# Count pairs (i, j) from given array such that i < j and arr[i] > K * arr[j]

• Difficulty Level : Medium
• Last Updated : 13 Jun, 2022

Given an array arr[] of length N and an integer K, the task is to count the number of pairs (i, j) such that i < j and arr[i] > K * arr[j].

Examples:

Input: arr[] = {5, 6, 2, 5}, K = 2
Output: 2
Explanation: The array consists of two such pairs:
(5, 2): Index of 5 and 2 are 0, 2 respectively. Therefore, the required conditions (0 < 2 and 5 > 2 * 2) are satisfied.
(6, 2): Index of 6 and 2 are 1, 2 respectively. Therefore, the required conditions (0 < 2 and 6 > 2 * 2) are satisfied.

Input: arr[] = {4, 6, 5, 1}, K = 2
Output: 3

Naive Approach: The simplest approach to solve the problem is to traverse the array and for every index, find numbers having indices greater than it, such that the element in it when multiplied by K is less than the element at the current index.

Follow the below steps to solve the problem:

1. Initialize a variable, say cnt, with 0 to count the total number of required pairs.
2. Traverse the array from left to right.
3. For each possible index, say i, traverse the indices i + 1 to N – 1 and increase the value of cnt by 1 if any element, say arr[j], is found such that arr[j] * K is less than arr[i].
4. After complete traversal of the array, print cnt as the required count of pairs.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the count required pairs``void` `getPairs(``int` `arr[], ``int` `N, ``int` `K)``{``    ``// Stores count of pairs``    ``int` `count = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``for` `(``int` `j = i + 1; j < N; j++) {` `            ``// Check if the condition``            ``// is satisfied or not``            ``if` `(arr[i] > K * arr[j])``                ``count++;``        ``}``    ``}``    ``cout << count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 5, 6, 2, 5 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 2;` `    ``// Function Call``    ``getPairs(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG``{` `// Function to find the count required pairs``static` `void` `getPairs(``int` `arr[], ``int` `N, ``int` `K)``{``    ``// Stores count of pairs``    ``int` `count = ``0``;` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``for` `(``int` `j = i + ``1``; j < N; j++)``        ``{` `            ``// Check if the condition``            ``// is satisfied or not``            ``if` `(arr[i] > K * arr[j])``                ``count++;``        ``}``    ``}``    ``System.out.print(count);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``5``, ``6``, ``2``, ``5` `};``    ``int` `N = arr.length;``    ``int` `K = ``2``;` `    ``// Function Call``    ``getPairs(arr, N, K);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program for the above approach` `# Function to find the count required pairs``def` `getPairs(arr, N, K):``    ` `    ``# Stores count of pairs``    ``count ``=` `0` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(i ``+` `1``, N):``            ` `            ``# Check if the condition``            ``# is satisfied or not``            ``if` `(arr[i] > K ``*` `arr[j]):``                ``count ``+``=` `1``                ` `    ``print``(count)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``5``, ``6``, ``2``, ``5` `]``    ``N   ``=` `len``(arr)``    ``K ``=` `2` `    ``# Function Call``    ``getPairs(arr, N, K)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG``{` `// Function to find the count required pairs``static` `void` `getPairs(``int` `[]arr, ``int` `N, ``int` `K)``{``    ``// Stores count of pairs``    ``int` `count = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``for` `(``int` `j = i + 1; j < N; j++)``        ``{` `            ``// Check if the condition``            ``// is satisfied or not``            ``if` `(arr[i] > K * arr[j])``                ``count++;``        ``}``    ``}``    ``Console.Write(count);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 5, 6, 2, 5 };``    ``int` `N = arr.Length;``    ``int` `K = 2;` `    ``// Function Call``    ``getPairs(arr, N, K);``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output

`2`

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The idea is to use the concept of merge sort and then count pairs according to the given conditions. Follow the steps below to solve the problem:

• Initialize a variable, say answer, to count the number of pairs satisfying the given condition.
• Repeatedly partition the array into two equal halves or almost equal halves until one element is left in each partition.
• Call a recursive function that counts the number of times the condition arr[i] > K * arr[j] and i < j is satisfied after merging the two partitions.
• Perform it by initializing two variables, say i and j, for the indices of the first and second half respectively.
• Increment j till arr[i] > K * arr[j] and j < size of the second half. Add (j – (mid + 1)) to the answer and increment i.
• After completing the above steps, print the value of answer as the required number of pairs.

