# For each element in 1st array count elements less than or equal to it in 2nd array

Given two unsorted arrays arr1[] and arr2[]. They may contain duplicates. For each element in arr1[] count elements less than or equal to it in array arr2[].
Source: Amazon Interview Experience | Set 354 (For SDE-2)

Examples:

`Input : arr1[] = [1, 2, 3, 4, 7, 9]        arr2[] = [0, 1, 2, 1, 1, 4]Output : [4, 5, 5, 6, 6, 6]So the number of elements less than or equal to 1 is 4, 2 is 5, 3 is 5, 4 is 6, 7 is 6 and 9 is 6.Input : arr1[] = [5, 10, 2, 6, 1, 8, 6, 12]        arr2[] = [6, 5, 11, 4, 2, 3, 7]Output : [4, 6, 1, 5, 0, 6, 5, 7]So the number of elements less than or equal to 5 is 4, 10 is 6, 2 is 1, 6 is 5, 1 is 0, 8 is 6, 6 is 5 and 12 is 7 `

Naive Approach:

Approach: The idea is to use two loops, the outer loop for elements of array arr1[] and an inner loop for elements of array arr2[]. Then for each element of arr1[], count elements less than or equal to it in arr2[].

Algorithm :

1. Traverse through the elements of the first array from start to end.
2. For every element in the first array.
3. Traverse through the elements in the second array and find the count of elements that are less than or equal to the element of the first array.
4. Print the count for every index.

Implementation:

## C++

 `// C++ code for the above algorithm` `#include ` `using` `namespace` `std;`   `// Function to count for each` `// element in 1st array,` `// elements less than or equal` `// to it in 2nd array` `void` `countEleLessThanOrEqual(``int` `arr1[], ``int` `arr2[],` `                             ``int` `m, ``int` `n)` `{` `    ``// Run two loops to count` `    ``// First loop to traverse the first array` `    ``// Second loop to traverse the second array` `    ``for` `(``int` `i = 0; i < m; i++) {` `        ``int` `count = 0;`   `        ``// Traverse through second array` `        ``for` `(``int` `j = 0; j < n; j++)` `            ``if` `(arr2[j] <= arr1[i])` `                ``count++;`   `        ``cout << count << ``" "``;` `    ``}` `}`   `// Driver program to test above` `int` `main()` `{` `    ``int` `arr1[] = { 1, 2, 3, 4, 7, 9 };` `    ``int` `arr2[] = { 0, 1, 2, 1, 1, 4 };` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `    ``return` `0;` `}`

## Java

 `import` `java.util.Arrays;`   `class` `GFG {`   `    ``// function to count for each` `    ``// element in 1st array,` `    ``// elements less than or equal` `    ``// to it in 2nd array` `    ``static` `int` `countEleLessThanOrEqual(` `        ``int` `arr1[], ``int` `arr2[],` `        ``int` `m, ``int` `n)` `    ``{` `        ``// Run two loops to count` `        ``// First loop to traverse the first array` `        ``// Second loop to traverse the second array` `        ``for` `(``int` `i = ``0``; i < m; i++) {` `            ``int` `count = ``0``;`   `            ``// Traverse through second array` `            ``for` `(``int` `j = ``0``; j < n; j++)` `                ``if` `(arr2[j] <= arr1[i])` `                    ``count++;` `            ``System.out.print(count + ``" "``);` `        ``}` `        ``return` `m;` `    ``}`   `    ``// Driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `arr1[] = { ``1``, ``2``, ``3``, ``4``, ``7``, ``9` `};` `        ``int` `arr2[] = { ``0``, ``1``, ``2``, ``1``, ``1``, ``4` `};` `        ``countEleLessThanOrEqual(` `            ``arr1, arr2, arr1.length, arr2.length);` `    ``}` `}`   `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 code for the above algorithm`   `# function to count for each element in 1st array,` `# elements less than or equal to it in 2nd array` `def` `countEleLessThanOrEqual(arr1, arr2, m, n):` `    `  `    ``# Run two loops to count ` `    ``# First loop to traverse the first array` `    ``# Second loop to traverse the second array` `    ``for` `i ``in` `range``(m):` `        `  `        ``count ``=` `0` `        ``# Traverse through second array` `        ``for` `j ``in` `range``(n):` `            ``if` `(arr2[j] <``=` `arr1[i]):` `                ``count``+``=` `1` `            `  `        ``print``(count, end ``=``" "``)`   `# Driver program to test above` `arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``]` `arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``]` `m ``=` `len``(arr1)` `n ``=` `len``(arr2)` `countEleLessThanOrEqual(arr1, arr2, m, n)` ` `  `# This code is contributed by shubhamsingh10`

