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Number of pairs in an array with the sum greater than 0
  • Difficulty Level : Medium
  • Last Updated : 01 Feb, 2021

Given an array arr[] of size N, the task is to find the number of distinct pairs in the array whose sum is > 0.

Examples: 

Input: arr[] = { 3, -2, 1 } 
Output:
Explanation: 
There are two pairs of elements in the array whose sum is positive. They are: 
{3, -2} = 1 
{3, 1} = 4

Input: arr[] = { -1, -1, -1, 0 } 
Output:
Explanation: 
There are no pairs of elements in the array whose sum is positive. 
 

Naive Approach: The naive approach for this problem is to consider all the unique pairs of elements in the array. For every pair, check if the sum is positive or not. 
Time Complexity: O(N2)



Efficient Approach: 

  • The idea is to use the concept of sorting and two pointer technique.
  • For this problem, sorting is used because for the sum arr[i] + arr[j] > 0 where i, j are some random indices in the array, either arr[i] > 0 or arr[j] > 0 or both arr[i] and arr[j] > 0.
  • Therefore, once the array is sorted, since we need to find the unique pairs. For every ‘i’ such that arr[i] > 0, we need to find the number of j’s such that arr[j] + arr[j] > 0.
  • Here, it is easy to find the count of pairs by using two pointer technique because the array is sorted. We just need to find the leftmost position of ‘j’ for which the condition holds true. This is found using the lower_bound of -arr[i] + 1.
  • For example, let the array arr[] = {-4, 4, -5, 5, 3, -2, -3, -1, 2, 1}. This array is sorted. Therefore, the array becomes, {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}. For some random i, lets assume arr[i] = 4. Therefore, the index of -3 is found in the array which is 2. Now, we can be sure that for all the values between the indices 2 and 8, the value of arr[i] + arr[j] > 0.

Below is the implementation of the above approach: 

CPP




// C++ program to find the
// number of pairs in the
// array with the sum > 0
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number
// of pairs in the array with
// sum > 0
int findNumOfPair(int* a, int n)
{
 
    // Sorting the given array
    sort(a, a + n);
 
    // Variable to store the count of pairs
    int ans = 0;
 
    // Loop to iterate through the array
    for (int i = 0; i < n; ++i) {
 
        // Ignore if the value is negative
        if (a[i] <= 0)
            continue;
 
        // Finding the index using lower_bound
        int j = lower_bound(a, a + n, -a[i] + 1) - a;
 
        // Finding the number of pairs between
        // two indices i and j
        ans += i - j;
    }
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 3, -2, 1 };
    int n = sizeof(a) / sizeof(a[0]);
 
    int ans = findNumOfPair(a, n);
    cout << ans << endl;
 
    return 0;
}

Java




// Java program to find the
// number of pairs in the
// array with the sum > 0
import java.util.*;
 
class GFG {
 
    // Function to find the number
    // of pairs in the array with
    // sum > 0
    static int findNumOfPair(int arr[], int n)
    {
 
        // Sorting the given array
        Arrays.sort(arr);
 
        // Variable to store the count of pairs
        int ans = 0;
 
        // Loop to iterate through the array
        for (int i = 0; i < n; ++i) {
 
            // Ignore if the value is negative
            if (arr[i] <= 0)
                continue;
           
            /*
            minReqVal val is the min value ,which will
            give >=1 after adding with the arr[i]
            */
            int minReqVal = -arr[i] + 1;
            int j = lower_bound(arr, minReqVal);
 
            if (j >= 0)
                ans += i - j;
        }
        return ans;
    }
 
    /*
         it return the index of a minimum Number in the
         array which is just >= val
      */
    static int lower_bound(int arr[], int val)
    {
        int start = 0, end = arr.length;
 
        /*
          using the Binary search technique , since our
          array is sorted
        */
        while (start < end) {
            int mid = (start + end) >> 1;
 
            if (val > arr[mid])
                start = mid + 1;
            else
                end = mid;
        }
 
        // when we dont find the answer return -1
        if (start == arr.length)
            return -1;
 
        return start;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = {-2,-1,-1,-1,-1,0 ,1,2,3};
        int n = a.length;
 
        int ans = findNumOfPair(a, n);
        System.out.println(ans);
    }
}
 
// This code is contributed by Pradeep Mondal P

Python3




# Python3 program to find the
# number of pairs in the
# array with the sum > 0
from bisect import bisect_left as lower_bound
 
# Function to find the number
# of pairs in the array with
# sum > 0
 
 
def findNumOfPair(a, n):
 
    # Sorting the given array
    a = sorted(a)
 
    # Variable to store the count of pairs
    ans = 0
 
    # Loop to iterate through the array
    for i in range(n):
 
        # Ignore if the value is negative
        if (a[i] <= 0):
            continue
 
        # Finding the index using lower_bound
        j = lower_bound(a, -a[i] + 1)
 
        # Finding the number of pairs between
        # two indices i and j
        ans += i - j
    return ans
 
 
# Driver code
if __name__ == '__main__':
    a = [3, -2, 1]
    n = len(a)
 
    ans = findNumOfPair(a, n)
    print(ans)
 
# This code is contributed by mohit kumar 29

C#




// C# program to find the
// number of pairs in the
// array with the sum > 0
using System;
 
class GFG {
 
    // Function to find the number
    // of pairs in the array with
    // sum > 0
    static int findNumOfPair(int[] arr, int n)
    {
 
        // Sorting the given array
        Array.Sort(arr);
 
        // Variable to store the count of pairs
        int ans = 0;
 
        // Loop to iterate through the array
        for (int i = 0; i < n; ++i) {
 
            // Ignore if the value is negative
            if (arr[i] <= 0)
                continue;
 
            /*
            minReqVal val is the min value ,which will
            give >=1 after adding with the arr[i]
            */
            int minReqVal = -arr[i] + 1;
            int j = lower_bound(arr, minReqVal);
 
            if (j >= 0)
                ans += i - j;
        }
        return ans;
    }
 
    /*
           it return the index of a minimum Number in the
           array which is just >= val
        */
    static int lower_bound(int[] arr, int val)
    {
        int start = 0, end = arr.Length;
 
        /*
          using the Binary search technique , since our
          array is sorted
        */
        while (start < end) {
            int mid = (start + end) >> 1;
 
            if (val > arr[mid])
                start = mid + 1;
            else
                end = mid;
        }
 
        // when we dont find the answer return -1
        if (start == arr.Length)
            return -1;
 
        return start;
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { -2, 1, 3 };
        int n = a.Length;
 
        int ans = findNumOfPair(a, n);
        Console.Write(ans);
    }
}
 
// This code is contributed by Pradeep Mondal P
Output
2

Time Complexity: O(N * log(N))
 

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