Given an array containing N elements and a number K. The task is to find the sum of all such elements which are divisible by K.
Examples:
Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
K = 3
Output : 30
Explanation: As 15, 9, 6 are divisible by 3. So, sum of elements divisible by K = 15 + 9 + 6 = 30.
Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
K = 2
Output : 20
The idea is to traverse the array and check the elements one by one. If an element is divisible by K then add that element’s value with the sum so far and continue this process while the end of the array reached.
Below is the implementation of the above approach:
C++
// C++ program to find sum of all the elements// in an array divisible by a given number K#include <iostream>using namespace std;// Function to find sum of all the elements// in an array divisible by a given number Kint findSum(int arr[], int n, int k){ int sum = 0; // Traverse the array for (int i = 0; i < n; i++) { // If current element is divisible by k // add it to sum if (arr[i] % k == 0) { sum += arr[i]; } } // Return calculated sum return sum;}// Driver codeint main(){ int arr[] = { 15, 16, 10, 9, 6, 7, 17 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; cout << findSum(arr, n, k); return 0;} |
C
// C program to find sum of all the elements// in an array divisible by a given number K#include <stdio.h>// Function to find sum of all the elements// in an array divisible by a given number Kint findSum(int arr[], int n, int k){ int sum = 0; // Traverse the array for (int i = 0; i < n; i++) { // If current element is divisible by k // add it to sum if (arr[i] % k == 0) { sum += arr[i]; } } // Return calculated sum return sum;}// Driver codeint main(){ int arr[] = { 15, 16, 10, 9, 6, 7, 17 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; printf("%d",findSum(arr, n, k)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java program to find sum of all the elements// in an array divisible by a given number Kimport java.io.*;class GFG {// Function to find sum of all the elements// in an array divisible by a given number Kstatic int findSum(int arr[], int n, int k){ int sum = 0; // Traverse the array for (int i = 0; i < n; i++) { // If current element is divisible by k // add it to sum if (arr[i] % k == 0) { sum += arr[i]; } } // Return calculated sum return sum;} // Driver code public static void main (String[] args) { int arr[] = { 15, 16, 10, 9, 6, 7, 17 }; int n = arr.length; int k = 3; System.out.println( findSum(arr, n, k)); }}// this code is contributed by anuj_67.. |
Python3
# Python3 program to find sum of # all the elements in an array# divisible by a given number K # Function to find sum of all # the elements in an array # divisible by a given number Kdef findSum(arr, n, k) : sum = 0 # Traverse the array for i in range(n) : # If current element is divisible # by k add it to sum if arr[i] % k == 0 : sum += arr[i] # Return calculated sum return sum# Driver codeif __name__ == "__main__" : arr = [ 15, 16, 10, 9, 6, 7, 17] n = len(arr) k = 3 print(findSum(arr, n, k))# This code is contributed by ANKITRAI1 |
C#
// C# program to find sum of all the elements// in an array divisible by a given number Kusing System;public class GFG{// Function to find sum of all the elements// in an array divisible by a given number Kstatic int findSum(int []arr, int n, int k){ int sum = 0; // Traverse the array for (int i = 0; i < n; i++) { // If current element is divisible by k // add it to sum if (arr[i] % k == 0) { sum += arr[i]; } } // Return calculated sum return sum;} // Driver code static public void Main (){ int []arr = { 15, 16, 10, 9, 6, 7, 17 }; int n = arr.Length; int k = 3; Console.WriteLine( findSum(arr, n, k)); }}//This code is contributed by anuj_67.. |
PHP
<?php// PHP program to find sum of all // the elements in an array divisible// by a given number K// Function to find sum of all // the elements in an array // divisible by a given number Kfunction findSum($arr, $n, $k){ $sum = 0; // Traverse the array for ($i = 0; $i < $n; $i++) { // If current element is divisible // by k add it to sum if ($arr[$i] % $k == 0) { $sum += $arr[$i]; } } // Return calculated sum return $sum;}// Driver code$arr = array(15, 16, 10, 9, 6, 7, 17);$n = sizeof($arr);$k = 3;echo findSum($arr, $n, $k);// This code is contributed // by Akanksha Rai(Abby_akku) |
Javascript
<script> // Javascript program to find sum of all the elements // in an array divisible by a given number K // Function to find sum of all the elements // in an array divisible by a given number K function findSum(arr, n, k) { let sum = 0; // Traverse the array for (let i = 0; i < n; i++) { // If current element is divisible by k // add it to sum if (arr[i] % k == 0) { sum += arr[i]; } } // Return calculated sum return sum; } let arr = [ 15, 16, 10, 9, 6, 7, 17 ]; let n = arr.length; let k = 3; document.write(findSum(arr, n, k));</script> |
Output
30
complexity Analysis:
- Time Complexity: O(N), where N is the number of elements in the array.
- Auxiliary Space: O(1)
Using Recursion :
Start traversing the array using recursion and check if the current element is divisible by K. If yes then update the answer. Print the answer when all the elements get traversed.
Method 1 : (Pass By Reference)
Implementation :
- Make a variable sum and pass that address of that variable
- We passing address of sum variable so that no change occur in it after every another function call.
- We just increment our pointer and make another function call.
- Add the value at pointer into our sum and make another function call until the base condition hits.