## C#

 `// C# implementation of For each element` `using` `System;`   `class` `GFG {`   `    ``// function to count for each element in 1st array,` `    ``// elements less than or equal to it in 2nd array` `    ``static` `void` `countEleLessThanOrEqual(` `        ``int``[] arr1, ``int``[] arr2,` `        ``int` `m, ``int` `n)` `    ``{` `        ``// Run two loops to count` `        ``// First loop to traverse the first array` `        ``// Second loop to traverse the second array` `        ``for` `(``int` `i = 0; i < m; i++) {` `            ``int` `count = 0;`   `            ``// Traverse through second array` `            ``for` `(``int` `j = 0; j < n; j++)` `                ``if` `(arr2[j] <= arr1[i])` `                    ``count++;` `            ``Console.Write((count) + ``" "``);` `        ``}` `    ``}`   `    ``// Driver method` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr1 = { 1, 2, 3, 4, 7, 9 };` `        ``int``[] arr2 = { 0, 1, 2, 1, 1, 4 };`   `        ``countEleLessThanOrEqual(arr1,` `                                ``arr2, arr1.Length, arr2.Length);` `    ``}` `}`   `// This code is contributed by mohit kumar 29.`

## Javascript

 ``

Output

```4 5 5 6 6 6

```

Complexity Analysis:

• Time complexity: O(m * n).
Considering arr1[] and arr2[] are of sizes m and n respectively.
• Space Complexity: O(1).
As no extra space is required

Efficient Solution:

Approach: Sort the elements of 2nd array, i.e., array arr2[]. Then perform a modified binary search on array arr2[]. For each element x of array arr1[], find the last index of the largest element smaller than or equal to x in sorted array arr2[]. The index of the largest element will give the count of elements.

Algorithm:

1. Sort the second array.
2. Traverse through the elements of the first array from start to end.
3. For every element in the first array.
4. Do a binary search on the second array and find the index of the largest element smaller than or equal to the element of the first array.
5. The index of the largest element will give the count of elements. Print the count for every index.

## C++

 `// C++ implementation of For each element in 1st` `// array count elements less than or equal to it` `// in 2nd array` `#include `   `using` `namespace` `std;`   `// function returns the index of largest element` `// smaller than equal to 'x' in 'arr'. For duplicates` `// it returns the last index of occurrence of required` `// element. If no such element exits then it returns -1.` `int` `binary_search(``int` `arr[], ``int` `l, ``int` `h, ``int` `x)` `{` `    ``while` `(l <= h) {` `        ``int` `mid = (l + h) / 2;`   `        ``// if 'x' is greater than or equal to arr[mid],` `        ``// then search in arr[mid+1...h]` `        ``if` `(arr[mid] <= x)` `            ``l = mid + 1;`   `        ``// else search in arr[l...mid-1]` `        ``else` `            ``h = mid - 1;` `    ``}`   `    ``// required index` `    ``return` `h;` `}`   `// function to count for each element in 1st array,` `// elements less than or equal to it in 2nd array` `void` `countEleLessThanOrEqual(` `    ``int` `arr1[], ``int` `arr2[],` `    ``int` `m, ``int` `n)` `{` `    ``// sort the 2nd array` `    ``sort(arr2, arr2 + n);`   `    ``// for each element of 1st array` `    ``for` `(``int` `i = 0; i < m; i++) {` `        ``// last index of largest element` `        ``// smaller than or equal to x` `        ``int` `index = binary_search(` `            ``arr2, 0, n - 1, arr1[i]);`   `        ``// required count for the element arr1[i]` `        ``cout << (index + 1) << ``" "``;` `    ``}` `}`   `// Driver program to test above` `int` `main()` `{` `    ``int` `arr1[] = { 1, 2, 3, 4, 7, 9 };` `    ``int` `arr2[] = { 0, 1, 2, 1, 1, 4 };` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `    ``return` `0;` `}`

