# Sum of elements in 1st array such that number of elements less than or equal to them in 2nd array is maximum

Given two unsorted arrays arr1[] and arr2[], the task is to find the sum of elements of arr1[] such that number of elements less than or equal to them in arr2[] is maximum.

Examples:

Input: arr1[] = {1, 2, 3, 4, 7, 9}, arr2[] = {0, 1, 2, 1, 1, 4}
Output: 20
Below table shows the count of elements in arr2[] which are ≤ the elements of arr1[]

arr1[i] Count
1 4
2 5
3 5
4 6
7 6
9 6

Count for 4, 7 and 9 is maximum.
Hence, the resultant sum is 4 + 7 + 9 = 20.

Input:arr1[] = {5, 10, 2, 6, 1, 8, 6, 12}, arr2[] = {6, 5, 11, 4, 2, 3, 7}
Output: 12

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

Approach: Idea behind efficient solution for the above problem is to use hashing of second array and then finding the cumulative sum of hashed array. After that the count of elements in second array less than or equal to elements of 1st array can easily be calculated. This will give a frequency array which represents the count of elements in second array less than or equal to elements of 1st array from where the sum of elements of first array can be calculated corresponding to maximum frequency in frequency array.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MAX 100000 ` ` `  `// Function to return the required sum ` `int` `findSumofEle(``int` `arr1[], ``int` `m, ` `                 ``int` `arr2[], ``int` `n) ` `{ ` `    ``// Creating hash array initially ` `    ``// filled with zero ` `    ``int` `hash[MAX] = { 0 }; ` ` `  `    ``// Calculate the frequency ` `    ``// of elements of arr2[] ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``hash[arr2[i]]++; ` ` `  `    ``// Running sum of hash array ` `    ``// such that hash[i] will give count of ` `    ``// elements less than or equal to i in arr2[] ` `    ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``hash[i] = hash[i] + hash[i - 1]; ` ` `  `    ``// To store the maximum value of ` `    ``// the number of elements in arr2[] which are ` `    ``// smaller than or equal to some element of arr1[] ` `    ``int` `maximumFreq = 0; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``maximumFreq = max(maximumFreq, hash[arr1[i]]); ` ` `  `    ``// Calculate the sum of elements from arr1[] ` `    ``// corresponding to maximum frequency ` `    ``int` `sumOfElements = 0; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : 0; ` ` `  `    ``// Return the required sum ` `    ``return` `sumOfElements; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr1[] = { 2, 5, 6, 8 }; ` `    ``int` `arr2[] = { 4, 10 }; ` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]); ` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]); ` ` `  `    ``cout << findSumofEle(arr1, m, arr2, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = ``100000``; ` ` `  `    ``// Function to return the required sum ` `    ``static` `int` `findSumofEle(``int` `arr1[], ``int` `m, ` `                            ``int` `arr2[], ``int` `n) ` `    ``{ ` `        ``// Creating hash array initially ` `        ``// filled with zero ` `        ``int` `hash[] = ``new` `int``[MAX]; ` ` `  `        ``// Calculate the frequency ` `        ``// of elements of arr2[] ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``hash[arr2[i]]++; ` `        ``} ` ` `  `        ``// Running sum of hash array ` `        ``// such that hash[i] will give count of ` `        ``// elements less than or equal to i in arr2[] ` `        ``for` `(``int` `i = ``1``; i < MAX; i++) ` `        ``{ ` `            ``hash[i] = hash[i] + hash[i - ``1``]; ` `        ``} ` ` `  `        ``// To store the maximum value of ` `        ``// the number of elements in arr2[] which are ` `        ``// smaller than or equal to some element of arr1[] ` `        ``int` `maximumFreq = ``0``; ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``{ ` `            ``maximumFreq = Math.max(maximumFreq, hash[arr1[i]]); ` `        ``} ` ` `  `        ``// Calculate the sum of elements from arr1[] ` `        ``// corresponding to maximum frequency ` `        ``int` `sumOfElements = ``0``; ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``{ ` `            ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : ``0``; ` `        ``} ` ` `  `        ``// Return the required sum ` `        ``return` `sumOfElements; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr1[] = {``2``, ``5``, ``6``, ``8``}; ` `        ``int` `arr2[] = {``4``, ``10``}; ` `        ``int` `m = arr1.length; ` `        ``int` `n = arr2.length; ` ` `  `        ``System.out.println(findSumofEle(arr1, m, arr2, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

