Given an array arr[] of size N, the task is to count the number of array elements whose sum of digits is equal to K.
Examples:
Input: arr[] = {23, 54, 87, 29, 92, 62}, K = 11
Output: 2
Explanation:
29 = 2 + 9 = 11
92 = 9 + 2 = 11Input: arr[]= {11, 04, 57, 99, 98, 32}, K = 18
Output: 1
Approach: Follow the steps below to solve the problem:
- Initialize a variable, say N, to store the size of the array.
- Initialize a variable, say count, to store the elements having sum of digits equal to K.
- Declare a function, sumOfDigits() to calculate the sum of digits of a number.
- Traverse the array arr[] and for each array element, check if the sum of digits is equal to K or not. If found to be true, then increment count by 1.
- Print the value of count as the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the// sum of digits of the number Nint sumOfDigits(int N){ // Stores the sum of digits int sum = 0; while (N != 0) { sum += N % 10; N /= 10; } // Return the sum return sum;}// Function to count array elementsint elementsHavingDigitSumK(int arr[], int N, int K){ // Store the count of array // elements having sum of digits K int count = 0; // Traverse the array for (int i = 0; i < N; ++i) { // If sum of digits is equal to K if (sumOfDigits(arr[i]) == K) { // Increment the count count++; } } // Print the count cout << count;}// Driver Codeint main(){ // Given array int arr[] = { 23, 54, 87, 29, 92, 62 }; // Given value of K int K = 11; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Function call to count array elements // having sum of digits equal to K elementsHavingDigitSumK(arr, N, K); return 0;} |
Java
// Java program for the above approach public class GFG{ // Function to calculate the // sum of digits of the number N static int sumOfDigits(int N) { // Stores the sum of digits int sum = 0; while (N != 0) { sum += N % 10; N /= 10; } // Return the sum return sum; } // Function to count array elements static void elementsHavingDigitSumK(int[] arr, int N, int K) { // Store the count of array // elements having sum of digits K int count = 0; // Traverse the array for (int i = 0; i < N; ++i) { // If sum of digits is equal to K if (sumOfDigits(arr[i]) == K) { // Increment the count count++; } } // Print the count System.out.println(count); } // Driver code public static void main(String args[]) { // Given array int[] arr = { 23, 54, 87, 29, 92, 62 }; // Given value of K int K = 11; // Size of the array int N = arr.length; // Function call to count array elements // having sum of digits equal to K elementsHavingDigitSumK(arr, N, K); }}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to calculate the# sum of digits of the number Ndef sumOfDigits(N) : # Stores the sum of digits sum = 0 while (N != 0) : sum += N % 10 N //= 10 # Return the sum return sum# Function to count array elementsdef elementsHavingDigitSumK(arr, N, K) : # Store the count of array # elements having sum of digits K count = 0 # Traverse the array for i in range(N): # If sum of digits is equal to K if (sumOfDigits(arr[i]) == K) : # Increment the count count += 1 # Print the count print(count)# Driver Code# Given arrayarr = [ 23, 54, 87, 29, 92, 62 ]# Given value of KK = 11# Size of the arrayN = len(arr) # Function call to count array elements# having sum of digits equal to KelementsHavingDigitSumK(arr, N, K)# This code is contributed by souravghosh0416. |
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG { // Function to calculate the // sum of digits of the number N static int sumOfDigits(int N) { // Stores the sum of digits int sum = 0; while (N != 0) { sum += N % 10; N /= 10; } // Return the sum return sum; } // Function to count array elements static void elementsHavingDigitSumK(int[] arr, int N, int K) { // Store the count of array // elements having sum of digits K int count = 0; // Traverse the array for (int i = 0; i < N; ++i) { // If sum of digits is equal to K if (sumOfDigits(arr[i]) == K) { // Increment the count count++; } } // Print the count Console.WriteLine(count); } // Driver code static void Main() { // Given array int[] arr = { 23, 54, 87, 29, 92, 62 }; // Given value of K int K = 11; // Size of the array int N = arr.Length; // Function call to count array elements // having sum of digits equal to K elementsHavingDigitSumK(arr, N, K); }}// This code is contributed by divyeshrabadiya07. |
Javascript
<script> // JavaScript program for the above approach // Function to calculate the // sum of digits of the number N function sumOfDigits(N) { // Stores the sum of digits let sum = 0; while (N != 0) { sum += N % 10; N = parseInt(N / 10, 10); } // Return the sum return sum; } // Function to count array elements function elementsHavingDigitSumK(arr, N, K) { // Store the count of array // elements having sum of digits K let count = 0; // Traverse the array for (let i = 0; i < N; ++i) { // If sum of digits is equal to K if (sumOfDigits(arr[i]) == K) { // Increment the count count++; } } // Print the count document.write(count); } // Given array let arr = [ 23, 54, 87, 29, 92, 62 ]; // Given value of K let K = 11; // Size of the array let N = arr.length; // Function call to count array elements // having sum of digits equal to K elementsHavingDigitSumK(arr, N, K); </script> |
Output:
2
Time Complexity: O(N * logN)
Auxiliary Space: O(1)
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