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++
// C++ implementation of the approach#include <iostream>using namespace std;// Function for finding sum of elements// whose diff with mean is not more than kint 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 codeint 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;} |
Java
// 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 |
Python3
# Python3 implementation of the approach# Function for finding sum of elements# whose diff with mean is not more than kdef 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 codearr1 = [ 1, 2, 3, 4, 7, 9]arr2 = [ 0, 1, 2, 1, 1, 4]k = 2m = len(arr1)n = len(arr2)print(findSumofEle(arr1, m, arr2, n, k))# This code is contributed by mohit kumar |
C#
// 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 |
PHP
<?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 ?> |
Javascript
<script>// Javascript implementation of the approach// Function for finding sum of elements// whose diff with mean is not more than kfunction findSumofEle(arr1, m, arr2, n, k){ var arraySum = 0; // Find the mean of second array for (var i = 0; i < n; i++) arraySum += arr2[i]; var mean = (arraySum / n); // Find sum of elements from array1 // whose difference with mean in not more than k var sumOfElements = 0; var difference; for (var 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 codevar arr1 = [ 1, 2, 3, 4, 7, 9 ];var arr2 = [ 0, 1, 2, 1, 1, 4 ];var k = 2;var m, n;m = arr1.length;n = arr2.length;document.write( findSumofEle(arr1, m, arr2, n, k));</script> |
Output:
6
Time Complexity: O(n + m)
Auxiliary Space: O(1)
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