C++
#include <iostream>using namespace std;void solve(int arr[], int k, int i, int& ans){ if (i < 0) { // boundry condition to stop the recurion return; } if (arr[i] % k == 0) { // current element is divisible // so add that element to our ans ans += arr[i]; } solve(arr, k, i - 1, ans); // further iterate array using recursion}int main(){ int n = 6; int arr[n] = { 2, 6, 7, 12, 14, 18 }; int k = 2; int sum = 0; solve(arr, k, n - 1, sum); cout << sum << " "; return 0;}//this code is contributed by bhardwajji |
Java
import java.util.*;public class Main { // function to recursively compute the sum of elements in arr that are divisible by k public static void solve(int arr[], int k, int i, int[] ans) { if (i < 0) { // base case to stop recursion return; } if (arr[i] % k == 0) { // current element is divisible by k ans[0] += arr[i]; // add current element to the sum } solve(arr, k, i - 1, ans); // recursively call the function with the previous index } public static void main(String[] args) { int n = 6; int arr[] = {2, 6, 7, 12, 14, 18}; int k = 2; int[] sum = new int[1]; // store the sum in an array to avoid pass-by-value solve(arr, k, n - 1, sum); System.out.print(sum[0] + " "); }} |
Python3
def solve(arr, k, i, ans): # Base case to stop recursion if i < 0: return # If the current element is divisible by k, add it to the sum if arr[i] % k == 0: ans[0] += arr[i] # Recursively call the function with the previous index solve(arr, k, i - 1, ans)# Main functionif __name__ == '__main__': n = 6 arr = [2, 6, 7, 12, 14, 18] k = 2 # Store the sum in an array to avoid pass-by-value sum_arr = [0] # Call the recursive function solve(arr, k, n - 1, sum_arr) # Print the sum print(sum_arr[0], end=' ') |
C#
using System;public class Program{ public static void Solve(int[] arr, int k, int i, ref int ans) { if (i < 0) // boundry condition to stop the recursion return; if (arr[i] % k == 0) // current element is divisible { // so add that element to our ans ans += arr[i]; } Solve(arr, k, i - 1, ref ans); // further iterate array using recursion } public static void Main() { int n = 6; int[] arr = { 2, 6, 7, 12, 14, 18 }; int k = 2; int sum = 0; Solve(arr, k, n - 1, ref sum); Console.WriteLine(sum); }} |
Javascript
function solve(arr, k, i, ans) { // Base case to stop recursion if (i < 0) { return; } // If the current element is divisible by k, add it to the sum if (arr[i] % k === 0) { ans[0] += arr[i]; } // Recursively call the function with the previous index solve(arr, k, i - 1, ans);}// Main functionconst n = 6;const arr = [2, 6, 7, 12, 14, 18];const k = 2;// Store the sum in an array to avoid pass-by-valueconst sum_arr = [0];// Call the recursive functionsolve(arr, k, n - 1, sum_arr);// Print the sumconsole.log(sum_arr[0]); |
Output
52
Method 2 : (Pass By Value)
Implementation :
- Take a sum variable and only pass its value into the function solve.
- Function will return int value and we use that value to find our ans.
- If number is divisible by K then return the number ( arr[i] ) and if not divisible then return 0.
C++
#include <iostream>using namespace std;int solve(int arr[], int k, int i){ // returning the sum of elements divisible by k if (i < 0) { return 0; } if (arr[i] % k == 0) { return arr[i] + solve(arr, k, i - 1); // add current element and // further call the recursion } else { return 0 + solve(arr, k, i - 1); // nothing to add so just // call the futher recursion }}int main(){ int n = 6; int arr[n] = { 2, 6, 7, 12, 14, 18 }; int k = 2; int sum = solve(arr, k, n - 1); cout << sum << " "; return 0;}// this code is contributed by bhardwajji |
Java
/*package whatever //do not write package name here */import java.util.*;public class Main { // Function to calculate the sum of array elements that // are divisible by k static int solve(int[] arr, int k, int i) { // Base case: If the index is less than zero, return // zero if (i < 0) { return 0; } // If the current element is divisible by k, add it // to the sum and call the function recursively if (arr[i] % k == 0) { return arr[i] + solve(arr, k, i - 1); } // If the current element is not divisible by k, // just call the function recursively without adding // anything else { return 0 + solve(arr, k, i - 1); } } // Main function public static void main(String[] args) { int n = 6; int[] arr = { 2, 6, 7, 12, 14, 18 }; int k = 2; // Calculate the sum of array elements that are // divisible by k int sum = solve(arr, k, n - 1); // Print the sum System.out.println(sum); }} |
Python3
# This function returns the sum of elements divisible by kdef solve(arr, k, i): if i < 0: return 0 if arr[i] % k == 0: # add current element and further call the recursion return arr[i] + solve(arr, k, i - 1) else: # nothing to add so just call the further recursion return 0 + solve(arr, k, i - 1)n = 6arr = [2, 6, 7, 12, 14, 18]k = 2sum = solve(arr, k, n - 1)print(sum) |
C#
using System;public class MainClass { public static int Solve(int[] arr, int k, int i) { // returning the sum of elements divisible by k if (i < 0) { return 0; } if (arr[i] % k == 0) { // add current element and further call the // recursion return arr[i] + Solve(arr, k, i - 1); } else { // nothing to add so just call the futher // recursion return 0 + Solve(arr, k, i - 1); } } public static void Main() { int n = 6; int[] arr = { 2, 6, 7, 12, 14, 18 }; int k = 2; int sum = Solve(arr, k, n - 1); Console.WriteLine(sum); }} |
Javascript
// JavaScript program to find the sum of array elements// divisible by a given number kfunction solve(arr, k, i) {// returning the sum of elements divisible by kif (i < 0) {return 0;}if (arr[i] % k == 0) {return (arr[i] +solve(arr, k, i - 1) // add current element and further call the recursion);} else {return 0 + solve(arr, k, i - 1); // nothing to add so just call the further recursion}}// Driver codeconst arr = [2, 6, 7, 12, 14, 18];const k = 2;const sum = solve(arr, k, arr.length - 1);console.log(sum); |
Output
52
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
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