## Java

 `// Java implementation of For` `// each element in 1st` `// array count elements less` `// than or equal to it` `// in 2nd array`   `import` `java.util.Arrays;`   `class` `GFG {` `    ``// method returns the index` `    ``// of largest element` `    ``// smaller than equal to 'x'` `    ``// in 'arr'. For duplicates` `    ``// it returns the last index` `    ``// of occurrence of required` `    ``// element. If no such element` `    ``// exits then it returns -1.` `    ``static` `int` `binary_search(` `        ``int` `arr[], ``int` `l,` `        ``int` `h, ``int` `x)` `    ``{` `        ``while` `(l <= h) {` `            ``int` `mid = (l + h) / ``2``;`   `            ``// if 'x' is greater than or equal` `            ``// to arr[mid], then search in` `            ``// arr[mid+1...h]` `            ``if` `(arr[mid] <= x)` `                ``l = mid + ``1``;`   `            ``// else search in arr[l...mid-1]` `            ``else` `                ``h = mid - ``1``;` `        ``}`   `        ``// Required index` `        ``return` `h;` `    ``}`   `    ``// Method to count for each` `    ``// element in 1st array,` `    ``// elements less than or equal` `    ``// to it in 2nd array` `    ``static` `void` `countEleLessThanOrEqual(` `        ``int` `arr1[], ``int` `arr2[],` `        ``int` `m, ``int` `n)` `    ``{` `        ``// Sort the 2nd array` `        ``Arrays.sort(arr2);`   `        ``// for each element of 1st array` `        ``for` `(``int` `i = ``0``; i < m; i++) {` `            ``// Last index of largest element` `            ``// smaller than or equal to x` `            ``int` `index = binary_search(` `                ``arr2, ``0``, n - ``1``, arr1[i]);`   `            ``// Required count for the element arr1[i]` `            ``System.out.print((index + ``1``) + ``" "``);` `        ``}` `    ``}`   `    ``// Driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr1[] = { ``1``, ``2``, ``3``, ``4``, ``7``, ``9` `};` `        ``int` `arr2[] = { ``0``, ``1``, ``2``, ``1``, ``1``, ``4` `};`   `        ``countEleLessThanOrEqual(` `            ``arr1, arr2, arr1.length,` `            ``arr2.length);` `    ``}` `}`

## Python3

 `# python implementation of For each element in 1st ` `# array count elements less than or equal to it` `# in 2nd array`   `# function returns the index of largest element ` `# smaller than equal to 'x' in 'arr'. For duplicates` `# it returns the last index of occurrence of required` `# element. If no such element exits then it returns -1` `def` `bin_search(arr, n, x):    ` `  ``l ``=` `0` `  ``h ``=` `n ``-` `1` `  ``while``(l <``=` `h):` `      ``mid ``=` `int``((l ``+` `h) ``/``/` `2``)` `      ``# if 'x' is greater than or equal to arr[mid], ` `      ``# then search in arr[mid + 1...h]` `      ``if``(arr[mid] <``=` `x):` `          ``l ``=` `mid ``+` `1``;` `      ``else``:` `          ``# else search in arr[l...mid-1]` `          ``h ``=` `mid ``-` `1` `  ``# required index` `  ``return` `h`   `# function to count for each element in 1st array,` `# elements less than or equal to it in 2nd array` `def` `countElements(arr1, arr2, m, n):` `  ``# sort the 2nd array` `  ``arr2.sort()`   `  ``# for each element in first array` `  ``for` `i ``in` `range``(m):` `      ``# last index of largest element ` `      ``# smaller than or equal to x` `      ``index ``=` `bin_search(arr2, n, arr1[i])` `      ``# required count for the element arr1[i]` `      ``print``(index ``+` `1``,end``=``" "``)`   `# driver program to test above function` `arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``]` `arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``]` `m ``=` `len``(arr1)` `n ``=` `len``(arr2)` `countElements(arr1, arr2, m, n)`   `# This code is contributed by Aditi Sharma`

## C#

 `// C# implementation of For each element` `// in 1st array count elements less than` `// or equal to it in 2nd array` `using` `System;`   `public` `class` `GFG {`   `    ``// method returns the index of` `    ``// largest element smaller than` `    ``// equal to 'x' in 'arr'. For` `    ``// duplicates it returns the last` `    ``// index of occurrence of required` `    ``// element. If no such element` `    ``// exits then it returns -1.` `    ``static` `int` `binary_search(``int``[] arr,` `                             ``int` `l, ``int` `h, ``int` `x)` `    ``{` `        ``while` `(l <= h) {` `            ``int` `mid = (l + h) / 2;`   `            ``// if 'x' is greater than or` `            ``// equal to arr[mid], then` `            ``// search in arr[mid+1...h]` `            ``if` `(arr[mid] <= x)` `                ``l = mid + 1;`   `            ``// else search in` `            ``// arr[l...mid-1]` `            ``else` `                ``h = mid - 1;` `        ``}`   `        ``// required index` `        ``return` `h;` `    ``}`   `    ``// method to count for each element` `    ``// in 1st array, elements less than` `    ``// or equal to it in 2nd array` `    ``static` `void` `countEleLessThanOrEqual(` `        ``int``[] arr1, ``int``[] arr2,` `        ``int` `m, ``int` `n)` `    ``{`   `        ``// sort the 2nd array` `        ``Array.Sort(arr2);`   `        ``// for each element of 1st array` `        ``for` `(``int` `i = 0; i < m; i++) {` `            ``// last index of largest` `            ``// element smaller than or` `            ``// equal to x` `            ``int` `index = binary_search(` `                ``arr2, 0, n - 1, arr1[i]);`   `            ``// required count for the` `            ``// element arr1[i]` `            ``Console.Write((index + 1) + ``" "``);` `        ``}` `    ``}`   `    ``// Driver method` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr1 = { 1, 2, 3, 4, 7, 9 };` `        ``int``[] arr2 = { 0, 1, 2, 1, 1, 4 };`   `        ``countEleLessThanOrEqual(arr1,` `                                ``arr2, arr1.Length, arr2.Length);` `    ``}` `}`   `// This code is contributed by Sam007.`