 `# Python 3 implementation of the approach ` `MAX` `=` `100000` ` `  `# Function to return the required sum ` `def` `findSumofEle(arr1, m, arr2, n): ` `     `  `    ``# Creating hash array initially ` `    ``# filled with zero ` `    ``hash` `=` `[``0` `for` `i ``in` `range``(``MAX``)] ` ` `  `    ``# Calculate the frequency ` `    ``# of elements of arr2[] ` `    ``for` `i ``in` `range``(n): ` `        ``hash``[arr2[i]] ``+``=` `1` ` `  `    ``# Running sum of hash array ` `    ``# such that hash[i] will give count of ` `    ``# elements less than or equal to i in arr2[] ` `    ``for` `i ``in` `range``(``1``, ``MAX``, ``1``): ` `        ``hash``[i] ``=` `hash``[i] ``+` `hash``[i ``-` `1``] ` ` `  `    ``# To store the maximum value of ` `    ``# the number of elements in arr2[]  ` `    ``# which are smaller than or equal  ` `    ``# to some element of arr1[] ` `    ``maximumFreq ``=` `0` `    ``for` `i ``in` `range``(m): ` `        ``maximumFreq ``=` `max``(maximumFreq,  ` `                          ``hash``[arr1[i]]) ` ` `  `    ``# Calculate the sum of elements from arr1[] ` `    ``# corresponding to maximum frequency ` `    ``sumOfElements ``=` `0` `    ``for` `i ``in` `range``(m): ` `        ``if` `(maximumFreq ``=``=` `hash``[arr1[i]]): ` `            ``sumOfElements ``+``=` `arr1[i] ` ` `  `    ``# Return the required sum ` `    ``return` `sumOfElements ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr1 ``=` `[``2``, ``5``, ``6``, ``8``] ` `    ``arr2 ``=` `[``4``, ``10``] ` `    ``m ``=` `len``(arr1) ` `    ``n ``=` `len``(arr2)  ` `    ``print``(findSumofEle(arr1, m, arr2, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = 100000; ` ` `  `    ``// Function to return the required sum ` `    ``static` `int` `findSumofEle(``int``[] arr1, ``int` `m, ` `                            ``int``[] arr2, ``int` `n) ` `    ``{ ` `        ``// Creating hash array initially ` `        ``// filled with zero ` `        ``int``[] hash = ``new` `int``[MAX]; ` ` `  `        ``// Calculate the frequency ` `        ``// of elements of arr2[] ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``hash[arr2[i]]++; ` `        ``} ` ` `  `        ``// Running sum of hash array ` `        ``// such that hash[i] will give count of ` `        ``// elements less than or equal to i in arr2[] ` `        ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``{ ` `            ``hash[i] = hash[i] + hash[i - 1]; ` `        ``} ` ` `  `        ``// To store the maximum value of ` `        ``// the number of elements in arr2[] which are ` `        ``// smaller than or equal to some element of arr1[] ` `        ``int` `maximumFreq = 0; ` `        ``for` `(``int` `i = 0; i < m; i++) ` `        ``{ ` `            ``maximumFreq = Math.Max(maximumFreq, hash[arr1[i]]); ` `        ``} ` ` `  `        ``// Calculate the sum of elements from arr1[] ` `        ``// corresponding to maximum frequency ` `        ``int` `sumOfElements = 0; ` `        ``for` `(``int` `i = 0; i < m; i++) ` `        ``{ ` `            ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : 0; ` `        ``} ` ` `  `        ``// Return the required sum ` `        ``return` `sumOfElements; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int``[] arr1 = {2, 5, 6, 8}; ` `        ``int``[] arr2 = {4, 10}; ` `        ``int` `m = arr1.Length; ` `        ``int` `n = arr2.Length; ` ` `  `        ``Console.WriteLine(findSumofEle(arr1, m, arr2, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by Code_Mech. `

 ` `

Output:
```19
```

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Article Tags :
The text passage describes a problem involving two unsorted arrays and the task of finding the sum of elements from the first array such that the number of elements less than or equal to them in the second array is maximum. It includes examples and a detailed explanation of the approach, followed by code implementations in C++, Java, Python, and C#.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Sum of elements in 1st array such that number of elements less than or equal to them in 2nd array is maximum

Given two unsorted arrays arr1[] and arr2[], the task is to find the sum of elements of arr1[] such that number of elements less than or equal to them in arr2[] is maximum.

