# Number of indices pair such that element pair sum from first Array is greater than second Array

• Difficulty Level : Hard
• Last Updated : 24 Dec, 2021

Given two integer arrays A[] and B[] of equal sizes, the task is to find the number of pairs of indices {i, j} in the arrays such that A[i] + A[j] > B[i] + B[j] and i < j.
Examples:

Input: A[] = {4, 8, 2, 6, 2}, B[] = {4, 5, 4, 1, 3}
Output:
Explanation:
There are a total of 7 pairs of indices {i, j} in the array following the condition. They are:
{0, 1}: A + A > B + B
{0, 3}: A + A > B + B
{1, 2}: A + A > B + B
{1, 3}: A + A > B + B
{1, 4}: A + A > B + B
{2, 3}: A + A > B + B
{3, 4}: A + A > B + B
Input: A[] = {1, 3, 2, 4}, B[] = {1, 3, 2, 4}
Output:
Explanation:
No such possible pairs of {i, j} can be found that satisfies the given condition

Naive Approach: The naive approach is to consider all the possible pairs of {i, j} in the given arrays and check if A[i] + A[j] > B[i] + B[j]. This can be done by using the concept of nested loops.
Time Complexity: O(N2)
Efficient Approach: The key observation from the problem is that the given condition can also be visualised as (ai-bi) + (aj-bj)> 0 so we can make another array to store the difference of both arrays. let this array be D . Therefore, the problem reduces to finding pairs with Di+Dj>0. Now we can sort D array and for each corresponding element Di we will find the no of good pairs that Di can make and add this no of pairs to a count variable.For each element Di to find the no of good pairs it can make we can use the upper_bound function of the standard template library to find the upper bound of -Di. since the array is sorted so all elements present after -Di will also make good pair with Di .thus,if upper bound of -Di is x and n be the total size of array then total pairs corresponding to Di will be n-x. This approach takes O(NlogN) time.