## Javascript

 ``

## PHP

 ``

Output

```4 5 5 6 6 6

```

Complexity Analysis:

• Time Complexity: O(mlogn + nlogn).
Considering arr1[] and arr2[] of sizes m and n respectively.
• Space Complexity: O(1).
As no extra space is required

Another way of solving the problem is to sort the second array and use the upper_bound() inbuilt function for each value of first array.

Implementation:

## C++

 `// C++ implementation of For each element in 1st` `// array count elements less than or equal to it` `// in 2nd array` `#include ` `using` `namespace` `std;`   `// function to count for each element in 1st array,` `// elements less than or equal to it in 2nd array` `void` `countEleLessThanOrEqual(``int` `arr1[], ``int` `arr2[], ``int` `m, ``int` `n)` `{` `    ``// sort the 2nd array` `    ``sort(arr2, arr2 + n);`   `    ``// for each element of 1st array` `    ``for` `(``int` `i = 0; i < m; i++) {` `        ``int` `x = upper_bound(arr2, arr2 + n, arr1[i]) - arr2;` `        ``cout << x << ``" "``;` `    ``}` `}`   `// Driver program to test above` `int` `main()` `{` `    ``int` `arr1[] = { 1, 2, 3, 4, 7, 9 };` `    ``int` `arr2[] = { 0, 1, 2, 1, 1, 4 };` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// JAVA implementation of For each element in 1st` `// array count elements less than or equal to it` `// in 2nd array` `import` `java.util.*;` `class` `GFG` `{` `  ``public` `static` `int` `upper_bound(``int` `arr[], ``int` `key)` `  ``{` `    ``int` `upperBound = ``0``;`   `    ``while` `(upperBound < arr.length)` `    ``{`   `      ``// If current value is lesser than or equal to` `      ``// key` `      ``if` `(arr[upperBound] <= key)` `        ``upperBound++;`   `      ``// This value is just greater than key` `      ``else` `{` `        ``return` `upperBound;` `      ``}` `    ``}` `    ``return` `upperBound;` `  ``}`   `  ``// function to count for each element in 1st array,` `  ``// elements less than or equal to it in 2nd array` `  ``public` `static` `void` `countEleLessThanOrEqual(``int` `arr1[],` `                                             ``int` `arr2[],` `                                             ``int` `m, ``int` `n)` `  ``{`   `    ``// sort the 2nd array` `    ``Arrays.sort(arr2);`   `    ``// for each element of 1st array` `    ``for` `(``int` `i = ``0``; i < m; i++) {` `      ``int` `x = upper_bound(arr2, arr1[i]);` `      ``System.out.print(x + ``" "``);` `    ``}` `  ``}`   `  ``// Driver program to test above` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int` `arr1[] = { ``1``, ``2``, ``3``, ``4``, ``7``, ``9` `};` `    ``int` `arr2[] = { ``0``, ``1``, ``2``, ``1``, ``1``, ``4` `};` `    ``int` `m = arr1.length;` `    ``int` `n = arr2.length;` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `  ``}` `}`   `// This code is contributed by Taranpreet`

## Python3

 `# Python3 implementation of For each element in 1st` `# array count elements less than or equal to it` `# in 2nd array` `def` `upper_bound(arr, key):`   `    ``upperBound ``=` `0`   `    ``while` `(upperBound < ``len``(arr)):`   `        ``# If current value is lesser than or equal to` `        ``# key` `        ``if``(arr[upperBound] <``=` `key):` `            ``upperBound ``+``=` `1`   `        ``# This value is just greater than key` `        ``else``:` `            ``return` `upperBound`   `    ``return` `upperBound`   `# function to count for each element in 1st array,` `# elements less than or equal to it in 2nd array` `def` `countEleLessThanOrEqual(arr1, arr2, m, n):`   `    ``# sort the 2nd array` `    ``arr2.sort()`   `    ``# for each element of 1st array` `    ``for` `i ``in` `range``(m):` `        ``x ``=` `upper_bound(arr2, arr1[i])` `        ``print``(x, end``=``" "``)`   `# Driver program to test above` `arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``]` `arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``]` `m ``=` `len``(arr1)` `n ``=` `len``(arr2)` `countEleLessThanOrEqual(arr1, arr2, m, n)`   `# This code is contributed by Abhijeet Kumar(abhijeet19403)`

