explanation:
Initially the array is {1, 3, 2}.
Operation 1: Swap element at index 0 and 1. The array modifies to {3, 1, 2}.
Operation 2: Swap element at index 1 and 2. The array modifies to {3, 2, 1}.
Therefore, the minimum number of operation required is 2.



Input: arr[] = {100, 95, 100, 100, 88}
Output: 0

Approach: Follow the steps below to solve the given problem:

  • Initialize a variable, say, count, to store the total number of swaps required.
  • Find the minimum and the maximum element of the array and store it in a variable, say minElement and maxElement respectively.
  • If the minimum and maximum elements are found to be the same, the array consists of a single distinct element. Therefore, print 0 as both the elements are at their correct positions respectively.
  • Otherwise, traverse the array and find the index of the first occurrence of maxElement and the last occurrence of minElement and store it in a variable, say minIndex and maxIndex respectively.
  • Update the value of count as the sum of maxIndex and (N – 1 – minIndex) as the number of swap operations required to make the largest and the smallest element as the first and the last elements of the array respectively.
  • If minIndex is less than maxIndex, then decrease count by 1, as one swap operation will overlap.
  • After completing the above steps, print the value of count as the minimum number of swaps required.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of swaps required to make the first
// and the last elements the largest
// and smallest element in the array
int minimum_swaps(int arr[], int n)
{
    // Stores the count of swaps
    int count = 0;
 
    // Stores the maximum element
    int max_el = *max_element(arr,
                              arr + n);
 
    // Stores the minimum element
    int min_el = *min_element(arr,
                              arr + n);
 
    // If the array contains
    // a single distinct element
    if (min_el == max_el)
        return 0;
 
    // Stores the indices of the
    // maximum and minimum elements
    int index_max = -1;
    int index_min = -1;
 
    for (int i = 0; i < n; i++) {
 
        // If the first index of the
        // maximum element is found
        if (arr[i] == max_el
            && index_max == -1) {
            index_max = i;
        }
 
        // If current index has
        // the minimum element
        if (arr[i] == min_el) {
            index_min = i;
        }
    }
 
    // Update the count of operations to
    // place largest element at the first
    count += index_max;
 
    // Update the count of operations to
    // place largest element at the last
    count += (n - 1 - index_min);
 
    // If smallest element is present
    // before the largest element initially
    if (index_min < index_max)
        count -= 1;
 
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 1, 6, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << minimum_swaps(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.Arrays;
import java.util.Collections;
 
class GFG{
 
// Function to find the minimum number
// of swaps required to make the first
// and the last elements the largest
// and smallest element in the array
static int minimum_swaps(Integer arr[], int n)
{
     
    // Stores the count of swaps
    int count = 0;
     
    // Stores the maximum element
    int max_el = Collections.max(Arrays.asList(arr));
 
    // Stores the minimum element
    int min_el = Collections.min(Arrays.asList(arr));
 
    // If the array contains
    // a single distinct element
    if (min_el == max_el)
        return 0;
 
    // Stores the indices of the
    // maximum and minimum elements
    int index_max = -1;
    int index_min = -1;
 
    for(int i = 0; i < n; i++)
    {
         
        // If the first index of the
        // maximum element is found
        if (arr[i] == max_el &&
            index_max == -1)
        {
            index_max = i;
        }
 
        // If current index has
        // the minimum element
        if (arr[i] == min_el)
        {
            index_min = i;
        }
    }
 
    // Update the count of operations to
    // place largest element at the first
    count += index_max;
 
    // Update the count of operations to
    // place largest element at the last
    count += (n - 1 - index_min);
 
    // If smallest element is present
    // before the largest element initially
    if (index_min < index_max)
        count -= 1;
 
    return count;
}
 
// Driver Code
public static void main (String[] args)
{
    Integer arr[] = { 2, 4, 1, 6, 5 };
    int N = arr.length;
     
    System.out.println(minimum_swaps(arr, N));
}
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
 
# Function to find the minimum number
# of swaps required to make the first
# and the last elements the largest
# and smallest element in the array
def minimum_swaps(arr, n):
     
