差异至多 K 且没有元素重复的对数

给定一个数组arr[]和一个数字K ，任务是计算差值小于或等于 K 的对的数量，这样一个元素只能被认为是一对。

例子：

Input: arr[] = {1, 3, 3, 9, 4}, K = 2 
Output: 2 
Explanation: 
There are only two pairs whose difference is atmost 2 
(1, 3), (3, 4)

Input: arr[] = {1, 4, 3, 7, 5}, K = 2 
Output: 2 
Explanation: 
There are five pairs in the array whose difference is atmost 2, 
(1, 3), (3, 4), (4, 5), (3, 5), (5, 7) 
But only two of them can be considered at a time because one element 
can be taken in a pair only once. 
 

 

方法：
思路是对数组进行排序，找出相邻元素的差值，如果差值最多为K，则考虑成对并增加计数，然后根据条件，任何元素只能在一对中，然后如果找到一对，则将计数器增加 2，这样任何元素都只出现在一对中。

例如：

Given Array - {1, 4, 3, 7, 5}, K = 2 
After Sorting Array will be - {1, 3, 4, 5, 7}
Step 1 - i = 0, count = 0
Consider the pair of elements for i and i + 1
Pair - (1, 3), Difference = 3 - 1 = 2
As the Difference is less than equal to 2
count = 1 and i = 2

Step 2 - i = 2, count = 1
Consider the pair of elements for i and i + 1
Pair - (4, 5), Difference = 5 - 4 = 1
As the Difference is less than equal to 2
count = 2 and i = 4
As i is greater than length-2,
there will be no more possible pairs.

算法：

• 使用任何排序算法对数组进行排序，以便连续元素在一起。
• 将索引计数器(例如i )初始化为零并运行 while 循环，直到索引计数器小于(长度 – 1)
  1. 检查索引i和i + 1处元素的差异。
  2. 如果差值小于或等于 K，则将索引增加 2，并将计数器增加 1，以考虑立即增加元素。
  3. 否则，将索引增加 1 以考虑由下一个元素形成的对。

下面是上述方法的实现：

C++

// C++ implementation to count the
// number of pairs whose difference
// is atmost K in an array
#include 
#include
 
using namespace std;
 
    // Function to count the
    // number of pairs whose difference
    // is atmost K in an array
    int countPairs(int arr[], int k, int n)
    {
         
        // Variable to store the count of pairs
        // whose difference is atmost K
        int pair = 0;
        int index = 0;
         
        //int n = sizeof(arr)/sizeof(arr[0]);
         
        // Sorting the Array
        sort(arr,arr + n) ;
                 
        // Loop to consider the consecutive
        // pairs of the array
        while(index < n -1)
        {
             
            // if Pair found increment
            // the index by 2
            if (arr[index + 1] - arr[index] <= k){
                pair += 1 ;
                index += 2 ;
            }
            else{
                index += 1;
            }
        }
        return pair ;
     
    }
     
// Driver Code
int main() {
    int arr[] = {1, 4, 3, 7, 5} ;
    int k = 2;
     
     int n = sizeof(arr)/sizeof(arr[0]);
    // Function Call
    int count = countPairs(arr, k,n) ;
    cout << count << endl;;
}
 
// This code is contributed by AnkitRai01

Java

// Java implementation to count the
// number of pairs whose difference
// is atmost K in an array
import java.util.*;
 
class GFG
{
 
    // Function to count the
    // number of pairs whose difference
    // is atmost K in an array
    static int countPairs(int arr[], int k)
    {
         
        // Sorting the Array
        Arrays.sort(arr) ;
         
        // Variable to store the count of pairs
        // whose difference is atmost K
        int pair = 0;
        int index = 0;
         
        // Loop to consider the consecutive
        // pairs of the array
        while(index < arr.length -1)
        {
             
            // if Pair found increment
            // the index by 2
            if (arr[index + 1] - arr[index] <= k){
                pair += 1 ;
                index += 2 ;
            }
            else{
                index += 1;
            }
        }
        return pair ;
     
    }
     
    // Driver Code
    public static void main (String[] args) {
        int arr[] = {1, 4, 3, 7, 5} ;
        int k = 2;
         
        // Function Call
        int count = countPairs(arr, k) ;
        System.out.println(count);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to count the
# number of pairs whose difference
# is atmost K in an array
 
 
# Function to count the
# number of pairs whose difference
# is atmost K in an array
def countPairs(arr, k):
     
    # Sorting the Array
    arr.sort()
     
    # Variable to store the count of pairs
    # whose difference is atmost K
    pair = 0
    index = 0
     
    # Loop to consider the consecutive
    # pairs of the array
    while(index < len(arr)-1):
         
        # if Pair found increment
        # the index by 2
        if arr[index + 1] - arr[index] <= k:
            pair += 1
            index += 2
        else:
            index += 1
             
    return pair
 
# Driver Code
if __name__ == "__main__":
    arr = [1, 4, 3, 7, 5]
    k = 2
    # Function Call
    count = countPairs(arr, k)
    print(count)

C#

// C# implementation to count the
// number of pairs whose difference
// is atmost K in an array
using System;
 
class GFG
{
 
    // Function to count the
    // number of pairs whose difference
    // is atmost K in an array
    static int countPairs(int []arr, int k)
    {
         
        // Sorting the Array
        Array.Sort(arr) ;
         
        // Variable to store the count of pairs
        // whose difference is atmost K
        int pair = 0;
        int index = 0;
         
        // Loop to consider the consecutive
        // pairs of the array
        while(index < arr.Length - 1)
        {
             
            // if Pair found increment
            // the index by 2
            if (arr[index + 1] - arr[index] <= k)
            {
                pair += 1 ;
                index += 2 ;
            }
            else
            {
                index += 1;
            }
        }
        return pair ;
    }
     
    // Driver Code
    public static void Main ()
    {
        int []arr = {1, 4, 3, 7, 5} ;
        int k = 2;
         
        // Function Call
        int count = countPairs(arr, k) ;
        Console.WriteLine(count);
    }
}
 
// This code is contributed by AnkitRai01

Javascript

    // Javascript implementation to count the
    // number of pairs whose difference
    // is atmost K in an array
 
 
    // Function to count the
    // number of pairs whose difference
    // is atmost K in an array
    function countPairs(arr, k, n)
    {
         
        // Variable to store the count of pairs
        // whose difference is atmost K
        let pair = 0;
        let index = 0;
         
        //int n = sizeof(arr)/sizeof(arr[0]);
         
        // Sorting the Array
        arr.sort((a, b) => a - b) ;
                 
        // Loop to consider the consecutive
        // pairs of the array
        while(index < n -1)
        {
             
            // if Pair found increment
            // the index by 2
            if (arr[index + 1] - arr[index] <= k){
                pair += 1 ;
                index += 2 ;
            }
            else{
                index += 1;
            }
        }
        return pair ;
     
    }
     
// Driver Code
 
let arr = [1, 4, 3, 7, 5];
let k = 2;
     
let n = arr.length;
 
// Function Call
let count = countPairs(arr, k,n);
 
document.write(count + "<br>");
 
// This code is contributed by _saurabh_jaiswal

输出：

2

性能分析：

• 时间复杂度：在上面给出的方法中，对需要 O(N logN) 的数组进行排序，并且还有一次迭代来计算 O(N) 的对数。
  因此，该方法的整体复杂度为O(N logN + N) 。
• 空间复杂度：在上面给出的方法中，没有额外的空间。因此，该方法的整体空间复杂度将为O(1)

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