## C#

 `// C# implementation of For each element in 1st array count` `// elements less than or equal to it in 2nd array`   `using` `System;` `using` `System.Collections;`   `public` `class` `GFG {`   `    ``public` `static` `int` `upper_bound(``int``[] arr, ``int` `key)` `    ``{` `        ``int` `upperBound = 0;`   `        ``while` `(upperBound < arr.Length) {`   `            ``// If current value is lesser than or equal to` `            ``// key` `            ``if` `(arr[upperBound] <= key)` `                ``upperBound++;`   `            ``// This value is just greater than key` `            ``else` `{` `                ``return` `upperBound;` `            ``}` `        ``}` `        ``return` `upperBound;` `    ``}`   `    ``// function to count for each element in 1st array,` `    ``// elements less than or equal to it in 2nd array` `    ``public` `static` `void` `countEleLessThanOrEqual(``int``[] arr1,` `                                               ``int``[] arr2,` `                                               ``int` `m, ``int` `n)` `    ``{`   `        ``// sort the 2nd array` `        ``Array.Sort(arr2);`   `        ``// for each element of 1st array` `        ``for` `(``int` `i = 0; i < m; i++) {` `            ``int` `x = upper_bound(arr2, arr1[i]);` `            ``Console.Write(x + ``" "``);` `        ``}` `    ``}`   `    ``static` `public` `void` `Main()` `    ``{`   `        ``// Code` `        ``int``[] arr1 = { 1, 2, 3, 4, 7, 9 };` `        ``int``[] arr2 = { 0, 1, 2, 1, 1, 4 };` `        ``int` `m = arr1.Length;` `        ``int` `n = arr2.Length;` `        ``countEleLessThanOrEqual(arr1, arr2, m, n);` `    ``}` `}`   `// This code is contributed by lokeshmvs21.`

## Javascript

 `function` `upper_bound(arr, key)` `  ``{` `    ``let upperBound = 0;`   `    ``while` `(upperBound < arr.length)` `    ``{`   `      ``// If current value is lesser than or equal to` `      ``// key` `      ``if` `(arr[upperBound] <= key)` `        ``upperBound++;`   `      ``// This value is just greater than key` `      ``else` `{` `        ``return` `upperBound;` `      ``}` `    ``}` `    ``return` `upperBound;` `  ``}`   `  ``// function to count for each element in 1st array,` `  ``// elements less than or equal to it in 2nd array` `  ``function` `countEleLessThanOrEqual(arr1, arr2, m, n)` `  ``{`   `    ``// sort the 2nd array` `    ``arr2.sort();`   `    ``// for each element of 1st array` `    ``for` `(let i = 0; i < m; i++) {` `      ``let x = upper_bound(arr2, arr1[i]);` `      ``console.log(x + ``" "``);` `    ``}` `  ``}`     ` ``// Driver program to test above` `    ``let arr1 = [ 1, 2, 3, 4, 7, 9 ];` `    ``let arr2 = [ 0, 1, 2, 1, 1, 4 ];` `    ``let m = arr1.length;` `    ``let n = arr2.length;` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);`   `// This code is contributed by aadityaburujwale.`

Output

```4 5 5 6 6 6

```

Time Complexity: O(n logn + m log n), where m and n represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Another Approach: We can also use two pointers to find our solution.  first, sort both the arrays. after sorting put i pointer for arr1 and pointer for arr2 at the beginning. we will traverse over the elements of arr1 by using i pointer & inside this loop, we will go over each element of arr2 by j pointer. wherever arr2[j] <= arr1[i] we will increase j . if the condition fails print j

Note:- this will give answer in sorted manner

## C++

 `// C++ implementation of For each element in 1st` `// array count elements less than or equal to it` `// in 2nd array` `#include ` `using` `namespace` `std;`   `void` `countEleLessThanOrEqual(``int` `arr1[], ``int` `arr2[], ``int` `m, ``int` `n)` `{` `    ``// sorting both the arrays` `    ``sort(arr1, arr1 + m);` `    ``sort(arr2, arr2 + n);`   `    ``// pointer for arr2` `    ``int` `j = 0;`   `    ``// traversing  each element of 1st array` `    ``for` `(``int` `i = 0; i < m; i++)` `    ``{` `        ``// checking the condition for each element` `        ``while` `(j < n && arr2[j] <= arr1[i])` `            ``j++;`   `        ``cout << j << ``" "``;` `    ``}` `}`   `// Driver program to test above` `int` `main()` `{` `    ``int` `arr1[] = {1, 2, 3, 4, 7, 9};` `    ``int` `arr2[] = {0, 1, 2, 1, 1, 4};` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `    ``return` `0;` `}` `// This code is contributed by Naveen Shah`

