Dada una array arr[] de tamaño N , la tarea es contar el número de elementos de la array cuya suma de dígitos es igual a K.
Ejemplos:
Entrada: arr[] = {23, 54, 87, 29, 92, 62}, K = 11
Salida: 2
Explicación:
29 = 2 + 9 = 11
92 = 9 + 2 = 11Entrada: arr[]= {11, 04, 57, 99, 98, 32}, K = 18
Salida: 1
Enfoque: siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos N , para almacenar el tamaño de la array.
- Inicialice una variable, digamos count , para almacenar los elementos que tengan una suma de dígitos igual a K .
- Declare una función, sumOfDigits() para calcular la suma de los dígitos de un número .
- Recorra la array arr[] y para cada elemento de la array, verifique si la suma de los dígitos es igual a K o no. Si se encuentra que es cierto, entonces incremente el conteo en 1 .
- Imprime el valor de count como la respuesta requerida.
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
2
Complejidad de tiempo: O(N * logN)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por patelajeet y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA