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# Count pairs in a sorted array whose product is less than k

Given a sorted integer array and number k, the task is to count pairs in an array whose product is less than x.

Examples:

Input: A = {2, 3, 5, 6}, k = 16
Output:
Pairs having product less than 16: (2, 3), (2, 5), (2, 6), (3, 5)

Input: A = {2, 3, 4, 6, 9}, k = 20
Output:
Pairs having product less than 20: (2, 3), (2, 4), (2, 6), (2, 9), (3, 4), (3, 6)

A simple solution of this problem run two loops to generate all pairs and one by one and check if current pairâ€™s product is less than x or not.

An Efficient solution of this problem is take initial and last value of index in l and r variable. Consider below two cases:

• Case-I:
• Lets consider i < j and A[i]*A[j] < k then we can say that A[i]*A[j-1] < k as A[j-1] < A[j] for a sorted array,
• Similarly A[i]*A[j-2] < k, A[i]*A[j-3] < k, ….., A[i]*A[i+1] < k.
• Case-II:
• Lets consider i k then we can say that A[i]*A[j+1] > k as A[j+1] > A[j] for a sorted array,
• similarly A[i]*A[j+2] > k, A[i]*A[j+3] > k, ….., A[i]*A[n-1] > k.

Above problem is similar to Count pairs in a sorted array whose sum is less than x, the only thing that is different is to find the product of pairs instead of sum.

Below is the algorithm to solve this problem:

```1) Initialize two variables l and r to find the candidate
elements in the sorted array.
(a) l = 0
(b) r = n - 1
2) Initialize : result = 0
2) Loop while l < r.

// If current left and current
// right have product smaller than x,
// the all elements from l+1 to r
// form a pair with current
(a) If (arr[l] * arr[r] < x)
result = result + (r - l)
l++;

(b) Else
r--;

3) Return result```

Below is the implementation of the above algorithm:

## C++

 `// C++ program to find number of pairs with` `// product less than k in a sorted array` `#include ` `using` `namespace` `std;`   `// Function to count the pairs` `int` `fun(``int` `A[], ``int` `n, ``int` `k)` `{` `    ``// count to keep count of` `    ``// number of pairs with product` `    ``// less than k` `    ``int` `count = 0;` `    ``int` `i = 0;` `    ``int` `j = n - 1;`   `    ``// Traverse the array` `    ``while` `(i < j) {`   `        ``// If product is less than k` `        ``// then count that pair` `        ``// and increment 'i'` `        ``if` `(A[i] * A[j] < k) {` `            ``count += (j - i);` `            ``i++;` `        ``}`   `        ``// Else decrement 'j'` `        ``else` `{` `            ``j--;` `        ``}` `    ``}`   `    ``// Return count of pairs` `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `A[] = { 2, 3, 4, 6, 9 };` `    ``int` `n = ``sizeof``(A) / ``sizeof``(``int``);` `    ``int` `k = 20;` `    ``cout << ``"Number of pairs with product less than "` `         ``<< k << ``" = "` `<< fun(A, n, k) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java program to find number ` `// of pairs with product less ` `// than k in a sorted array` `class` `GFG` `{`   `// Function to count the pairs` `static` `int` `fun(``int` `A[], ` `               ``int` `n, ``int` `k)` `{` `    ``// count to keep count of` `    ``// number of pairs with ` `    ``// product less than k` `    ``int` `count = ``0``;` `    ``int` `i = ``0``;` `    ``int` `j = n - ``1``;`   `    ``// Traverse the array` `    ``while` `(i < j) ` `    ``{`   `        ``// If product is less than ` `        ``// k then count that pair` `        ``// and increment 'i'` `        ``if` `(A[i] * A[j] < k) ` `        ``{` `            ``count += (j - i);` `            ``i++;` `        ``}`   `        ``// Else decrement 'j'` `        ``else` `        ``{` `            ``j--;` `        ``}` `    ``}`   `    ``// Return count of pairs` `    ``return` `count;` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `A[] = {``2``, ``3``, ``4``, ``6``, ``9``};` `    ``int` `n = A.length;` `    ``int` `k = ``20``;` `    `  `    ``System.out.println(``"Number of pairs with "` `+` `                     ``"product less than 20 = "` `+ ` `                                  ``fun(A, n, k));` `}` `}`   `// This code is contributed ` `// by Kirti_Mangal`

## Python

 `# Python program to find number of pairs with` `# product less than k in a sorted array `   `def` `fun(A, k):` `    ``# count to keep count of number ` `    ``# of pairs with product less than k` `    ``count ``=` `0` `    ``n ``=` `len``(A)` `    ``# Left pointer pointing to leftmost part` `    ``i ``=` `0` `    `  `    ``# Right pointer pointing to rightmost part` `    ``j ``=` `n``-``1` `    `  `    ``# While left and right pointer don't meet` `    ``while` `i < j:` `        ``if` `A[i]``*``A[j] < k:` `            ``count ``+``=` `(j``-``i)` `            ``# Increment the left pointer` `            ``i``+``=` `1` `        ``else``:` `            ``# Decrement the right pointer` `            ``j``-``=` `1` `    ``return` `count`   `# Driver code to test above function` `A ``=` `[``2``, ``3``, ``4``, ``6``, ``9``]` `k ``=` `20` `print``(``"Number of pairs with product less than "``,` `k, ``" = "``, fun(A, k))`

## C#

 `// C# program to find number ` `// of pairs with product less ` `// than k in a sorted array` `using` `System;`   `class` `GFG` `{`   `// Function to count the pairs` `static` `int` `fun(``int` `[]A, ` `               ``int` `n, ``int` `k)` `{` `    ``// count to keep count of` `    ``// number of pairs with ` `    ``// product less than k` `    ``int` `count = 0;` `    ``int` `i = 0;` `    ``int` `j = n - 1;`   `    ``// Traverse the array` `    ``while` `(i < j) ` `    ``{`   `        ``// If product is less than ` `        ``// k then count that pair` `        ``// and increment 'i'` `        ``if` `(A[i] * A[j] < k) ` `        ``{` `            ``count += (j - i);` `            ``i++;` `        ``}`   `        ``// Else decrement 'j'` `        ``else` `        ``{` `            ``j--;` `        ``}` `    ``}`   `    ``// Return count of pairs` `    ``return` `count;` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `[]A = {2, 3, 4, 6, 9};` `    ``int` `n = A.Length;` `    ``int` `k = 20;` `    `  `    ``Console.WriteLine(``"Number of pairs with "` `+` `                    ``"product less than 20 = "` `+ ` `                                 ``fun(A, n, k));` `}` `}`   `// This code is contributed ` `// by Subhadeep`

## PHP

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## Javascript

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Output

`Number of pairs with product less than 20 = 6`

Complexity Analysis:

• Time Complexity: O(N), where N is the size of the given array.
• Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach 2: Binary Search:

This problem can be solved using binary search. We can fix one element of the pair and then use binary search to find the maximum index of the second element such that their product is less than k. We can repeat this process for all elements of the array and sum up the counts to get the total number of pairs with product less than k.

