Count array elements that divide the sum of all other elements
Given an array arr[], the task is to count the number of elements from the array which divide the sum of all other elements.
Examples:
Input: arr[] = {3, 10, 4, 6, 7}
Output: 3
3 divides (10 + 4 + 6 + 7) i.e. 27
10 divides (3 + 4 + 6 + 7) i.e. 20
6 divides (3 + 10 + 4 + 7) i.e. 24Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 2
Naive Approach: Run two-loop from 0 to N, calculate the sum of all elements except the current element and if this element divides that sum, then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the count// of the required numbersint countNum(int N, int arr[]){ // To store the count of required numbers int count = 0; for (int i = 0; i < N; i++) { // Initialize sum to 0 int sum = 0; for (int j = 0; j < N; j++) { // If current element and the // chosen element are same if (i == j) continue; // Add all other numbers of array else sum += arr[j]; } // If sum is divisible by the chosen element if (sum % arr[i] == 0) count++; } // Return the count return count;}// Driver codeint main(){ int arr[] = { 3, 10, 4, 6, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countNum(n, arr); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the count// of the required numbersstatic int countNum(int N, int arr[]){ // To store the count of required numbers int count = 0; for (int i = 0; i < N; i++) { // Initialize sum to 0 int sum = 0; for (int j = 0; j < N; j++) { // If current element and the // chosen element are same if (i == j) continue; // Add all other numbers of array else sum += arr[j]; } // If sum is divisible by the chosen element if (sum % arr[i] == 0) count++; } // Return the count return count;}// Driver codepublic static void main(String[] args){ int arr[] = { 3, 10, 4, 6, 7 }; int n = arr.length; System.out.println(countNum(n, arr));}}// This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approach# Function to return the count# of the required numbersdef countNum(N, arr): # To store the count of # required numbers count = 0 for i in range(N): # Initialize sum to 0 Sum = 0 for j in range(N): # If current element and the # chosen element are same if (i == j): continue # Add all other numbers of array else: Sum += arr[j] # If Sum is divisible by the # chosen element if (Sum % arr[i] == 0): count += 1 # Return the count return count# Driver codearr = [3, 10, 4, 6, 7]n = len(arr)print(countNum(n, arr))# This code is contributed# by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the count// of the required numbersstatic int countNum(int N, int []arr){ // To store the count of required numbers int count = 0; for (int i = 0; i < N; i++) { // Initialize sum to 0 int sum = 0; for (int j = 0; j < N; j++) { // If current element and the // chosen element are same if (i == j) continue; // Add all other numbers of array else sum += arr[j]; } // If sum is divisible by the chosen element if (sum % arr[i] == 0) count++; } // Return the count return count;}// Driver codepublic static void Main(){ int []arr = { 3, 10, 4, 6, 7 }; int n = arr.Length; Console.WriteLine(countNum(n, arr));}}// This code is contributed by inder_verma.. |
PHP
<?php// Php implementation of the approach// Function to return the count// of the required numbersfunction countNum($N, $arr){// To store the count of// required numbers$count = 0;for ($i=0;$i<$N;$i++) {// Initialize sum to 0 $Sum = 0; for ($j = 0; $j < $N; $j++) { // If current element and the // chosen element are same if ($i == $j) continue; // Add all other numbers of array else $Sum += $arr[$j]; } // If Sum is divisible by the // chosen element if ($Sum % $arr[$i] == 0) $count += 1;}// Return the count return $count;}// Driver code$arr = array(3, 10, 4, 6, 7);$n = count($arr);echo countNum($n, $arr);// This code is contributed// by Srathore?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of the required numbers function countNum(N, arr) { // To store the count of required numbers let count = 0; for (let i = 0; i < N; i++) { // Initialize sum to 0 let sum = 0; for (let j = 0; j < N; j++) { // If current element and the // chosen element are same if (i == j) continue; // Add all other numbers of array else sum += arr[j]; } // If sum is divisible by the chosen element if (sum % arr[i] == 0) count++; } // Return the count return count; } let arr = [ 3, 10, 4, 6, 7 ]; let n = arr.length; document.write(countNum(n, arr)); // This code is contributed by vaibhavrabadiya117</script> |
Output:
3
Time Complexity: O(N2)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach: Run a single loop from 0 to N, calculate the sum of all the elements. Now run another loop from 0 to N and if (sum – arr[i]) % arr[i] = 0 then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the count// of the required numbersint countNum(int N, int arr[]){ // Initialize sum and count to 0 int sum = 0, count = 0; // Calculate sum of all // the array elements for (int i = 0; i < N; i++) sum += arr[i]; for (int i = 0; i < N; i++) // If current element satisfies the condition if ((sum - arr[i]) % arr[i] == 0) count++; // Return the count of required elements return count;}// Driver codeint main(){ int arr[] = { 3, 10, 4, 6, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countNum(n, arr); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the count // of the required numbers static int countNum(int N, int arr[]) { // Initialize sum and count to 0 int sum = 0, count = 0; // Calculate sum of all // the array elements for (int i = 0; i < N; i++) { sum += arr[i]; } // If current element satisfies the condition for (int i = 0; i < N; i++) { if ((sum - arr[i]) % arr[i] == 0) { count++; } } // Return the count of required elements return count; } // Driver code public static void main(String[] args) { int arr[] = {3, 10, 4, 6, 7}; int n = arr.length; System.out.println(countNum(n, arr)); }}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function to return the count# of the required numbersdef countNum(N, arr): # Initialize Sum and count to 0 Sum, count = 0, 0 # Calculate Sum of all the # array elements for i in range(N): Sum += arr[i] for i in range(N): # If current element satisfies # the condition if ((Sum - arr[i]) % arr[i] == 0): count += 1 # Return the count of required # elements return count# Driver codearr = [ 3, 10, 4, 6, 7 ]n = len(arr)print(countNum(n, arr))# This code is contributed # by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG { // Function to return the count // of the required numbers static int countNum(int N, int []arr) { // Initialize sum and count to 0 int sum = 0, count = 0; // Calculate sum of all // the array elements for (int i = 0; i < N; i++) { sum += arr[i]; } // If current element satisfies the condition for (int i = 0; i < N; i++) { if ((sum - arr[i]) % arr[i] == 0) { count++; } } // Return the count of required elements return count; } // Driver code public static void Main() { int []arr = {3, 10, 4, 6, 7}; int n = arr.Length; Console.WriteLine(countNum(n, arr)); }}/* This code contributed by PrinciRaj1992 */ |
PHP
<?php// PHP implementation of the approach// Function to return the count// of the required numbersfunction countNum($N, $arr){ // Initialize sum and count to 0 $sum = 0; $count = 0; // Calculate sum of all // the array elements for ($i = 0; $i < $N; $i++) $sum += $arr[$i]; for ($i = 0; $i < $N; $i++) // If current element satisfies // the condition if (($sum - $arr[$i]) % $arr[$i] == 0) $count++; // Return the count of required elements return $count;}// Driver code$arr = array(3, 10, 4, 6, 7);$n = count($arr);echo countNum($n, $arr);// This code contributed by Rajput-Ji ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of the required numbers function countNum(N, arr) { // Initialize sum and count to 0 let sum = 0, count = 0; // Calculate sum of all // the array elements for (let i = 0; i < N; i++) { sum += arr[i]; } // If current element satisfies the condition for (let i = 0; i < N; i++) { if ((sum - arr[i]) % arr[i] == 0) { count++; } } // Return the count of required elements return count; } let arr = [3, 10, 4, 6, 7]; let n = arr.length; document.write(countNum(n, arr)); </script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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