Count pairs from two arrays with difference exceeding K

Given two integer arrays arr[] and brr[] consisting of distinct elements of size N and M respectively and an integer K, the task is to find the count of pairs(arr[i], brr[j]) such that (brr[j] – arr[i]) > K.

Examples:

Input: arr[] = {5, 9, 1, 8}, brr[] {10, 12, 7, 4, 2, 3}, K = 3 
Output: 6 
Explanation:
Possible pairs that satisfy the given conditions are: { (5, 10), (5, 12), (1, 10), (1, 12), (1, 7), (8, 12) }. 
Therefore, the required output is 6.

Input: arr[] = {2, 10}, brr[] = {5, 7}, K = 2 
Output: 2 
Explanation:
Possible pairs that satisfy the given conditions are: { (2, 5), (2, 7) }. 
Therefore, the required output is 2.

Naive approach: The simplest approach to solve this problem is to traverse the array and generate all possible pairs of the given array and for each pair, check if (brr[j] – arr[i]) > K or not. If found to be true then increment the counter. Finally, print the value of the counter.

Algorithm

Initialize a variable sum to zero.
Loop through each element of the input array:
a. If the current element is even (i.e., its value is divisible by 2 with no remainder), add it to the sum.
Return the value of sum.

C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> arr = {5, 9, 1, 8};
    vector<int> brr = {10, 12, 7, 4, 2, 3};
    int K = 3;
    int count = 0;

    // Traverse the array and generate all possible pairs
    for (int i = 0; i < arr.size(); i++) {
        for (int j = 0; j < brr.size(); j++) {
            // Check if (brr[j] - arr[i]) > K
            if (brr[j] - arr[i] > K) {
                // Increment the counter
                count++;
            }
        }
    }

    // Print the count
    cout << count << endl;

    return 0;
}

Java

public class GFG {
    public static void main(String[] args) {
        int[] arr = {5, 9, 1, 8};
        int[] brr = {10, 12, 7, 4, 2, 3};
        int K = 3;
        int count = 0;

        // Traverse the array and generate all possible pairs
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < brr.length; j++) {
                // Check if (brr[j] - arr[i]) > K
                if (brr[j] - arr[i] > K) {
                    // Increment the counter
                    count++;
                }
            }
        }

        // Print the count
        System.out.println(count);
    }
}

Python

arr = [5, 9, 1, 8]
brr = [10, 12, 7, 4, 2, 3]
K = 3

count = 0

# Traverse the array and generate all possible pairs
for i in range(len(arr)):
    for j in range(len(brr)):
        # Check if (brr[j] - arr[i]) > K
        if brr[j] - arr[i] > K:
            # Increment the counter
            count += 1

# Print the count
print(count)

C#

// C# code for above approach
using System;

public class GFG {
    static public void Main() {
        int[] arr = {5, 9, 1, 8};
        int[] brr = {10, 12, 7, 4, 2, 3};
        int K = 3;
        int count = 0;

        // Traverse the array and generate all possible pairs
        for (int i = 0; i < arr.Length; i++) {
            for (int j = 0; j < brr.Length; j++) {
                // Check if (brr[j] - arr[i]) > K
                if (brr[j] - arr[i] > K) {
                    // Increment the counter
                    count++;
                }
            }
        }

        // Print the count
        Console.WriteLine(count);
    }
}

// This code is contributed by Utkarsh Kumar

Javascript

function main() {
  let arr = [5, 9, 1, 8];
  let brr = [10, 12, 7, 4, 2, 3];
  let K = 3;
  let count = 0;

  // Traverse the array and generate all possible pairs
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < brr.length; j++) {
      // Check if (brr[j] - arr[i]) > K
      if (brr[j] - arr[i] > K) {
        // Increment the counter
        count++;
      }
    }
  }

  // Print the count
  console.log(count);
}

main();
// This code is contributed by shivhack999

Output

6

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

Efficient approach: To optimize the above approach the idea is to first sort the array and then use two pointer techniques. Follow the steps below to solve the problem:

• Initialize a variable, say cntPairs to store the count of pairs that satisfy the given conditions.
• Sort the given array.
• Initialize two variables, say i = 0 and j = 0 to store the index of left and right pointers respectively.
• Traverse both the array and check if (brr[j] – arr[i]) > K or not. If found to be true then update the value of cntPairs += (M – j) and increment the value of i pointer variable.
• Otherwise, increment the value of j pointer variable.
• Finally, print the value of cntPrint.