Examples:

Input: arr1[] = {1, 2, 3, 4, 7, 9}, arr2[] = {0, 1, 2, 1, 1, 4}
Output: 20
Below table shows the count of elements in arr2[] which are ≤ the elements of arr1[]

arr1[i] Count
1 4
2 5
3 5
4 6
7 6
9 6

Count for 4, 7 and 9 is maximum.
Hence, the resultant sum is 4 + 7 + 9 = 20.

Input:arr1[] = {5, 10, 2, 6, 1, 8, 6, 12}, arr2[] = {6, 5, 11, 4, 2, 3, 7}
Output: 12

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

Approach: Idea behind efficient solution for the above problem is to use hashing of second array and then finding the cumulative sum of hashed array. After that the count of elements in second array less than or equal to elements of 1st array can easily be calculated. This will give a frequency array which represents the count of elements in second array less than or equal to elements of 1st array from where the sum of elements of first array can be calculated corresponding to maximum frequency in frequency array.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MAX 100000 ` ` `  `// Function to return the required sum ` `int` `findSumofEle(``int` `arr1[], ``int` `m, ` `                 ``int` `arr2[], ``int` `n) ` `{ ` `    ``// Creating hash array initially ` `    ``// filled with zero ` `    ``int` `hash[MAX] = { 0 }; ` ` `  `    ``// Calculate the frequency ` `    ``// of elements of arr2[] ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``hash[arr2[i]]++; ` ` `  `    ``// Running sum of hash array ` `    ``// such that hash[i] will give count of ` `    ``// elements less than or equal to i in arr2[] ` `    ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``hash[i] = hash[i] + hash[i - 1]; ` ` `  `    ``// To store the maximum value of ` `    ``// the number of elements in arr2[] which are ` `    ``// smaller than or equal to some element of arr1[] ` `    ``int` `maximumFreq = 0; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``maximumFreq = max(maximumFreq, hash[arr1[i]]); ` ` `  `    ``// Calculate the sum of elements from arr1[] ` `    ``// corresponding to maximum frequency ` `    ``int` `sumOfElements = 0; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : 0; ` ` `  `    ``// Return the required sum ` `    ``return` `sumOfElements; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr1[] = { 2, 5, 6, 8 }; ` `    ``int` `arr2[] = { 4, 10 }; ` `    ``int` `m = ``sizeof``(arr1) / ``sizeof``(arr1[0]); ` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2[0]); ` ` `  `    ``cout << findSumofEle(arr1, m, arr2, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = ``100000``; ` ` `  `    ``// Function to return the required sum ` `    ``static` `int` `findSumofEle(``int` `arr1[], ``int` `m, ` `                            ``int` `arr2[], ``int` `n) ` `    ``{ ` `        ``// Creating hash array initially ` `        ``// filled with zero ` `        ``int` `hash[] = ``new` `int``[MAX]; ` ` `  `        ``// Calculate the frequency ` `        ``// of elements of arr2[] ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``hash[arr2[i]]++; ` `        ``} ` ` `  `        ``// Running sum of hash array ` `        ``// such that hash[i] will give count of ` `        ``// elements less than or equal to i in arr2[] ` `        ``for` `(``int` `i = ``1``; i < MAX; i++) ` `        ``{ ` `            ``hash[i] = hash[i] + hash[i - ``1``]; ` `        ``} ` ` `  `        ``// To store the maximum value of ` `        ``// the number of elements in arr2[] which are ` `        ``// smaller than or equal to some element of arr1[] ` `        ``int` `maximumFreq = ``0``; ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``{ ` `            ``maximumFreq = Math.max(maximumFreq, hash[arr1[i]]); ` `        ``} ` ` `  `        ``// Calculate the sum of elements from arr1[] ` `        ``// corresponding to maximum frequency ` `        ``int` `sumOfElements = ``0``; ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``{ ` `            ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : ``0``; ` `        ``} ` ` `  `        ``// Return the required sum ` `        ``return` `sumOfElements; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr1[] = {``2``, ``5``, ``6``, ``8``}; ` `        ``int` `arr2[] = {``4``, ``10``}; ` `        ``int` `m = arr1.length; ` `        ``int` `n = arr2.length; ` ` `  `        ``System.out.println(findSumofEle(arr1, m, arr2, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

