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

• Difficulty Level : Hard
• Last Updated : 26 Oct, 2021

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:
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:

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 ``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;``}`

## Java

 `// 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`

## 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 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`

## C#

 `// 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`

## Javascript

 ``

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:
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.

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 ``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;``}`

## Java

 `// Java program of the above approach``import` `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 approach``from` `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 code``if` `__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 approach``using` `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 Code``static` `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

 ``

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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