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 count pairs that satisfy
// the given conditions
int count_pairs(int arr[], int brr[],
                int N, int M, int K)
{
    // Stores index of
    // the left pointer.
    int i = 0;

    // Stores index of
    // the right pointer
    int j = 0;

    // Stores count of total pairs
    // that satisfy the conditions
    int cntPairs = 0;

    // Sort arr[] array
    sort(arr, arr + N);

    // Sort brr[] array
    sort(brr, brr + M);

    // Traverse both the array
    // and count then pairs
    while (i < N && j < M) {

        // If the value of
        // (brr[j] - arr[i]) exceeds K
        if (brr[j] - arr[i] > K) {

            // Update cntPairs
            cntPairs += (M - j);

            // Update
            i++;
        }
        else {

            // Update j
            j++;
        }
    }

    return cntPairs;
}

// Driver Code
int main()
{
    int arr[] = { 5, 9, 1, 8 };
    int brr[] = { 10, 12, 7, 4, 2, 3 };
    int K = 3;
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = sizeof(brr) / sizeof(brr[0]);
    cout << count_pairs(arr, brr, N, M, K);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to count pairs that satisfy
// the given conditions
static int count_pairs(int arr[], int brr[],
                       int N, int M, int K)
{

    // Stores index of
    // the left pointer.
    int i = 0;

    // Stores index of
    // the right pointer
    int j = 0;

    // Stores count of total pairs
    // that satisfy the conditions
    int cntPairs = 0;

    // Sort arr[] array
    Arrays.sort(arr);

    // Sort brr[] array
    Arrays.sort(brr);

    // Traverse both the array
    // and count then pairs
    while (i < N && j < M)
    {

        // If the value of
        // (brr[j] - arr[i]) exceeds K
        if (brr[j] - arr[i] > K)
        {

            // Update cntPairs
            cntPairs += (M - j);

            // Update
            i++;
        }
        else
        {

            // Update j
            j++;
        }
    }
    return cntPairs;
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 9, 1, 8 };
    int brr[] = { 10, 12, 7, 4, 2, 3 };
    int K = 3;

    int N = arr.length;
    int M = brr.length;

    System.out.println(count_pairs(arr, brr, N, M, K));
}
}

// This code is contributed by SURENDRA_GANGWAR

Python3

# Python3 program to implement
# the above approach

# Function to count pairs that satisfy
# the given conditions
def count_pairs(arr, brr, N, M, K):

    # Stores index of
    # the left pointer.
    i = 0

    # Stores index of
    # the right pointer
    j = 0

    # Stores count of total pairs
    # that satisfy the conditions
    cntPairs = 0

    # Sort arr[] array
    arr = sorted(arr)

    # Sort brr[] array
    brr = sorted(brr)

    # Traverse both the array
    # and count then pairs
    while (i < N and j < M):

        # If the value of
        # (brr[j] - arr[i]) exceeds K
        if (brr[j] - arr[i] > K):

            # Update cntPairs
            cntPairs += (M - j)

            # Update
            i += 1
        else:

            # Update j
            j += 1

    return cntPairs

# Driver Code
if __name__ == '__main__':

    arr = [ 5, 9, 1, 8 ]
    brr = [ 10, 12, 7, 4, 2, 3 ]
    K = 3

    N = len(arr)
    M = len(brr)

    print(count_pairs(arr, brr, N, M, K))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{

// Function to count pairs
// that satisfy the given
// conditions
static int count_pairs(int[] arr, int[] brr,
                       int N, int M, int K)
{
  // Stores index of
  // the left pointer.
  int i = 0;

  // Stores index of
  // the right pointer
  int j = 0;

  // Stores count of total pairs
  // that satisfy the conditions
  int cntPairs = 0;

  // Sort arr[] array
  Array.Sort(arr);

  // Sort brr[] array
  Array.Sort(brr);

  // Traverse both the array
  // and count then pairs
  while (i < N && j < M)
  {
    // If the value of
    // (brr[j] - arr[i])
    // exceeds K
    if (brr[j] - arr[i] > K)
    {
      // Update cntPairs
      cntPairs += (M - j);

      // Update
      i++;
    }
    else
    {
      // Update j
      j++;
    }
  }
  return cntPairs;
}

// Driver code
static void Main()
{
  int[] arr = {5, 9, 1, 8};
  int[] brr = {10, 12,
               7, 4, 2, 3};
  int K = 3;
  int N = arr.Length;
  int M = brr.Length;
  Console.WriteLine(
  count_pairs(arr, brr,
              N, M, K));
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>
// Javascript program to implement
// the above approach

// Function to count pairs that satisfy
// the given conditions
function count_pairs(arr, brr, N, M, K)
{

    // Stores index of
    // the left pointer.
    let i = 0;

    // Stores index of
    // the right pointer
    let j = 0;

    // Stores count of total pairs
    // that satisfy the conditions
    let cntPairs = 0;

    // Sort arr[] array
    (arr).sort(function(a,b){return a-b;});

    // Sort brr[] array
    (brr).sort(function(a,b){return a-b;});

    // Traverse both the array
    // and count then pairs
    while (i < N && j < M)
    {

        // If the value of
        // (brr[j] - arr[i]) exceeds K
        if (brr[j] - arr[i] > K)
        {

            // Update cntPairs
            cntPairs += (M - j);

            // Update
            i++;
        }
        else
        {

            // Update j
            j++;
        }
    }
    return cntPairs;
}

// Driver Code
let arr = [5, 9, 1, 8];
let brr = [ 10, 12, 7, 4, 2, 3];
let K = 3;
let N = arr.length;
let M = brr.length;
document.write(count_pairs(arr, brr, N, M, K));

// This code is contributed by unknown2108
</script>

Output

6

Time Complexity: O(N * log(N) + M * log(M))
Auxiliary Space: O(1)