Given two arrays a[] and b[], both of size N. The task is to count the number of distinct pairs such that (a[i] + a[j] ) > ( b[i] + b[j] ) subject to condition that (j > i).
Examples:
Input: N = 5, a[] = {1, 2, 3, 4, 5}, b[] = {2, 5, 6, 1, 9}
Output: 1
Explanation:
Only one such pair exists and that is (0, 3).
a[0] = 1, a[3] = 4 and a[0] + a[3] = 1 + 4 = 5.
b[0] = 2, b[3] = 1 and b[0] + b[3] = 2 + 1 = 3.
Clearly, 5 > 3 and j > i. Thus (0, 3) is a possible pair.Input: N = 5, a[] = {2, 4, 2, 7, 8}, b[] = {1, 3, 6, 4, 5}
Output: 6
Naive Approach:
The simplest way is to iterate through every possible pair and if it satisfies the condition, then increment the count. Then, return the count as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above problem#include <bits/stdc++.h>using namespace std;// function to find the number of// pairs satisfying the given cond.int count_pairs(int a[], int b[], int N){ // variables used for traversal int i, j; // count variable to store the // count of possible pairs int count = 0; // Nested loop to find out the // possible pairs for (i = 0; i < (N - 1); i++) { for (j = (i + 1); j < N; j++) { // Check if the given // condition is satisfied // or not. If yes then // increment the count. if ((a[i] + a[j]) > (b[i] + b[j])) { count++; } } } // Return the count value return count;}// Driver Codeint main(){ // Size of the arrays int N = 5; // Initialise the arrays int a[N] = { 1, 2, 3, 4, 5 }; int b[N] = { 2, 5, 6, 1, 9 }; // function call that returns // the count of possible pairs cout << count_pairs(a, b, N) << endl; return 0;} |
Java
// Java program for the above problemclass GFG{// function to find the number of// pairs satisfying the given cond.static int count_pairs(int []a, int b[], int N){ // variables used for traversal int i, j; // count variable to store the // count of possible pairs int count = 0; // Nested loop to find out the // possible pairs for (i = 0; i < (N - 1); i++) { for (j = (i + 1); j < N; j++) { // Check if the given // condition is satisfied // or not. If yes then // increment the count. if ((a[i] + a[j]) > (b[i] + b[j])) { count++; } } } // Return the count value return count;}// Driver Codepublic static void main(String[] args){ // Size of the arrays int N = 5; // Initialise the arrays int a[] = new int[]{ 1, 2, 3, 4, 5 }; int b[] = new int[]{ 2, 5, 6, 1, 9 }; // function call that returns // the count of possible pairs System.out.println(count_pairs(a, b, N));}}// This code is contributed by rock_cool |
Python3
# Python3 program for the above problem# function to find the number of# pairs satisfying the given cond.def count_pairs(a, b, N): # count variable to store the # count of possible pairs count = 0; # Nested loop to find out the # possible pairs for i in range(0, N - 1): for j in range(i + 1, N): # Check if the given # condition is satisfied # or not. If yes then # increment the count. if ((a[i] + a[j]) > (b[i] + b[j])): count += 1; # Return the count value return count;# Driver Code# Size of the arraysN = 5;# Initialise the arraysa = [ 1, 2, 3, 4, 5 ];b = [ 2, 5, 6, 1, 9 ];# function call that returns# the count of possible pairsprint(count_pairs(a, b, N)# This code is contributed by Code_Mech |
C#