Below is the implementation of the above approach:

## C++14

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to merge two sorted arrays``int` `merge(``int` `arr[], ``int` `temp[],``          ``int` `l, ``int` `m, ``int` `r, ``int` `K)``{``    ``// i: index to left subarray``    ``int` `i = l;` `    ``// j: index to right subarray``    ``int` `j = m + 1;` `    ``// Stores count of pairs that``    ``// satisfy the given condition``    ``int` `cnt = 0;` `    ``for` `(``int` `i = l; i <= m; i++) {``        ``bool` `found = ``false``;` `        ``// Traverse to check for the``        ``// valid conditions``        ``while` `(j <= r) {` `            ``// If condition satisfies``            ``if` `(arr[i] >= K * arr[j]) {``                ``found = ``true``;``                  ``j++;``            ``}``            ``else``                ``break``;``        ``}` `        ``// While a[i] > K*a[j] satisfies``        ``// increase j` `        ``// All elements in the right``        ``// side of the left subarray``        ``// also satisfies``        ``if` `(found) {``            ``cnt += j - (m + 1);``            ``j--;``        ``}``    ``}` `    ``// Sort the two given arrays and``    ``// store in the resultant array``    ``int` `k = l;``    ``i = l;``    ``j = m + 1;` `    ``while` `(i <= m && j <= r) {` `        ``if` `(arr[i] <= arr[j])``            ``temp[k++] = arr[i++];``        ``else``            ``temp[k++] = arr[j++];``    ``}` `    ``// Elements which are left``    ``// in the left subarray``    ``while` `(i <= m)``        ``temp[k++] = arr[i++];` `    ``// Elements which are left``    ``// in the right subarray``    ``while` `(j <= r)``        ``temp[k++] = arr[j++];` `    ``for` `(``int` `i = l; i <= r; i++)``        ``arr[i] = temp[i];` `    ``// Return the count obtained``    ``return` `cnt;``}` `// Function to partition array into two halves``int` `mergeSortUtil(``int` `arr[], ``int` `temp[],``                  ``int` `l, ``int` `r, ``int` `K)``{``    ``int` `cnt = 0;``    ``if` `(l < r) {` `        ``// Same as (l + r) / 2, but avoids``        ``// overflow for large l and h``        ``int` `m = (l + r) / 2;` `        ``// Sort first and second halves``        ``cnt += mergeSortUtil(arr, temp,``                             ``l, m, K);``        ``cnt += mergeSortUtil(arr, temp,``                             ``m + 1, r, K);` `        ``// Call the merging function``        ``cnt += merge(arr, temp, l,``                     ``m, r, K);``    ``}` `    ``return` `cnt;``}` `// Function to print the count of``// required pairs using Merge Sort``void` `mergeSort(``int` `arr[], ``int` `N, ``int` `K)``{``    ``int` `temp[N];` `    ``cout << mergeSortUtil(arr, temp, 0,``                          ``N - 1, K);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 6, 2, 5 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 2;` `    ``// Function Call``    ``mergeSort(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG``{``    ` `    ``// Function to merge two sorted arrays``    ``static` `int` `merge(``int` `arr[], ``int` `temp[],``              ``int` `l, ``int` `m, ``int` `r, ``int` `K)``    ``{``      ` `        ``// i: index to left subarray``        ``int` `i = l;``    ` `        ``// j: index to right subarray``        ``int` `j = m + ``1``;``    ` `        ``// Stores count of pairs that``        ``// satisfy the given condition``        ``int` `cnt = ``0``;``    ` `        ``for` `(i = l; i <= m; i++)``        ``{``            ``boolean` `found = ``false``;``    ` `            ``// Traverse to check for the``            ``// valid conditions``            ``while` `(j <= r)``            ``{``    ` `                ``// If condition satisfies``                ``if` `(arr[i] >= K * arr[j])``                ``{``                    ``found = ``true``;``                ``}``                ``else``                    ``break``;``                ``j++;``            ``}``    ` `            ``// While a[i] > K*a[j] satisfies``            ``// increase j``    ` `            ``// All elements in the right``            ``// side of the left subarray``            ``// also satisfies``            ``if` `(found == ``true``)``            ``{``                ``cnt += j - (m + ``1``);``                ``j--;``            ``}``        ``}``    ` `        ``// Sort the two given arrays and``        ``// store in the resultant array``        ``int` `k = l;``        ``i = l;``        ``j = m + ``1``;``    ` `        ``while` `(i <= m && j <= r)``        ``{``    ` `            ``if` `(arr[i] <= arr[j])``                ``temp[k++] = arr[i++];``            ``else``                ``temp[k++] = arr[j++];``        ``}``    ` `        ``// Elements which are left``        ``// in the left subarray``        ``while` `(i <= m)``            ``temp[k++] = arr[i++];``    ` `        ``// Elements which are left``        ``// in the right subarray``        ``while` `(j <= r)``            ``temp[k++] = arr[j++];``    ` `        ``for` `(i = l; i <= r; i++)``            ``arr[i] = temp[i];``    ` `        ``// Return the count obtained``        ``return` `cnt;``    ``}``    ` `    ``// Function to partition array into two halves``    ``static` `int` `mergeSortUtil(``int` `arr[], ``int` `temp[],``                      ``int` `l, ``int` `r, ``int` `K)``    ``{``        ``int` `cnt = ``0``;``        ``if` `(l < r)``        ``{``    ` `            ``// Same as (l + r) / 2, but avoids``            ``// overflow for large l and h``            ``int` `m = (l + r) / ``2``;``    ` `            ``// Sort first and second halves``            ``cnt += mergeSortUtil(arr, temp,``                                 ``l, m, K);``            ``cnt += mergeSortUtil(arr, temp,``                                 ``m + ``1``, r, K);``    ` `            ``// Call the merging function``            ``cnt += merge(arr, temp, l,``                         ``m, r, K);``        ``}   ``        ``return` `cnt;``    ``}``    ` `    ``// Function to print the count of``    ``// required pairs using Merge Sort``    ``static` `void` `mergeSort(``int` `arr[], ``int` `N, ``int` `K)``    ``{``        ``int` `temp[] = ``new` `int``[N];``        ``System.out.print(mergeSortUtil(arr, temp, ``0``, N - ``1``, K));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``5``, ``6``, ``2``, ``5` `};``        ``int` `N = arr.length;``        ``int` `K = ``2``;``    ` `        ``// Function Call``        ``mergeSort(arr, N, K);   ``    ``}``}` `// This code is contributed by AnkThon`