    # Stores the count of swaps
    count = 0
 
    # Stores the maximum element
    max_el = max(arr)
 
    # Stores the minimum element
    min_el = min(arr)
 
    # If the array contains
    # a single distinct element
    if (min_el == max_el):
        return 0
 
    # Stores the indices of the
    # maximum and minimum elements
    index_max = -1
    index_min = -1
 
    for i in range(n):
         
        # If the first index of the
        # maximum element is found
        if (arr[i] == max_el and
            index_max == -1):
            index_max = i
 
        # If current index has
        # the minimum element
        if (arr[i] == min_el):
            index_min = i
 
    # Update the count of operations to
    # place largest element at the first
    count += index_max
 
    # Update the count of operations to
    # place largest element at the last
    count += (n - 1 - index_min)
 
    # If smallest element is present
    # before the largest element initially
    if (index_min < index_max):
        count -= 1
 
    return count
 
# Driver Code
if __name__ == '__main__':
     
    arr = [2, 4, 1, 6, 5]
    N = len(arr)
     
    print(minimum_swaps(arr, N))
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
using System.Linq;
 
class GFG{
 
// Function to find the minimum number
// of swaps required to make the first
// and the last elements the largest
// and smallest element in the array
static int minimum_swaps(int[] arr, int n)
{
     
    // Stores the count of swaps
    int count = 0;
 
    // Stores the maximum element
    int max_el = arr.Max();
 
    // Stores the minimum element
    int min_el = arr.Min();
 
    // If the array contains
    // a single distinct element
    if (min_el == max_el)
        return 0;
 
    // Stores the indices of the
    // maximum and minimum elements
    int index_max = -1;
    int index_min = -1;
 
    for(int i = 0; i < n; i++)
    {
         
        // If the first index of the
        // maximum element is found
        if (arr[i] == max_el &&
            index_max == -1)
        {
            index_max = i;
        }
 
        // If current index has
        // the minimum element
        if (arr[i] == min_el)
        {
            index_min = i;
        }
    }
 
    // Update the count of operations to
    // place largest element at the first
    count += index_max;
 
    // Update the count of operations to
    // place largest element at the last
    count += (n - 1 - index_min);
 
    // If smallest element is present
    // before the largest element initially
    if (index_min < index_max)
        count -= 1;
 
    return count;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 2, 4, 1, 6, 5 };
    int N = arr.Length;
 
    Console.WriteLine(minimum_swaps(arr, N));
}
}
 
// This code is contributed by ukasp

Javascript




<script>
      // JavaScript program for the above approach
      // Function to find the minimum number
      // of swaps required to make the first
      // and the last elements the largest
      // and smallest element in the array
      function minimum_swaps(arr, n) {
        // Stores the count of swaps
        var count = 0;
 
        // Stores the maximum element
        var max_el = Math.max(...arr);
 
        // Stores the minimum element
        var min_el = Math.min(...arr);
 
        // If the array contains
        // a single distinct element
        if (min_el === max_el) return 0;
 
        // Stores the indices of the
        // maximum and minimum elements
        var index_max = -1;
        var index_min = -1;
 
        for (var i = 0; i < n; i++) {
          // If the first index of the
          // maximum element is found
          if (arr[i] === max_el && index_max === -1) {
            index_max = i;
          }
 
          // If current index has
          // the minimum element
          if (arr[i] === min_el) {
            index_min = i;
          }
        }
 
        // Update the count of operations to
        // place largest element at the first
        count += index_max;
 
        // Update the count of operations to
        // place largest element at the last
        count += n - 1 - index_min;
 
        // If smallest element is present
        // before the largest element initially
        if (index_min < index_max) count -= 1;
 
        return count;
      }
 
      // Driver Code
      var arr = [2, 4, 1, 6, 5];
      var N = arr.length;
 
      document.write(minimum_swaps(arr, N));
    </script>
Output: 
4

 

Time Complexity: O(N)
Auxiliary Space: O(1)




My Personal Notes arrow_drop_up
Recommended Articles
Page :