// C# program for the above problemusing System;class GFG{// function to find the number of// pairs satisfying the given cond.static int count_pairs(int []a, int []b, int N){ // variables used for traversal int i, j; // count variable to store the // count of possible pairs int count = 0; // Nested loop to find out the // possible pairs for (i = 0; i < (N - 1); i++) { for (j = (i + 1); j < N; j++) { // Check if the given // condition is satisfied // or not. If yes then // increment the count. if ((a[i] + a[j]) > (b[i] + b[j])) { count++; } } } // Return the count value return count;}// Driver Codepublic static void Main(){ // Size of the arrays int N = 5; // Initialise the arrays int []a = new int[]{ 1, 2, 3, 4, 5 }; int []b = new int[]{ 2, 5, 6, 1, 9 }; // function call that returns // the count of possible pairs Console.Write(count_pairs(a, b, N));}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript program for the above problem// function to find the number of// pairs satisfying the given cond.function count_pairs(a, b, N){ // Variables used for traversal let i, j; // count variable to store the // count of possible pairs let count = 0; // Nested loop to find out the // possible pairs for(i = 0; i < (N - 1); i++) { for(j = (i + 1); j < N; j++) { // Check if the given // condition is satisfied // or not. If yes then // increment the count. if ((a[i] + a[j]) > (b[i] + b[j])) { count++; } } } // Return the count value return count;} // Driver code// Size of the arrayslet N = 5;// Initialise the arrayslet a = [ 1, 2, 3, 4, 5 ];let b = [ 2, 5, 6, 1, 9 ];// Function call that returns// the count of possible pairsdocument.write(count_pairs(a, b, N));// This code is contributed by divyesh072019</script> |
Output:
1
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps below to solve the problem:
- Re-arrange the given inequality as:
=> a[i] + a[j] > b[i] + b[j]
=> a[i] – b[i] + a[j] – b[j] > 0
- Initialize another array c[ ] of size N that stores the values of a[i] – b[i].
- Sort the array c[ ].
- Initialise an answer variable to 0. Iterate over the array c[ ].
- For every index i in the array c[ ] do the following operations:
- If c[i] <= 0, simply continue.
- If c[i] > 0, calculate the minimum index position and store the value in a variable pos such that c[pos] + c[i] > 0. The value of pos can be easily found using lower_bound function in C++ STL.
- Add (i – pos) to the answer.
Illustration:
- N = 5, a[] = {1, 2, 3, 4, 5}, b[] = {2, 5, 6, 1, 9}
- The array c[] for this example will be c[] = {-1, -3, -3, 3, -4}
- After sorting, array c[] = {-4, -3, -3, -1, 3}
- The first and only positive value of array c[] is found at index 4.
- pos = lower_bound(-c[4] + 1) = lower_bound(-2) = 3.
- Count of possible pairs = i – pos = 4 – 3 = 1.
Note: Had there been more than one positive number found in the array c[], then find out the pos for each and every positive value in c[] and add the value of i – pos to the count.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number// of pairs.int numberOfPairs(int* a, int* b, int n){ // Array c[] where // c[i] = a[i] - b[i] int c[n]; for (int i = 0; i < n; i++) { c[i] = a[i] - b[i]; } // Sort the array c sort(c, c + n); // Initialise answer as 0 int answer = 0; // Iterate from index // 0 to n - 1 for (int i = 1; i < n; i++) { // If c[i] <= 0 then // in the sorted array // c[i] + c[pos] can never // greater than 0 // where pos < i if (c[i] <= 0) continue; // Find the minimum index // such that c[i] + c[j] > 0 // which is equivalent to // c[j] >= - c[i] + 1 int pos = lower_bound(c, c + n, -c[i] + 1) - c; // Add ( i - pos) to answer answer += (i - pos); } // return the answer return answer;}// Driver codeint32_t main(){ // Number of elements // in a and b int n = 5; // array a int a[] = { 1, 2, 3, 4, 5 }; // array b int b[] = { 2, 5, 6, 1, 9 }; cout << numberOfPairs(a, b, n) << endl; return 0;} |
Java