## Python3

 `# Python3 program for the above approach` `# Function to merge two sorted arrays``def` `merge(arr, temp, l, m, r, K) :` `    ``# i: index to left subarray``    ``i ``=` `l`` ` `    ``# j: index to right subarray``    ``j ``=` `m ``+` `1`` ` `    ``# Stores count of pairs that``    ``# satisfy the given condition``    ``cnt ``=` `0``    ``for` `l ``in` `range``(m ``+` `1``) :``        ``found ``=` `False`` ` `        ``# Traverse to check for the``        ``# valid conditions``        ``while` `(j <``=` `r) :`` ` `            ``# If condition satisfies``            ``if` `(arr[i] >``=` `K ``*` `arr[j]) :``                ``found ``=` `True`        `            ``else` `:``                ``break``            ``j ``+``=` `1`` ` `        ``# While a[i] > K*a[j] satisfies``        ``# increase j`` ` `        ``# All elements in the right``        ``# side of the left subarray``        ``# also satisfies``        ``if` `(found) :``            ``cnt ``+``=` `j ``-` `(m ``+` `1``)``            ``j ``-``=` `1`` ` `    ``# Sort the two given arrays and``    ``# store in the resultant array``    ``k ``=` `l``    ``i ``=` `l``    ``j ``=` `m ``+` `1`` ` `    ``while` `(i <``=` `m ``and` `j <``=` `r) :``        ``if` `(arr[i] <``=` `arr[j]) :``            ``temp[k] ``=` `arr[i]``            ``k ``+``=` `1``            ``i ``+``=` `1``        ``else` `:``            ``temp[k] ``=` `arr[j]``            ``k ``+``=` `1``            ``j ``+``=` `1`` ` `    ``# Elements which are left``    ``# in the left subarray``    ``while` `(i <``=` `m) :``        ``temp[k] ``=` `arr[i]``        ``k ``+``=` `1``        ``i ``+``=` `1`` ` `    ``# Elements which are left``    ``# in the right subarray``    ``while` `(j <``=` `r) :``        ``temp[k] ``=` `arr[j]``        ``k ``+``=` `1``        ``j ``+``=` `1``    ``for` `i ``in` `range``(l, r ``+` `1``) :``        ``arr[i] ``=` `temp[i]`` ` `    ``# Return the count obtained``    ``return` `cnt`` ` `# Function to partition array into two halves``def` `mergeSortUtil(arr, temp, l, r, K) :``    ``cnt ``=` `0``    ``if` `(l < r) :`` ` `        ``# Same as (l + r) / 2, but avoids``        ``# overflow for large l and h``        ``m ``=` `(l ``+` `r) ``/``/` `2`` ` `        ``# Sort first and second halves``        ``cnt ``+``=` `mergeSortUtil(arr, temp, l, m, K)``        ``cnt ``+``=` `mergeSortUtil(arr, temp, m ``+` `1``, r, K)`` ` `        ``# Call the merging function``        ``cnt ``+``=` `merge(arr, temp, l, m, r, K)``    ``return` `cnt`` ` `# Function to print the count of``# required pairs using Merge Sort``def` `mergeSort(arr, N, K) :``    ``temp ``=` `[``0``]``*``N``    ``print``(mergeSortUtil(arr, temp, ``0``, N ``-` `1``, K))` `  ``# Driver code``arr ``=` `[ ``5``, ``6``, ``2``, ``5` `]``N ``=` `len``(arr)``K ``=` `2` `# Function Call``mergeSort(arr, N, K)` `# This code is contributed by divyeshrabadiya07.