• The given condition in the question can be rewritten as:

```A[i] + A[j] > B[i] + B[j]
A[i] + A[j] - B[i] - B[j] > 0
(A[i] - B[i]) + (A[j] - B[j]) > 0```
• Create another array, say D, to store the difference between elements at the corresponding index in both array, i.e.

`D[i] = A[i] - B[i]`
• Now to make sure that the constraint i < j is satisfied, sort the difference array D, so that each element i is smaller than elements to its right.
• If at some index i, the value in the difference array D is negative, then we only need to find the nearest position ‘j’ at which the value is just greater than -D[i], so that on summation the value becomes > 0.
Inorder to find such index ‘j’, upper_bound() function or Binary Search can be used, since the array is sorted.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the number of indices pair``// such that pair sum from first Array``// is greater than second Array` `#include ``using` `namespace` `std;` `// Function to get the number of pairs of indices``// {i, j} in the given two arrays A and B such that``// A[i] + A[j] > B[i] + B[j]``int` `getPairs(vector<``int``> A, vector<``int``> B, ``int` `n)``{``    ``// Initializing the difference array D``    ``vector<``int``> D(n);` `    ``// Computing the difference between the``    ``// elements at every index and storing``    ``// it in the array D``    ``for` `(``int` `i = 0; i < n; i++) {``        ``D[i] = A[i] - B[i];``    ``}` `    ``// Sort the array D``    ``sort(D.begin(), D.end());` `    ``// Variable to store the total``    ``// number of pairs that satisfy``    ``// the given condition``    ``long` `long` `total = 0;` `    ``// Loop to iterate through the difference``    ``// array D and find the total number``    ``// of pairs of indices that follow the``    ``// given condition``    ``for` `(``int` `i = n - 1; i >= 0; i--) {` `        ``// If the value at the index i is positive,``        ``// then it remains positive for any pairs``        ``// with j such that j > i.``        ``if` `(D[i] > 0) {``            ``total += n - i - 1;``        ``}` `        ``// If the value at that index is negative``        ``// then we need to find the index of the``        ``// value just greater than -D[i]``        ``else` `{``            ``int` `k = upper_bound(D.begin(),``                                ``D.end(), -D[i])``                    ``- D.begin();``            ``total += n - k;``        ``}``    ``}``    ``return` `total;``}` `// Driver code``int` `main()``{``    ``int` `n = 5;``    ``vector<``int``> A;``    ``vector<``int``> B;` `    ``A.push_back(4);``    ``A.push_back(8);``    ``A.push_back(2);``    ``A.push_back(6);``    ``A.push_back(2);` `    ``B.push_back(4);``    ``B.push_back(5);``    ``B.push_back(4);``    ``B.push_back(1);``    ``B.push_back(3);` `    ``cout << getPairs(A, B, n);``}`

## Java

 `// Java program to find the number of indices pair``// such that pair sum from first Array``// is greater than second Array``import` `java.util.*;` `class` `GFG{` `// Function to get the number of pairs of indices``// {i, j} in the given two arrays A and B such that``// A[i] + A[j] > B[i] + B[j]``static` `long` `getPairs(Vector A, Vector B, ``int` `n)``{``    ``// Initializing the difference array D``    ``int` `[]D = ``new` `int``[n];` `    ``// Computing the difference between the``    ``// elements at every index and storing``    ``// it in the array D``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``D[i] = A.get(i) - B.get(i);``    ``}` `    ``// Sort the array D``    ``Arrays.sort(D);` `    ``// Variable to store the total``    ``// number of pairs that satisfy``    ``// the given condition``    ``long` `total = ``0``;` `    ``// Loop to iterate through the difference``    ``// array D and find the total number``    ``// of pairs of indices that follow the``    ``// given condition``    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) {` `        ``// If the value at the index i is positive,``        ``// then it remains positive for any pairs``        ``// with j such that j > i.``        ``if` `(D[i] > ``0``) {``            ``total += n - i - ``1``;``        ``}` `        ``// If the value at that index is negative``        ``// then we need to find the index of the``        ``// value just greater than -D[i]``        ``else` `{``            ``int` `k = upper_bound(D,``0``, D.length, -D[i]);``            ``total += n - k;``        ``}``    ``}``    ``return` `total;``}``static` `int` `upper_bound(``int``[] a, ``int` `low,``                        ``int` `high, ``int` `element)``{``    ``while``(low < high){``        ``int` `middle = low + (high - low)/``2``;``        ``if``(a[middle] > element)``            ``high = middle;``        ``else``            ``low = middle + ``1``;``    ``}``    ``return` `low;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``5``;``    ``Vector A = ``new` `Vector();``    ``Vector B= ``new` `Vector();` `    ``A.add(``4``);``    ``A.add(``8``);``    ``A.add(``2``);``    ``A.add(``6``);``    ``A.add(``2``);` `    ``B.add(``4``);``    ``B.add(``5``);``    ``B.add(``4``);``    ``B.add(``1``);``    ``B.add(``3``);` `    ``System.out.print(getPairs(A, B, n));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python 3 program to find the number of indices pair``# such that pair sum from first Array``# is greater than second Array``import` `bisect` `# Function to get the number of pairs of indices``# {i, j} in the given two arrays A and B such that``# A[i] + A[j] > B[i] + B[j]``def` `getPairs(A,  B, n):` `    ``# Initializing the difference array D``    ``D ``=` `[``0``]``*``(n)`` ` `    ``# Computing the difference between the``    ``# elements at every index and storing``    ``# it in the array D``    ``for` `i ``in` `range``(n):``        ``D[i] ``=` `A[i] ``-` `B[i]`` ` `    ``# Sort the array D``    ``D.sort()`` ` `    ``# Variable to store the total``    ``# number of pairs that satisfy``    ``# the given condition``    ``total ``=` `0`` ` `    ``# Loop to iterate through the difference``    ``# array D and find the total number``    ``# of pairs of indices that follow the``    ``# given condition``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):`` ` `        ``# If the value at the index i is positive,``        ``# then it remains positive for any pairs``        ``# with j such that j > i.``        ``if` `(D[i] > ``0``):``            ``total ``+``=` `n ``-` `i ``-` `1`` ` `        ``# If the value at that index is negative``        ``# then we need to find the index of the``        ``# value just greater than -D[i]``        ``else``:``            ``k ``=` `bisect.bisect_right(D, ``-``D[i], ``0``, ``len``(D))``            ``total ``+``=` `n ``-` `k``    ``return` `total`` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``n ``=` `5``    ``A ``=` `[]``    ``B ``=` `[]`` ` `    ``A.append(``4``);``    ``A.append(``8``);``    ``A.append(``2``);``    ``A.append(``6``);``    ``A.append(``2``);`` ` `    ``B.append(``4``);``    ``B.append(``5``);``    ``B.append(``4``);``    ``B.append(``1``);``    ``B.append(``3``);`` ` `    ``print``(getPairs(A, B, n))` `# This code is contributed by chitranayal`

## C#

 `// C# program to find the number of indices pair``// such that pair sum from first Array``// is greater than second Array``using` `System;``using` `System.Collections.Generic;` `class` `GFG{`` ` `// Function to get the number of pairs of indices``// {i, j} in the given two arrays A and B such that``// A[i] + A[j] > B[i] + B[j]``static` `long` `getPairs(List<``int``> A, List<``int``> B, ``int` `n)``{``    ``// Initializing the difference array D``    ``int` `[]D = ``new` `int``[n];`` ` `    ``// Computing the difference between the``    ``// elements at every index and storing``    ``// it in the array D``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``D[i] = A[i] - B[i];``    ``}`` ` `    ``// Sort the array D``    ``Array.Sort(D);`` ` `    ``// Variable to store the total``    ``// number of pairs that satisfy``    ``// the given condition``    ``long` `total = 0;`` ` `    ``// Loop to iterate through the difference``    ``// array D and find the total number``    ``// of pairs of indices that follow the``    ``// given condition``    ``for` `(``int` `i = n - 1; i >= 0; i--) {`` ` `        ``// If the value at the index i is positive,``        ``// then it remains positive for any pairs``        ``// with j such that j > i.``        ``if` `(D[i] > 0) {``            ``total += n - i - 1;``        ``}`` ` `        ``// If the value at that index is negative``        ``// then we need to find the index of the``        ``// value just greater than -D[i]``        ``else` `{``            ``int` `k = upper_bound(D,0, D.Length, -D[i]);``            ``total += n - k;``        ``}``    ``}``    ``return` `total;``}``static` `int` `upper_bound(``int``[] a, ``int` `low,``                        ``int` `high, ``int` `element)``{``    ``while``(low < high){``        ``int` `middle = low + (high - low)/2;``        ``if``(a[middle] > element)``            ``high = middle;``        ``else``            ``low = middle + 1;``    ``}``    ``return` `low;``}`` ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 5;``    ``List<``int``> A = ``new` `List<``int``>();``    ``List<``int``> B= ``new` `List<``int``>();`` ` `    ``A.Add(4);``    ``A.Add(8);``    ``A.Add(2);``    ``A.Add(6);``    ``A.Add(2);`` ` `    ``B.Add(4);``    ``B.Add(5);``    ``B.Add(4);``    ``B.Add(1);``    ``B.Add(3);`` ` `    ``Console.Write(getPairs(A, B, n));``}``}` `// This code is contributed by sapnasingh4991`

## Javascript

 ``
Output:
`7`

Time Complexity Analysis:

• The sorting of the array takes O(N * log(N)) time.
• The time taken to find the index which is just greater than a specific value is O(Log(N)). Since in the worst case, this can be executed for N elements in the array, the overall time complexity for this is O(N * log(N)).
• Therefore, the overall time complexity is O(N * log(N)).

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