In this code, we use a nested loop to traverse the array and fix one element of the pair. We then use binary search to find the maximum index of the second element such that their product is less than k. We count the number of pairs for each fixed element and sum up the counts to get the total number of pairs with product less than k.

Here is the code to solve this problem using binary search in C++:

## C++

 `#include ` `using` `namespace` `std;`   `// Function to count the pairs` `int` `fun(``int` `A[], ``int` `n, ``int` `k)` `{` `    ``int` `count = 0;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `lo = i + 1;` `        ``int` `hi = n - 1;` `        ``while` `(lo <= hi) {` `            ``int` `mid = lo + (hi - lo) / 2;` `            ``if` `(A[i] * A[mid] < k) {` `                ``// All elements from lo to mid` `                ``// will satisfy the condition` `                ``count += (mid - lo + 1);` `                ``lo = mid + 1;` `            ``}` `            ``else` `{` `                ``hi = mid - 1;` `            ``}` `        ``}` `    ``}`   `    ``// Return count of pairs` `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `A[] = { 2, 3, 4, 6, 9 };` `    ``int` `n = ``sizeof``(A) / ``sizeof``(``int``);` `    ``int` `k = 20;` `    ``cout << ``"Number of pairs with product less than "` `<< k` `         ``<< ``" = "` `<< fun(A, n, k) << endl;`   `    ``return` `0;` `}`

## Java

 `// GFG` `// Java code for this approach` `import` `java.util.*;`   `public` `class` `Main {`   `    ``// Function to count the pairs` `    ``public` `static` `int` `fun(``int``[] A, ``int` `n, ``int` `k)` `    ``{` `        ``int` `count = ``0``;`   `        ``// Traverse the array` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``int` `lo = i + ``1``;` `            ``int` `hi = n - ``1``;` `            ``while` `(lo <= hi) {` `                ``int` `mid = lo + (hi - lo) / ``2``;` `                ``if` `(A[i] * A[mid] < k) {` `                    ``// All elements from lo to mid` `                    ``// will satisfy the condition` `                    ``count += (mid - lo + ``1``);` `                    ``lo = mid + ``1``;` `                ``}` `                ``else` `{` `                    ``hi = mid - ``1``;` `                ``}` `            ``}` `        ``}`   `        ``// Return count of pairs` `        ``return` `count;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] A = { ``2``, ``3``, ``4``, ``6``, ``9` `};` `        ``int` `n = A.length;` `        ``int` `k = ``20``;` `        ``System.out.println(` `            ``"Number of pairs with product less than "` `+ k` `            ``+ ``" = "` `+ fun(A, n, k));` `    ``}` `}`   `// This code is written by sundaram`

## Python3

 `# Function to count the pairs` `def` `fun(A, n, k):` `    ``count ``=` `0`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(n):` `        ``lo ``=` `i ``+` `1` `        ``hi ``=` `n ``-` `1` `        ``while` `lo <``=` `hi:` `            ``mid ``=` `lo ``+` `(hi ``-` `lo) ``/``/` `2` `            ``if` `A[i] ``*` `A[mid] < k:` `                ``# All elements from lo to mid will satisfy the condition` `                ``count ``+``=` `(mid ``-` `lo ``+` `1``)` `                ``lo ``=` `mid ``+` `1` `            ``else``:` `                ``hi ``=` `mid ``-` `1`   `    ``# Return count of pairs` `    ``return` `count`     `# Driver code` `A ``=` `[``2``, ``3``, ``4``, ``6``, ``9``]` `n ``=` `len``(A)` `k ``=` `20` `print``(``"Number of pairs with product less than"``, k, ``"="``, fun(A, n, k))`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `MainClass {` `    ``// Function to count the pairs` `    ``public` `static` `int` `Fun(``int``[] A, ``int` `n, ``int` `k)` `    ``{` `        ``int` `count = 0;`   `        ``// Traverse the array` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``int` `lo = i + 1;` `            ``int` `hi = n - 1;` `            ``while` `(lo <= hi) {` `                ``int` `mid = lo + (hi - lo) / 2;` `                ``if` `(A[i] * A[mid] < k) {` `                    ``// All elements from lo to mid` `                    ``// will satisfy the condition` `                    ``count += (mid - lo + 1);` `                    ``lo = mid + 1;` `                ``}` `                ``else` `{` `                    ``hi = mid - 1;` `                ``}` `            ``}` `        ``}`   `        ``// Return count of pairs` `        ``return` `count;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] A = { 2, 3, 4, 6, 9 };` `        ``int` `n = A.Length;` `        ``int` `k = 20;` `        ``Console.WriteLine(` `            ``"Number of pairs with product less than "` `+ k` `            ``+ ``" = "` `+ Fun(A, n, k));` `    ``}` `}`

## Javascript

 `// Function to count the pairs` `function` `fun(A, n, k) {` `    ``let count = 0;`   `    ``// Traverse the array` `    ``for` `(let i = 0; i < n; i++) {` `        ``let lo = i + 1;` `        ``let hi = n - 1;` `        ``while` `(lo <= hi) {` `            ``let mid = lo + Math.floor((hi - lo) / 2);` `            ``if` `(A[i] * A[mid] < k) {` `                ``// All elements from lo to mid` `                ``// will satisfy the condition` `                ``count += (mid - lo + 1);` `                ``lo = mid + 1;` `            ``}` `            ``else` `{` `                ``hi = mid - 1;` `            ``}` `        ``}` `    ``}`   `    ``// Return count of pairs` `    ``return` `count;` `}`   `// Driver code` `let A = [2, 3, 4, 6, 9];` `let n = A.length;` `let k = 20;` `console.log(`Number of pairs ``with` `product less than \${k} = \${fun(A, n, k)}`);`

Output

`Number of pairs with product less than 20 = 6`

Complexity Analysis:

Time Complexity: O(NLogN), where N is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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