将所有小于或等于 k 的元素组合在一起所需的最小交换次数

给定一个包含 n 个正整数和一个数字k的数组。找出将所有小于或等于k的数字组合在一起所需的最小交换次数。

Input:  arr[] = {2, 1, 5, 6, 3}, k = 3
Output: 1

Explanation: 
To bring elements 2, 1, 3 together, swap 
element '5' with '3' such that final array
will be-
arr[] = {2, 1, 3, 6, 5}

Input:  arr[] = {2, 7, 9, 5, 8, 7, 4}, k = 5
Output: 2

一个简单的解决方案是首先计算所有小于或等于k的元素（比如“好”）。现在遍历每个子数组并交换那些值大于k的元素。这种方法的时间复杂度是 O(n ² )
一种有效的方法是使用两个指针技术和滑动窗口。这种方法的时间复杂度是O(n)

1. 查找小于或等于“k”的所有元素的计数。假设计数是'cnt'
2. 对长度为“cnt”的窗口使用两个指针技术，每次跟踪此范围内有多少元素大于“k”。假设总计数是“坏的”。
3. 对每个长度为“cnt”的窗口重复步骤 2，并在其中取最少的计数“坏”。这将是最终的答案。

C++

// C++ program to find minimum swaps required
// to club all elements less than or equals
// to k together
#include 
using namespace std;
 
// Utility function to find minimum swaps
// required to club all elements less than
// or equals to k together
int minSwap(int *arr, int n, int k) {
     
    // Find count of elements which are
    // less than equals to k
    int count = 0;
    for (int i = 0; i < n; ++i)
        if (arr[i] <= k)
            ++count;
     
    // Find unwanted elements in current
    // window of size 'count'
    int bad = 0;
    for (int i = 0; i < count; ++i)
        if (arr[i] > k)
            ++bad;
     
    // Initialize answer with 'bad' value of
    // current window
    int ans = bad;
    for (int i = 0, j = count; j < n; ++i, ++j) {
         
        // Decrement count of previous window
        if (arr[i] > k)
            --bad;
         
        // Increment count of current window
        if (arr[j] > k)
            ++bad;
         
        // Update ans if count of 'bad'
        // is less in current window
        ans = min(ans, bad);
    }
    return ans;
}
 
// Driver code
int main() {
     
    int arr[] = {2, 1, 5, 6, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << minSwap(arr, n, k) << "\n";
     
    int arr1[] = {2, 7, 9, 5, 8, 7, 4};
    n = sizeof(arr1) / sizeof(arr1[0]);
    k = 5;
    cout << minSwap(arr1, n, k);
    return 0;
}

Java

// Java program to find minimum
// swaps required to club all
// elements less than or equals
// to k together
import java.lang.*;
 
class GFG {
     
// Utility function to find minimum swaps
// required to club all elements less than
// or equals to k together
static int minSwap(int arr[], int n, int k) {
 
    // Find count of elements which are
    // less than equals to k
    int count = 0;
    for (int i = 0; i < n; ++i)
    if (arr[i] <= k)
        ++count;
 
    // Find unwanted elements in current
    // window of size 'count'
    int bad = 0;
    for (int i = 0; i < count; ++i)
    if (arr[i] > k)
        ++bad;
 
    // Initialize answer with 'bad' value of
    // current window
    int ans = bad;
    for (int i = 0, j = count; j < n; ++i, ++j) {
 
    // Decrement count of previous window
    if (arr[i] > k)
        --bad;
 
    // Increment count of current window
    if (arr[j] > k)
        ++bad;
 
    // Update ans if count of 'bad'
    // is less in current window
    ans = Math.min(ans, bad);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = {2, 1, 5, 6, 3};
    int n = arr.length;
    int k = 3;
    System.out.print(minSwap(arr, n, k) + "\n");
 
    int arr1[] = {2, 7, 9, 5, 8, 7, 4};
    n = arr1.length;
    k = 5;
    System.out.print(minSwap(arr1, n, k));
}
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find
# minimum swaps required
# to club all elements less
# than or equals to k together
 
# Utility function to find
# minimum swaps required to
# club all elements less than
# or equals to k together
def minSwap(arr, n, k) :
     
    # Find count of elements
    # which are less than
    # equals to k
    count = 0
    for i in range(0, n) :
        if (arr[i] <= k) :
            count = count + 1
     
    # Find unwanted elements
    # in current window of
    # size 'count'
    bad = 0
    for i in range(0, count) :
        if (arr[i] > k) :
            bad = bad + 1
     
