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

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[].

Examples:

Input : arr1[] = [1, 2, 3, 4, 7, 9]
arr2[] = [0, 1, 2, 1, 1, 4]
Output : [4, 5, 5, 6, 6, 6]
Explanation: There are 4 elements less than or equal to 1 in second array, similarly there are 5 elements less than 2 in second array, calculate the values similarly for other elements.

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]
Explanation: There are 4 elements less than or equal to 5 in second array, similarly there are 6 elements less than 10 in second array, calculate the values similarly for other elements.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem is already discussed in the previous post.

Solution: In this article, a more optimized linear time solution to the above problem is discussed. The approach discussed here works for arrays with values in a small range. A range of values that can be used as an index in an array.

• Approach :
The idea is to create a prefix map up to the maximum element of the second array. The prefix array will store the maximum element up to that index, for example, prefix[i] will store the count of elements up to i. Then traverse through the first array and find the count of elements less than or equal to that element from the prefix array.
The prefix array is created by traversing through the prefix array and updating the current element by adding the precious element, i.e. prefix[i]+ =prefix[i-1].
• Algorithm:
1. create an extra space (prefix) of size of maximum element of second and first array or a map.
2. Traverse the second array.
3. For every element of the second array increase the count of prefix array, i.e. prefix[arr2[i]]++
4. Traverse through the prefix array from 1 to MAX (maximum element of second and first array and upadte the ith element by adding sum of i-1th element
5. Now traverse through the first array and print the value of prefix[arr_1[i]]
• Implementation:

## C++

 `// C++ program for each element in 1st  ` `// array count elements less than or equal to it  ` `// in 2nd array  ` ` `  `#include   ` `using` `namespace` `std;  ` ` `  `#define MAX 100000  ` ` `  `// Function for each element in 1st  ` `// array count elements less than or equal to it  ` `// in 2nd array  ` `void` `countEleLessThanOrEqual(``int` `arr1[], ``int` `m,  ` `                                ``int` `arr2[], ``int` `n)  ` `{  ` `    ``// Creating hash array initially  ` `    ``// filled with zero  ` `    ``int` `hash[MAX] = {0};  ` `     `  `    ``// Insert element of arr2[] to hash  ` `    ``// such that hash[i] will give count of  ` `    ``// element i in arr2[]  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``hash[arr2[i]]++;  ` ` `  `    ``// Presum of hash array  ` `    ``// such that hash[i] will give count of  ` `    ``// element less than or equals to i in arr2[]  ` `    ``for` `(``int` `i=1; i

## Java

 `// Java program for each element  ` `// in 1st array count elements  ` `// less than or equal to it in  ` `// 2nd array  ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `static` `int` `MAX = ``100000``; ` ` `  `// Function for each element  ` `// in 1st array count elements  ` `// less than or equal to it  ` `// in 2nd array  ` `static` `void` `countEleLessThanOrEqual(``int` `arr1[], ``int` `m,  ` `                                    ``int` `arr2[], ``int` `n) ` `{  ` `    ``// Creating hash array initially ` `    ``// filled with zero ` `    ``int` `hash[] = ``new` `int``[MAX]; ` `     `  `    ``// Insert element of arr2[] to ` `    ``// hash such that hash[i] will ` `    ``// give count of element i in arr2[] ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``hash[arr2[i]]++; ` ` `  `    ``// Presum of hash array ` `    ``// such that hash[i] will  ` `    ``// give count of element  ` `    ``// less than or equals to  ` `    ``// i in arr2[] ` `    ``for``(``int` `i = ``1``; i < MAX; i++) ` `    ``{ ` `        ``hash[i] = hash[i] +  ` `                  ``hash[i - ``1``]; ` `    ``} ` `     `  `    ``// Traverse arr1[] and  ` `    ``// print hash[arr[i]] ` `    ``for` `(``int` `i = ``0``; i < m; i++)  ` `    ``{ ` `        ``System.out.print(hash[arr1[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, n; ` `     `  `    ``m = arr1.length; ` `    ``n = arr2.length; ` `     `  `    ``countEleLessThanOrEqual(arr1, m, arr2, n); ` `} ` `} ` ` `  `// This code is contributed ` `// by inder_verma `

## Python3

 `# Python 3 program for each element in 1st  ` `# array count elements less than or equal  ` `# to it in 2nd array  ` ` `  `MAX` `=` `100000` ` `  `# Function for each element in 1st  ` `# array count elements less than or  ` `# equal to it in 2nd array  ` `def` `countEleLessThanOrEqual(arr1, m, arr2, n): ` `     `  `    ``# Creating hash array initially  ` `    ``# filled with zero  ` `    ``hash` `=` `[``0` `for` `i ``in` `range``(``MAX``)]  ` `     `  `    ``# Insert element of arr2[] to hash  ` `    ``# such that hash[i] will give count  ` `    ``# of element i in arr2[]  ` `    ``for` `i ``in` `range``(n): ` `        ``hash``[arr2[i]] ``+``=` `1` ` `  `    ``# Presum of hash array such that  ` `    ``# hash[i] will give count of element  ` `    ``# less than or equals to i in arr2[]  ` `    ``for` `i ``in` `range``(``1``, ``MAX``, ``1``): ` `        ``hash``[i] ``=` `hash``[i] ``+` `hash``[i ``-` `1``]  ` `     `  `    ``# Traverse arr1[] and print hash[arr[i]]  ` `    ``for` `i ``in` `range``(m): ` `        ``print``(``hash``[arr1[i]], end ``=` `" "``)      ` ` `  `# 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) ` `     `  `    ``countEleLessThanOrEqual(arr1, m, arr2, n)  ` `     `  `# This code is contributed by ` `# Shashank_Sharma `