`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG {` `  ``// Function to merge two sorted arrays``  ``static` `int` `merge(``int``[] arr, ``int``[] temp, ``int` `l, ``int` `m,``                   ``int` `r, ``int` `K)``  ``{` `    ``// i: index to left subarray``    ``int` `i = l;` `    ``// j: index to right subarray``    ``int` `j = m + 1;` `    ``// Stores count of pairs that``    ``// satisfy the given condition``    ``int` `cnt = 0;` `    ``for` `(i = l; i <= m; i++) {``      ``bool` `found = ``false``;` `      ``// Traverse to check for the``      ``// valid conditions``      ``while` `(j <= r) {` `        ``// If condition satisfies``        ``if` `(arr[i] >= K * arr[j]) {``          ``found = ``true``;``        ``}``        ``else``          ``break``;``        ``j++;``      ``}` `      ``// While a[i] > K*a[j] satisfies``      ``// increase j` `      ``// All elements in the right``      ``// side of the left subarray``      ``// also satisfies``      ``if` `(found == ``true``) {``        ``cnt += j - (m + 1);``        ``j--;``      ``}``    ``}` `    ``// Sort the two given arrays and``    ``// store in the resultant array``    ``int` `k = l;``    ``i = l;``    ``j = m + 1;` `    ``while` `(i <= m && j <= r)``    ``{``      ``if` `(arr[i] <= arr[j])``        ``temp[k++] = arr[i++];``      ``else``        ``temp[k++] = arr[j++];``    ``}` `    ``// Elements which are left``    ``// in the left subarray``    ``while` `(i <= m)``      ``temp[k++] = arr[i++];` `    ``// Elements which are left``    ``// in the right subarray``    ``while` `(j <= r)``      ``temp[k++] = arr[j++];` `    ``for` `(i = l; i <= r; i++)``      ``arr[i] = temp[i];` `    ``// Return the count obtained``    ``return` `cnt;``  ``}` `  ``// Function to partition array into two halves``  ``static` `int` `mergeSortUtil(``int``[] arr, ``int``[] temp, ``int` `l,``                           ``int` `r, ``int` `K)``  ``{``    ``int` `cnt = 0;``    ``if` `(l < r) {` `      ``// Same as (l + r) / 2, but avoids``      ``// overflow for large l and h``      ``int` `m = (l + r) / 2;` `      ``// Sort first and second halves``      ``cnt += mergeSortUtil(arr, temp, l, m, K);``      ``cnt += mergeSortUtil(arr, temp, m + 1, r, K);` `      ``// Call the merging function``      ``cnt += merge(arr, temp, l, m, r, K);``    ``}``    ``return` `cnt;``  ``}` `  ``// Function to print the count of``  ``// required pairs using Merge Sort``  ``static` `void` `mergeSort(``int``[] arr, ``int` `N, ``int` `K)``  ``{``    ``int``[] temp = ``new` `int``[N];``    ``Console.WriteLine(``      ``mergeSortUtil(arr, temp, 0, N - 1, K));``  ``}` `  ``// Driver code``  ``static` `public` `void` `Main()``  ``{` `    ``int``[] arr = ``new` `int``[] { 5, 6, 2, 5 };``    ``int` `N = arr.Length;``    ``int` `K = 2;` `    ``// Function Call``    ``mergeSort(arr, N, K); ``  ``}``}` `// This code is contributed by Dharanendra L V`

## Javascript

 ``

Output

`2`

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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