    # Initialize answer with
    # 'bad' value of current
    # window
    ans = bad
    j = count
    for i in range(0, n) :
         
        if(j == n) :
            break
             
        # Decrement count of
        # previous window
        if (arr[i] > k) :
            bad = bad - 1
         
        # Increment count of
        # current window
        if (arr[j] > k) :
            bad = bad + 1
         
        # Update ans if count
        # of 'bad' is less in
        # current window
        ans = min(ans, bad)
 
        j = j + 1
 
    return ans
 
# Driver code
arr = [2, 1, 5, 6, 3]
n = len(arr)
k = 3
print (minSwap(arr, n, k))
 
arr1 = [2, 7, 9, 5, 8, 7, 4]
n = len(arr1)
k = 5
print (minSwap(arr1, n, k))
 
# This code is contributed by
# Manish Shaw(manishshaw1)

C#

// C# program to find minimum
// swaps required to club all
// elements less than or equals
// to k together
using System;
 
class GFG {
     
    // Utility function to find minimum swaps
    // required to club all elements less than
    // or equals to k together
    static int minSwap(int []arr, int n, int k) {
     
        // Find count of elements which are
        // less than equals to k
        int count = 0;
        for (int i = 0; i < n; ++i)
        if (arr[i] <= k)
            ++count;
     
        // Find unwanted elements in current
        // window of size 'count'
        int bad = 0;
        for (int i = 0; i < count; ++i)
        if (arr[i] > k)
            ++bad;
     
        // Initialize answer with 'bad' value of
        // current window
        int ans = bad;
        for (int i = 0, j = count; j < n; ++i, ++j) {
     
            // Decrement count of previous window
            if (arr[i] > k)
                --bad;
         
            // Increment count of current window
            if (arr[j] > k)
                ++bad;
         
            // Update ans if count of 'bad'
            // is less in current window
            ans = Math.Min(ans, bad);
        }
        return ans;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = {2, 1, 5, 6, 3};
        int n = arr.Length;
        int k = 3;
        Console.WriteLine(minSwap(arr, n, k));
     
        int []arr1 = {2, 7, 9, 5, 8, 7, 4};
        n = arr1.Length;
        k = 5;
        Console.WriteLine(minSwap(arr1, n, k));
    }
}
 
// This code is contributed by vt_m.

PHP

 $k)
            ++$bad;
     
    // Initialize answer
    // with 'bad' value of
    // current window
    $ans = $bad;
    for ($i = 0, $j = $count; $j < $n;
                        ++$i, ++$j)
    {
         
        // Decrement count of
        // previous window
        if ($arr[$i] > $k)
            --$bad;
         
        // Increment count of
        // current window
        if ($arr[$j] > $k)
            ++$bad;
         
        // Update ans if count of 'bad'
        // is less in current window
        $ans = min($ans, $bad);
    }
    return $ans;
}
 
// Driver code
$arr = array(2, 1, 5, 6, 3);
$n = sizeof($arr);
$k = 3;
echo(minSwap($arr, $n, $k) . "\n");
     
$arr1 = array(2, 7, 9, 5, 8, 7, 4);
$n = sizeof($arr1);
$k = 5;
echo(minSwap($arr1, $n, $k));
 
// This code is contributed by Ajit.
?>

Javascript

 
// Utility function to find minimum swaps
// required to club all elements less than
// or equals to k together
function minSwap(arr,  n,  k) {
     
    // Find count of elements which are
    // less than equals to k
    var count = 0;
    for (var i = 0; i < n; ++i)
        if (arr[i] <= k)
            ++count;
     
    // Find unwanted elements in current
    // window of size 'count'
    var bad = 0;
    for (var i = 0; i < count; ++i)
        if (arr[i] > k)
            ++bad;
     
    // Initialize answer with 'bad' value of
    // current window
    var ans = bad;
    for (var i = 0, j = count; j < n; ++i, ++j) {
         
        // Decrement count of previous window
        if (arr[i] > k)
            --bad;
         
        // Increment count of current window
        if (arr[j] > k)
            ++bad;
         
        // Update ans if count of 'bad'
        // is less in current window
        ans = Math.min(ans, bad);
    }
    return ans;
}
 
    // Driver code
    var arr=[2, 1, 5, 6, 3];
    var n =5;
    var k = 3;
    document.write(minSwap(arr, n, k) + "<br>");
     
    var arr1 = [2, 7, 9, 5, 8, 7, 4];
    n = 7;
    k = 5;
    document.write(minSwap(arr1, n, k));
 
// This code is by Akshit Nikita Saxena