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.
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 conditionsint 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 Codeint 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 approachimport java.util.*;class GFG{ // Function to count pairs that satisfy// the given conditionsstatic 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 Codepublic 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 conditionsdef 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 Codeif __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 approachusing System;class GFG{ // Function to count pairs // that satisfy the given // conditionsstatic 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 |
6
Time Complexity: O(N * log(N) + M * log(M))
Auxiliary Space: O(1)
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