## Java

 `// Java implementation of For each element in 1st` `// array count elements less than or equal to it` `// in 2nd array` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG` `{` `  ``static` `void` `countEleLessThanOrEqual(``int``[] arr1,` `                                      ``int``[] arr2, ``int` `m,` `                                      ``int` `n)` `  ``{` `    ``Arrays.sort(arr1);` `    ``Arrays.sort(arr2);`   `    ``// pointer for arr2` `    ``int` `j = ``0``;`   `    ``// traversing  each element of 1st array` `    ``for` `(``int` `i = ``0``; i < m; i++)` `    ``{`   `      ``// checking the condition for each element` `      ``while` `(j < n && arr2[j] <= arr1[i])` `        ``j++;`   `      ``System.out.print(j + ``" "``);` `    ``}` `  ``}`   `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[] arr1 = { ``1``, ``2``, ``3``, ``4``, ``7``, ``9` `};` `    ``int``[] arr2 = { ``0``, ``1``, ``2``, ``1``, ``1``, ``4` `};` `    ``int` `m = arr1.length;` `    ``int` `n = arr2.length;`   `    ``countEleLessThanOrEqual(arr1, arr2, m, n);` `  ``}` `}`   `// This code is contributed by lokeshmvs21.`

## Python3

 `# implementation of For each element in 1st` `# array count elements less than or equal to it` `#  in 2nd array` `def` `countEleLessThanOrEqual(arr1, arr2, m, n):` `  `  `    ``# sorting both the arrays` `    ``arr1.sort()` `    ``arr2.sort()` `    `  `    ``# pointer for arr2` `    ``j ``=` `0` `    `  `    ``# traversing  each element of 1st array` `    ``for` `i ``in` `range``(m):` `        ``# checking the condition for each element` `        ``while` `(j < n ``and` `arr2[j] <``=` `arr1[i]):` `            ``j ``+``=` `1` `        ``print``(j, end``=``" "``)`   `# Driver program to test above` `arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``]` `arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``]` `m ``=` `len``(arr1)` `n ``=` `len``(arr2)` `countEleLessThanOrEqual(arr1, arr2, m, n)` `" This code is contributed by rajatkumargla19"`

## C#

 `// C# implementation of For each element` `// in 1st array count elements less than` `// or equal to it in 2nd array` `using` `System;` `public` `class` `GFG` `{`   `  ``// function to count for each element in 1st array,` `  ``// elements less than or equal to it in 2nd array` `  ``static` `void` `countEleLessThanOrEqual(``int``[] arr1,` `                                      ``int``[] arr2, ``int` `m,` `                                      ``int` `n)` `  ``{`   `    ``// sorting both of the array` `    ``Array.Sort(arr1);` `    ``Array.Sort(arr2);` `    ``int` `j = 0;`   `    ``// traversing the array from 0 to m-1 index` `    ``for` `(``int` `i = 0; i < m; i++) {` `      ``while` `(j < n && arr2[j] <= arr1[i]) {` `        ``j++;` `      ``}` `      ``// printing the value of j` `      ``Console.Write(j + ``" "``);` `    ``}` `  ``}`   `  ``// Driver method` `  ``public` `static` `void` `Main()` `  ``{` `    ``int``[] arr1 = { 1, 2, 3, 4, 7, 9 };` `    ``int``[] arr2 = { 0, 1, 2, 1, 1, 4 };`   `    ``countEleLessThanOrEqual(arr1, arr2, arr1.Length,` `                            ``arr2.Length);` `  ``}` `}`   `// This code is contributed by rajatkumargla19...`

## Javascript

 `    ``// JavaScript implementation of For each element in 1st` `    ``// array count elements less than or equal to it` `    ``// in 2nd array` `    ``function` `countEleLessThanOrEqual(arr1, arr2, m, n){` `        ``arr1.sort();` `        ``arr2.sort();` `        `  `        ``// pointer for arr2` `        ``let j = 0;` `        `  `        ``// traversing each element of 1st array` `        ``for` `(let i = 0; i < m; i++)` `        ``{` ` `  `              ``// checking the condition for each element` `              ``while` `(j < n && arr2[j] <= arr1[i])` `                ``j++;` ` `  `              ``console.log(j + ``" "``);` `        ``}` `    ``}`   `    ``let arr1 = [ 1, 2, 3, 4, 7, 9 ];` `    ``let arr2 = [ 0, 1, 2, 1, 1, 4 ];` `    ``let m = arr1.length;` `    ``let n = arr2.length;` `    `  `    ``countEleLessThanOrEqual(arr1, arr2, m, n);`   `// This code is contributed by lokeshmvs21.`