// Java program of the above approachimport java.util.*;class GFG{ // Function to find the number// of pairs.static int numberOfPairs(int[] a, int[] b, int n){ // Array c[] where // c[i] = a[i] - b[i] int c[] = new int[n]; for(int i = 0; i < n; i++) { c[i] = a[i] - b[i]; } // Sort the array c Arrays.sort(c); // Initialise answer as 0 int answer = 0; // Iterate from index // 0 to n - 1 for(int i = 1; i < n; i++) { // If c[i] <= 0 then // in the sorted array // c[i] + c[pos] can never // greater than 0 // where pos < i if (c[i] <= 0) continue; // Which is equivalent to // c[j] >= - c[i] + 1 int pos = -1; for(int j = 0; j < n; j++) { if (c[i] + c[j] > 0) { pos = j; break; } } // Add (i - pos) to answer answer += (i - pos); } // Return the answer return answer;} // Driver code public static void main (String[] args) { // Number of elements // in a and b int n = 5; // array a int a[] = { 1, 2, 3, 4, 5 }; // array b int b[] = { 2, 5, 6, 1, 9 }; System.out.println(numberOfPairs(a, b, n));}}// This code is contributed by offbeat |
Python3
# Python3 program of the above approachfrom bisect import bisect_left# Function to find the number# of pairs.def numberOfPairs(a, b, n): # Array c[] where # c[i] = a[i] - b[i] c = [0 for i in range(n)] for i in range(n): c[i] = a[i] - b[i] # Sort the array c c = sorted(c) # Initialise answer as 0 answer = 0 # Iterate from index # 0 to n - 1 for i in range(1, n): # If c[i] <= 0 then in the # sorted array c[i] + c[pos] # can never greater than 0 # where pos < i if (c[i] <= 0): continue # Find the minimum index # such that c[i] + c[j] > 0 # which is equivalent to # c[j] >= - c[i] + 1 pos = bisect_left(c, -c[i] + 1) # Add ( i - pos) to answer answer += (i - pos) # Return the answer return answer# Driver codeif __name__ == '__main__': # Number of elements # in a and b n = 5 # Array a a = [ 1, 2, 3, 4, 5 ] # Array b b = [ 2, 5, 6, 1, 9 ] print(numberOfPairs(a, b, n))# This code is contributed by mohit kumar 29 |
C#
// C# program of the above approachusing System;class GFG{ // Function to find the number// of pairs.static int numberOfPairs(int[] a, int[] b, int n){ // Array c[] where // c[i] = a[i] - b[i] int[] c = new int[n]; for(int i = 0; i < n; i++) { c[i] = a[i] - b[i]; } // Sort the array c Array.Sort(c); // Initialise answer as 0 int answer = 0; // Iterate from index // 0 to n - 1 for(int i = 1; i < n; i++) { // If c[i] <= 0 then // in the sorted array // c[i] + c[pos] can never // greater than 0 // where pos < i if (c[i] <= 0) continue; // Which is equivalent to // c[j] >= - c[i] + 1 int pos = -1; for(int j = 0; j < n; j++) { if (c[i] + c[j] > 0) { pos = j; break; } } // Add (i - pos) to answer answer += (i - pos); } // Return the answer return answer;}// Driver Codestatic void Main(){ // Number of elements // in a and b int n = 5; // Array a int[] a = { 1, 2, 3, 4, 5 }; // Array b int[] b = { 2, 5, 6, 1, 9 }; Console.WriteLine(numberOfPairs(a, b, n));}}// This code is contributed by divyeshrabadiya07 |
Javascript
<script> // Javascript program of the above approach // Function to find the number of pairs. function numberOfPairs(a, b, n) { // Array c[] where // c[i] = a[i] - b[i] let c = new Array(n); for(let i = 0; i < n; i++) { c[i] = a[i] - b[i]; } // Sort the array c c.sort(function(a, b){return a - b}); // Initialise answer as 0 let answer = 0; // Iterate from index // 0 to n - 1 for(let i = 1; i < n; i++) { // If c[i] <= 0 then // in the sorted array // c[i] + c[pos] can never // greater than 0 // where pos < i if (c[i] <= 0) continue; // Which is equivalent to // c[j] >= - c[i] + 1 let pos = -1; for(let j = 0; j < n; j++) { if (c[i] + c[j] > 0) { pos = j; break; } } // Add (i - pos) to answer answer += (i - pos); } // Return the answer return answer; } // Number of elements // in a and b let n = 5; // Array a let a = [ 1, 2, 3, 4, 5 ]; // Array b let b = [ 2, 5, 6, 1, 9 ]; document.write(numberOfPairs(a, b, n)); </script> |
Output:
1
Time Complexity: O( N * log(N) ), where N is the number of elements in array.
Auxiliary Space: O(N)
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