Sum of elements in an array whose difference with the mean of another array is less than k

Given two unsorted arrays arr1[] and arr2[]. Find the sum of elements from arr1[] whose difference with the mean of arr2[] is < k.

Examples:

Input: arr1[] = {1, 2, 3, 4, 7, 9}, arr2[] = {0, 1, 2, 1, 1, 4}, k = 2
Output: 6
Mean of 2nd array is 1.5.
Hence, 1, 2, 3 are the only elements
whose difference with mean is less than 2



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

Approach: Calculate the mean of the second array and then traverse the first array and calculate the sum of those elements whose absolute difference with mean is < k.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function for finding sum of elements
// whose diff with mean is not more than k
int findSumofEle(int arr1[], int m,
                 int arr2[], int n, int k)
{
    float arraySum = 0;
  
    // Find the mean of second array
    for (int i = 0; i < n; i++)
        arraySum += arr2[i];
    float mean = arraySum / n;
  
    // Find sum of elements from array1
    // whose difference with mean in not more than k
    int sumOfElements = 0;
    float difference;
  
    for (int i = 0; i < m; i++) {
        difference = arr1[i] - mean;
        if ((difference < 0) && (k > (-1) * difference)) {
            sumOfElements += arr1[i];
        }
        if ((difference >= 0) && (k > difference)) {
            sumOfElements += arr1[i];
        }
    }
  
    // Return result
    return sumOfElements;
}
  
// Driver code
int main()
{
    int arr1[] = { 1, 2, 3, 4, 7, 9 };
    int arr2[] = { 0, 1, 2, 1, 1, 4 };
    int k = 2;
    int m, n;
  
    m = sizeof(arr1) / sizeof(arr1[0]);
    n = sizeof(arr2) / sizeof(arr2[0]);
  
    cout << findSumofEle(arr1, m, arr2, n, k);
  
    return 0;
}

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Java














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// Java implementation of the approach  


class GFG 


{ 


      


// Function for finding sum of elements  


// whose diff with mean is not more than k  


static int findSumofEle(int []arr1, int m,  


                int []arr2, int n, int k)  


{  


    float arraySum = 0;  


  


    // Find the mean of second array  


    for (int i = 0; i < n; i++)  


        arraySum += arr2[i];  


    float mean = arraySum / n;  


  


    // Find sum of elements from array1  


    // whose difference with mean in not more than k  


    int sumOfElements = 0;  


    float difference = 0;  


  


    for (int i = 0; i < m; i++)  


    {  


        difference = arr1[i] - mean;  


        if ((difference < 0) && (k > (-1) * difference))  


        {  


            sumOfElements += arr1[i];  


        }  


        if ((difference >= 0) && (k > difference))  


        {  


            sumOfElements += arr1[i];  


        }  


    }  


  


    // Return result  


    return sumOfElements;  


}  


  


// 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 k = 2;  


  


    int m = arr1.length;  


    int n = arr2.length;  


  


    System.out.println(findSumofEle(arr1, m, arr2, n, k));  


}  


} 


  


// This code is contributed by mits 



















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Python3














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# Python3 implementation of the approach 


  


# Function for finding sum of elements 


# whose diff with mean is not more than k 


def findSumofEle(arr1, m, arr2, n, k): 


    arraySum = 0


  


    # Find the mean of second array 


    for i in range(n): 


        arraySum += arr2[i] 


    mean = arraySum / n 


  


    # Find sum of elements from array1 


    # whose difference with mean  


    # is not more than k 


    sumOfElements = 0


    difference = 0


  


    for i in range(m): 


  


        difference = arr1[i] - mean 


  


        if ((difference < 0) and (k > (-1) * difference)): 


            sumOfElements += arr1[i] 


  


        if ((difference >= 0) and (k > difference)): 


            sumOfElements += arr1[i] 


  


    # Return result 


    return sumOfElements 


  


# Driver code 


arr1 = [ 1, 2, 3, 4, 7, 9] 


arr2 = [ 0, 1, 2, 1, 1, 4] 


k = 2


  


m = len(arr1) 


n = len(arr2) 


  


print(findSumofEle(arr1, m, arr2, n, k)) 


  


# This code is contributed by mohit kumar 



















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C#














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// C# implementation of the approach  


using System;  


  


class GFG 


{ 


      


// Function for finding sum of elements  


// whose diff with mean is not more than k  


static int findSumofEle(int []arr1, int m,  


                int []arr2, int n, int k)  


{  


    float arraySum = 0;  


  


    // Find the mean of second array  


    for (int i = 0; i < n; i++)  


        arraySum += arr2[i];  


    float mean = arraySum / n;  


  


    // Find sum of elements from array1  


    // whose difference with mean in not more than k  


    int sumOfElements = 0;  


    float difference = 0;  


  


    for (int i = 0; i < m; i++)  


    {  


        difference = arr1[i] - mean;  


        if ((difference < 0) && (k > (-1) * difference))  


        {  


            sumOfElements += arr1[i];  


        }  


        if ((difference >= 0) && (k > difference))  


        {  


            sumOfElements += arr1[i];  


        }  


    }  


  


    // Return result  


    return sumOfElements;  


}  


  


// Driver code  


static void Main()  


{  


    int []arr1 = { 1, 2, 3, 4, 7, 9 };  


    int []arr2 = { 0, 1, 2, 1, 1, 4 };  


    int k = 2;  


  


    int m = arr1.Length;  


    int n = arr2.Length;  


  


    Console.WriteLine(findSumofEle(arr1, m, arr2, n, k));  


}  


} 


  


// This code is contributed by mits 



















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PHP














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<?php 


// PHP implementation of the approach  


  


// Function for finding sum of elements  


// whose diff with mean is not more than k  


function findSumofEle($arr1, $m, $arr2, $n, $k)  


{  


    $arraySum = 0;  


  


    // Find the mean of second array  


    for ($i = 0; $i < $n; $i++)  


        $arraySum += $arr2[$i];  


          


    $mean = $arraySum / $n;  


  


    // Find sum of elements from array1  


    // whose difference with mean  


    // is not more than k  


    $sumOfElements = 0;  


  


    for ($i = 0; $i < $m; $i++)  


    {  


        $difference = $arr1[$i] - $mean;  


        if (($difference < 0) &&  


            ($k > (-1) * $difference)) 


        {  


            $sumOfElements += $arr1[$i];  


        }  


        if (($difference >= 0) &&  


            ($k > $difference)) 


        {  


            $sumOfElements += $arr1[$i];  


        }  


    }  


  


    // Return result  


    return $sumOfElements;  


}  


  


// Driver code  


$arr1 = array( 1, 2, 3, 4, 7, 9 );  


$arr2 = array( 0, 1, 2, 1, 1, 4 );  


$k = 2;  


  


$m = count($arr1); 


$n = count($arr2); 


  


print(findSumofEle($arr1, $m,  


                   $arr2, $n, $k));  


  


// This code is contributed by Ryuga  


?> 



















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Output:


6






          
          
          

          

    

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