Output

```4 5 5 6 6 6

```

Time Complexity: O(nlogn + mlogm)
Auxiliary Space: O(1)

Another Approach: prefix-sum technique and hashing

Follow the steps to implement the approach:

1. Create a hash table to store the frequency of elements in arr2. Initialize the hash table with zeros.
2. Iterate through arr2 and update the frequency of each element in the hash table.
3. Create a prefix sum array prefixSum of size 100010 (maximum element value + 10). Initialize prefixSum[0] with the frequency of the first element in arr2.
4. Calculate the prefix sum array by adding the current frequency with the previous prefix sum.
5. Iterate through arr1. For each element arr1[i], access prefixSum[arr1[i]] to get the count of elements less than or equal to arr1[i] in arr2.
6. Store the counts in a result array or directly print them.

Below is the implementation:

## C++

 `#include ` `#include ` `using` `namespace` `std;` `// Function to count elements less than or equal to each` `// element in arr1[]` `void` `countElements(``int` `arr1[], ``int` `arr2[], ``int` `m, ``int` `n)` `{` `    ``int` `result[m];` `    ``int` `frequency[100010] = {` `        ``0` `    ``}; ``// Hash table to store frequency of elements`   `    ``// Calculate frequency of elements in arr2` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``frequency[arr2[i]]++;` `    ``}`   `    ``int` `prefixSum[100010];` `    ``prefixSum[0] = frequency[0];`   `    ``// Calculate prefix sum array` `    ``for` `(``int` `i = 1; i < 100010; ++i) {` `        ``prefixSum[i] = prefixSum[i - 1] + frequency[i];` `    ``}`   `    ``// Count elements less than or equal to arr1[i]` `    ``for` `(``int` `i = 0; i < m; ++i) {` `        ``result[i] = prefixSum[arr1[i]];` `    ``}`   `    ``// print the counts` `    ``for` `(``int` `i = 0; i < m; ++i) {` `        ``cout << result[i] << ``" "``;` `    ``}` `}` `// Driver Code` `int` `main()` `{` `    ``int` `arr1[] = { 1, 2, 3, 4, 7, 9 };` `    ``int` `arr2[] = { 0, 1, 2, 1, 1, 4 };` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` `    ``//   Function Called` `    ``countElements(arr1, arr2, m, n);`   `    ``return` `0;` `}` `// This code is contributed by Veerendra_Singh_Rajpoot`

## Java

 `import` `java.util.Arrays;`   `public` `class` `GFG {` `  ``// Function to count elements less than or equal to each` ` ``// element in arr1[]` `    ``public` `static` `void` `countElements(``int``[] arr1, ``int``[] arr2,` `                                     ``int` `m, ``int` `n)` `    ``{` `        ``// Create an array to store the result` `        ``int``[] result = ``new` `int``[m];` `        ``// Create an array to store the frequency of` `        ``// elements in arr2` `        ``int``[] frequency = ``new` `int``[``100010``];`   `        ``// Calculate the frequency of each element in arr2` `        ``for` `(``int` `i = ``0``; i < n; ++i) {` `            ``frequency[arr2[i]]++;` `        ``}`   `        ``// Calculate the prefix sum of the frequency array` `        ``int``[] prefixSum = ``new` `int``[``100010``];` `        ``prefixSum[``0``] = frequency[``0``];`   `        ``for` `(``int` `i = ``1``; i < ``100010``; ++i) {` `            ``prefixSum[i] = prefixSum[i - ``1``] + frequency[i];` `        ``}`   `        ``// Calculate the count of elements in arr1 using the` `        ``// prefix sum array` `        ``for` `(``int` `i = ``0``; i < m; ++i) {` `            ``result[i] = prefixSum[arr1[i]];` `        ``}`   `        ``// Print the count of elements in arr1` `        ``for` `(``int` `i = ``0``; i < m; ++i) {` `            ``System.out.print(result[i] + ``" "``);` `        ``}` `    ``}` `    ``// Driver Codea` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int``[] arr1 = { ``1``, ``2``, ``3``, ``4``, ``7``, ``9` `};` `        ``int``[] arr2 = { ``0``, ``1``, ``2``, ``1``, ``1``, ``4` `};` `        ``int` `m = arr1.length;` `        ``int` `n = arr2.length;`   `        ``// Call the countElements function` `        ``countElements(arr1, arr2, m, n);` `    ``}` `}` `// This code is contributed by Veerendra_Singh_Rajpoot`