## C#

 `// C# program for each element  ` `// in 1st array count elements  ` `// less than or equal to it in  ` `// 2nd array  ` `using` `System; ` `public` `class` `GFG {  ` ` `  `static` `int` `MAX = 100000;  ` ` `  `// Function for each element  ` `// in 1st array count elements  ` `// less than or equal to it  ` `// in 2nd array  ` `static` `void` `countEleLessThanOrEqual(``int` `[]arr1, ``int` `m,  ` `                                    ``int` `[]arr2, ``int` `n)  ` `{  ` `    ``// Creating hash array initially  ` `    ``// filled with zero  ` `    ``int` `[]hash = ``new` `int``[MAX];  ` `     `  `    ``// Insert element of arr2[] to  ` `    ``// hash such that hash[i] will  ` `    ``// give count of element i in arr2[]  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``hash[arr2[i]]++;  ` ` `  `    ``// Presum of hash array  ` `    ``// such that hash[i] will  ` `    ``// give count of element  ` `    ``// less than or equals to  ` `    ``// i in arr2[]  ` `    ``for``(``int` `i = 1; i < MAX; i++)  ` `    ``{  ` `        ``hash[i] = hash[i] +  ` `                ``hash[i - 1];  ` `    ``}  ` `     `  `    ``// Traverse arr1[] and  ` `    ``// print hash[arr[i]]  ` `    ``for` `(``int` `i = 0; i < m; i++)  ` `    ``{  ` `        ``Console.Write(hash[arr1[i]] + ``" "``);  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main ()  ` `{  ` `    ``int` `[]arr1 = {1, 2, 3, 4, 7, 9};  ` `    ``int` `[]arr2 = {0, 1, 2, 1, 1, 4};  ` `    ``int` `m, n;  ` `     `  `    ``m = arr1.Length;  ` `    ``n = arr2.Length;  ` `     `  `    ``countEleLessThanOrEqual(arr1, m, arr2, n);  ` `}  ` `}  ` ` `  `// This code is contributed by Shikha Singh.  `

## PHP

 ` `

Output:

```4 5 5 6 6 6
```
• Alternate method:This method just replaces the array with the vector and the hashmap that is made using an array is replaced by the map. The approach and algorithm remain the same.

• Implementation:

## C++

 `// C++ program for each element in 1st  ` `// array count elements less than or equal to it  ` `// in 2nd array  ` `#include ` `#include ` `#include ` ` `  `// Function for each element in 1st  ` `// array count elements less than or equal to it  ` `// in 2nd array  ` `void` `countLessThanOrEqual(``const` `std::vector<``int``>& vec1,  ` `                        ``const` `std::vector<``int``>& vec2) { ` `    ``std::map<``int``, unsigned ``int``> countOfVec2; ` `    ``for` `(``const` `auto``& item : vec2) { ` `        ``++countOfVec2[item]; ` `    ``} ` ` `  `    ``unsigned ``int` `prev = 0; ` `    ``for` `(``auto``& pair : countOfVec2) { ` `        ``pair.second += prev; ` `        ``prev = pair.second; ` `    ``} ` `    ``// Traverse arr1[] and print result  ` `    ``for` `(``const` `auto``& item : vec1) { ` `        ``unsigned ``int` `result = (--countOfVec2.upper_bound(item))->second; ` `        ``std::cout << result << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``std::vector<``int``> arr1 = { 1, 2, 3, 4, 7, 9 }; ` `    ``std::vector<``int``> arr2 = { 0, 1, 2, 1, 1, 4 }; ` ` `  `    ``countLessThanOrEqual(arr1, arr2); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program for each element in 1st  ` `# array count elements less than or equal to it  ` `# in 2nd array  ` ` `  `# Function for each element in 1st  ` `# array count elements less than or equal to it  ` `# in 2nd array  ` `def` `countLessThanOrEqual(vec1, vec2): ` `     `  `    ``countOfVec2 ``=` `{} ` `    ``for` `item ``in` `vec2: ` `        ``if` `item ``not` `in` `countOfVec2: ` `            ``countOfVec2[item] ``=` `0` `        ``countOfVec2[item] ``+``=` `1` `         `  `    ``prev ``=` `0` `    ``for` `pair ``in` `countOfVec2: ` `        ``countOfVec2[pair] ``+``=` `prev ` `        ``prev ``=` `countOfVec2[pair] ` `         `  `    ``val ``=` `list``(countOfVec2) ` `     `  `    ``# Traverse arr1[] and print result  ` `    ``for` `item ``in` `vec1: ` `        ``i ``=` `0` `        ``v ``=` `0` `        ``for` `i ``in` `range``(``len``(val)): ` `            ``if` `item < val[i]: ` `                ``v ``=` `i ` `                ``break` `        ``v ``-``=` `1` `        ``if` `v ``=``=` `-``1``: ` `            ``v ``=` `val[``-``1``] ` `        ``result ``=` `countOfVec2[v] ` `        ``print``(result, end ``=` `" "``) ` ` `  `# Driver code ` `arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``7``, ``9``] ` `arr2 ``=` `[``0``, ``1``, ``2``, ``1``, ``1``, ``4``] ` ` `  `countLessThanOrEqual(arr1, arr2) ` ` `  `# This code is contributed by shubhamsingh10 `

Output:

```4 5 5 6 6 6
```
• Complexity Analysis:

• Time Complexity: O(max + n ) where max is the maximum element of both arrays and n is the length of the array.
A single traversal of both the arrays and the prefix array is needed.
• Space Complexity: O(max) where max is the maximum element of both arrays.
A prefix array of size max is needed.

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