Minimize swaps required to maximize the count of elements replacing a greater element in an Array
Last Updated :
15 Apr, 2023
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++
#include <bits/stdc++.h>
using namespace std;
int countSwaps(int A[], int n)
{
sort(A, A + n);
int ind = 1, res = 0;
for (int i = 0; i < n; i++) {
while (ind < n and A[ind] == A[i])
ind++;
if (ind < n and A[ind] > A[i]) {
res++;
ind++;
}
if (ind >= n)
break;
}
return res;
}
int main()
{
int A[] = { 4, 3, 3, 2, 5 };
cout << countSwaps(A, 5);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int countSwaps(int A[], int n)
{
Arrays.sort(A);
int ind = 1, res = 0;
for (int i = 0; i < n; i++)
{
while (ind < n && A[ind] == A[i])
ind++;
if (ind < n && A[ind] > A[i])
{
res++;
ind++;
}
if (ind >= n)
break;
}
return res;
}
public static void main(String[] args)
{
int A[] = { 4, 3, 3, 2, 5 };
System.out.print(countSwaps(A, 5));
}
}
|
Python3
def countSwaps(A, n):
A.sort()
ind, res = 1, 0
for i in range(n):
while (ind < n and A[ind] == A[i]):
ind += 1
if (ind < n and A[ind] > A[i]):
res += 1
ind += 1
if (ind >= n):
break
return res
A = [ 4, 3, 3, 2, 5 ]
print (countSwaps(A, 5))
|
C#
using System;
class GFG{
static int countSwaps(int []A, int n)
{
Array.Sort(A);
int ind = 1, res = 0;
for (int i = 0; i < n; i++)
{
while (ind < n && A[ind] == A[i])
ind++;
if (ind < n && A[ind] > A[i])
{
res++;
ind++;
}
if (ind >= n)
break;
}
return res;
}
public static void Main(String[] args)
{
int []A = { 4, 3, 3, 2, 5 };
Console.Write(countSwaps(A, 5));
}
}
|
Javascript
<script>
function countSwaps(A, n)
{
A.sort();
let ind = 1, res = 0;
for (let i = 0; i < n; i++)
{
while (ind < n && A[ind] == A[i])
ind++;
if (ind < n && A[ind] > A[i])
{
res++;
ind++;
}
if (ind >= n)
break;
}
return res;
}
let A = [ 4, 3, 3, 2, 5 ];
document.write(countSwaps(A, 5));
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
int countSwaps(int A[], int n)
{
map<int, int> mp;
int max_frequency = 0;
for (int i = 0; i < n; i++) {
mp[A[i]]++;
max_frequency
= max(max_frequency, mp[A[i]]);
}
return n - max_frequency;
}
int main()
{
int A[] = { 6, 5, 4, 3, 2, 1 };
cout << countSwaps(A, 6);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int countSwaps(int arr[], int n)
{
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
int max_frequency = 0;
for (int i = 0; i < n; i++)
{
if(mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.put(arr[i], 1);
}
max_frequency = Math.max(max_frequency,
mp.get(arr[i]));
}
return n - max_frequency;
}
public static void main(String[] args)
{
int A[] = { 6, 5, 4, 3, 2, 1 };
System.out.print(countSwaps(A, 6));
}
}
|
Python3
def countSwaps(A, n):
mp = {}
max_frequency = 0
for i in range(n):
if A[i] in mp:
mp[A[i]] += 1
else:
mp[A[i]] = 1
max_frequency = max(max_frequency,
mp[A[i]])
return n - max_frequency
if __name__ == "__main__":
A = [6, 5, 4, 3, 2, 1]
print(countSwaps(A, 6))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int countSwaps(int []arr, int n)
{
Dictionary<int,
int> mp = new Dictionary<int,
int>();
int max_frequency = 0;
for(int i = 0; i < n; i++)
{
if(mp.ContainsKey(arr[i]))
{
mp[arr[i]] = mp[arr[i]] + 1;
}
else
{
mp.Add(arr[i], 1);
}
max_frequency = Math.Max(max_frequency,
mp[arr[i]]);
}
return n - max_frequency;
}
public static void Main(String[] args)
{
int []A = { 6, 5, 4, 3, 2, 1 };
Console.Write(countSwaps(A, 6));
}
}
|
Javascript
<script>
function countSwaps(A, n)
{
var mp = new Map();
var max_frequency = 0;
for (var i = 0; i < n; i++) {
if(mp.has(A[i]))
mp.set(A[i], mp.get(A[i])+1)
else
mp.set(A[i], 1);
max_frequency
= Math.max(max_frequency, mp.get(A[i]));
}
return n - max_frequency;
}
var A = [6, 5, 4, 3, 2, 1 ];
document.write( countSwaps(A, 6));
</script>
|
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)}
arr.sort()
swaps = 0
for i in range(n):
if position[arr[i]] != i:
position[arr[i]], position[arr[position[arr[i]]]] = position[arr[position[arr[i]]]], position[arr[i]]
swaps += 1
count = 0
for i in range(n-1):
if arr[i] < arr[i+1]:
count += 1
return count, swaps
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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