## Python3

 `# Function to count elements less than or equal to each` `# element in arr1[]`   `def` `countElements(arr1, arr2, m, n):` `    ``result ``=` `[``0``] ``*` `m` `    ``frequency ``=` `[``0``] ``*` `100010`  `# Hash table to store frequency of elements`   `    ``# Calculate frequency of elements in arr2` `    ``for` `i ``in` `range``(n):` `        ``frequency[arr2[i]] ``+``=` `1`   `    ``prefixSum ``=` `[``0``] ``*` `100010` `    ``prefixSum[``0``] ``=` `frequency[``0``]`   `    ``# Calculate prefix sum array` `    ``for` `i ``in` `range``(``1``, ``100010``):` `        ``prefixSum[i] ``=` `prefixSum[i ``-` `1``] ``+` `frequency[i]`   `    ``# Count elements less than or equal to arr1[i]` `    ``for` `i ``in` `range``(m):` `        ``result[i] ``=` `prefixSum[arr1[i]]`   `    ``# Print the counts` `    ``for` `i ``in` `range``(m):` `        ``print``(result[i], end``=``" "``)` `    ``print``()`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``]` `    ``arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``]` `    ``m ``=` `len``(arr1)` `    ``n ``=` `len``(arr2)` `    ``# Function Call` `    ``countElements(arr1, arr2, m, n)`     `# This code is contributed by akshitaguprzj3`

## C#

 `using` `System;`   `public` `class` `GFG {` `    ``// Function to count elements less than or equal to each` `    ``// element in arr1[]` `    ``public` `static` `void` `CountElements(``int``[] arr1, ``int``[] arr2,` `                                     ``int` `m, ``int` `n)` `    ``{` `        ``int``[] result = ``new` `int``[m];` `        ``int``[] frequency` `            ``= ``new` `int``[100010]; ``// Hash table to store` `                               ``// frequency of elements`   `        ``// Calculate frequency of elements in arr2` `        ``for` `(``int` `i = 0; i < n; ++i) {` `            ``frequency[arr2[i]]++;` `        ``}`   `        ``int``[] prefixSum = ``new` `int``[100010];` `        ``prefixSum[0] = frequency[0];`   `        ``// Calculate prefix sum array` `        ``for` `(``int` `i = 1; i < 100010; ++i) {` `            ``prefixSum[i] = prefixSum[i - 1] + frequency[i];` `        ``}`   `        ``// Count elements less than or equal to arr1[i]` `        ``for` `(``int` `i = 0; i < m; ++i) {` `            ``result[i] = prefixSum[arr1[i]];` `        ``}`   `        ``// print the counts` `        ``for` `(``int` `i = 0; i < m; ++i) {` `            ``Console.Write(result[i] + ``" "``);` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr1 = { 1, 2, 3, 4, 7, 9 };` `        ``int``[] arr2 = { 0, 1, 2, 1, 1, 4 };` `        ``int` `m = arr1.Length;` `        ``int` `n = arr2.Length;`   `        ``// Function Called` `        ``CountElements(arr1, arr2, m, n);` `    ``}` `}`   `// This code is contributed by rambabuguphka`

## Javascript

 `// Function to count elements less than or equal to each element in arr1[]` `function` `countElements(arr1, arr2, m, n) {` `    ``let result = [];` `    ``let frequency = ``new` `Array(100010).fill(0); ``// Hash table to store frequency of elements`   `    ``// Calculate frequency of elements in arr2` `    ``for` `(let i = 0; i < n; ++i) {` `        ``frequency[arr2[i]]++;` `    ``}`   `    ``let prefixSum = [];` `    ``prefixSum[0] = frequency[0];`   `    ``// Calculate prefix sum array` `    ``for` `(let i = 1; i < 100010; ++i) {` `        ``prefixSum[i] = prefixSum[i - 1] + frequency[i];` `    ``}`   `    ``// Count elements less than or equal to arr1[i]` `    ``for` `(let i = 0; i < m; ++i) {` `        ``result[i] = prefixSum[arr1[i]];` `    ``}`   `    ``// print the counts` `    ``for` `(let i = 0; i < m; ++i) {` `        ``console.log(result[i] + ``" "``);` `    ``}` `}`   `// Driver Code` `    ``let arr1 = [1, 2, 3, 4, 7, 9];` `    ``let arr2 = [0, 1, 2, 1, 1, 4];` `    ``let m = arr1.length;` `    ``let n = arr2.length;`   `    ``// Function Called` `    ``countElements(arr1, arr2, m, n);`

Output

```4 5 5 6 6 6

```

Time Complexity: O(n + m)
Auxiliary Space: O(m)

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