 `# Python 3 implementation of the approach ` `MAX` `=` `100000` ` `  `# Function to return the required sum ` `def` `findSumofEle(arr1, m, arr2, n): ` `     `  `    ``# Creating hash array initially ` `    ``# filled with zero ` `    ``hash` `=` `[``0` `for` `i ``in` `range``(``MAX``)] ` ` `  `    ``# Calculate the frequency ` `    ``# of elements of arr2[] ` `    ``for` `i ``in` `range``(n): ` `        ``hash``[arr2[i]] ``+``=` `1` ` `  `    ``# Running sum of hash array ` `    ``# such that hash[i] will give count of ` `    ``# elements less than or equal to i in arr2[] ` `    ``for` `i ``in` `range``(``1``, ``MAX``, ``1``): ` `        ``hash``[i] ``=` `hash``[i] ``+` `hash``[i ``-` `1``] ` ` `  `    ``# To store the maximum value of ` `    ``# the number of elements in arr2[]  ` `    ``# which are smaller than or equal  ` `    ``# to some element of arr1[] ` `    ``maximumFreq ``=` `0` `    ``for` `i ``in` `range``(m): ` `        ``maximumFreq ``=` `max``(maximumFreq,  ` `                          ``hash``[arr1[i]]) ` ` `  `    ``# Calculate the sum of elements from arr1[] ` `    ``# corresponding to maximum frequency ` `    ``sumOfElements ``=` `0` `    ``for` `i ``in` `range``(m): ` `        ``if` `(maximumFreq ``=``=` `hash``[arr1[i]]): ` `            ``sumOfElements ``+``=` `arr1[i] ` ` `  `    ``# Return the required sum ` `    ``return` `sumOfElements ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr1 ``=` `[``2``, ``5``, ``6``, ``8``] ` `    ``arr2 ``=` `[``4``, ``10``] ` `    ``m ``=` `len``(arr1) ` `    ``n ``=` `len``(arr2)  ` `    ``print``(findSumofEle(arr1, m, arr2, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = 100000; ` ` `  `    ``// Function to return the required sum ` `    ``static` `int` `findSumofEle(``int``[] arr1, ``int` `m, ` `                            ``int``[] arr2, ``int` `n) ` `    ``{ ` `        ``// Creating hash array initially ` `        ``// filled with zero ` `        ``int``[] hash = ``new` `int``[MAX]; ` ` `  `        ``// Calculate the frequency ` `        ``// of elements of arr2[] ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``hash[arr2[i]]++; ` `        ``} ` ` `  `        ``// Running sum of hash array ` `        ``// such that hash[i] will give count of ` `        ``// elements less than or equal to i in arr2[] ` `        ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``{ ` `            ``hash[i] = hash[i] + hash[i - 1]; ` `        ``} ` ` `  `        ``// To store the maximum value of ` `        ``// the number of elements in arr2[] which are ` `        ``// smaller than or equal to some element of arr1[] ` `        ``int` `maximumFreq = 0; ` `        ``for` `(``int` `i = 0; i < m; i++) ` `        ``{ ` `            ``maximumFreq = Math.Max(maximumFreq, hash[arr1[i]]); ` `        ``} ` ` `  `        ``// Calculate the sum of elements from arr1[] ` `        ``// corresponding to maximum frequency ` `        ``int` `sumOfElements = 0; ` `        ``for` `(``int` `i = 0; i < m; i++) ` `        ``{ ` `            ``sumOfElements += (maximumFreq == hash[arr1[i]]) ? arr1[i] : 0; ` `        ``} ` ` `  `        ``// Return the required sum ` `        ``return` `sumOfElements; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int``[] arr1 = {2, 5, 6, 8}; ` `        ``int``[] arr2 = {4, 10}; ` `        ``int` `m = arr1.Length; ` `        ``int` `n = arr2.Length; ` ` `  `        ``Console.WriteLine(findSumofEle(arr1, m, arr2, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by Code_Mech. `

 ` `

Output:
```19
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Article Tags :
user
The text passage describes a problem involving two unsorted arrays and the task of finding the sum of elements from the first array such that the number of elements less than or equal to them in the second array is maximum. It includes examples and a detailed explanation of the approach, followed by code implementations in C++, Java, Python, and C#.

TYPE: SINGLE_QUESTION_WITH_ANSWER