Count array elements having sum of digits equal to K
Last Updated :
29 Jan, 2022
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 = 11
Input: 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 N
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
int 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 Code
int 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 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 program for the above approach
# Function to calculate the
# sum of digits of the number N
def 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 elements
def 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 array
arr = [ 23, 54, 87, 29, 92, 62 ]
# Given value of K
K = 11
# Size of the array
N = len(arr)
# Function call to count array elements
# having sum of digits equal to K
elementsHavingDigitSumK(arr, N, K)
# This code is contributed by souravghosh0416.
// 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.
<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>
Time Complexity: O(N * logN)
Auxiliary Space: O(1)
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