Count of distinct index pair (i, j) such that element sum of First Array is greater

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++

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// CPP 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 Code
int 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;
}

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Java

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// Java program for the above problem
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 Code
public 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

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Python3

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# 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 arrays
N = 5;
  
# Initialise the arrays
a = [ 1, 2, 3, 4, 5 ];
b = [ 2, 5, 6, 1, 9 ];
  
# function call that returns
# the count of possible pairs
print(count_pairs(a, b, N);
  
# This code is contributed by Code_Mech

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C#

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// C# program for the above problem
using 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 Code
public 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

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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 tha 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:
    1. If c[i] <= 0, simply continue.
    2. 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.
    3. 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 postive value in c[] and add the value of i – pos to the count.

Below is the implementation of the above approach:

C++

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// 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 code
int32_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;
}

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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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