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 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
// 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 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#
// 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)