计数位数总和等于 K 的数组元素
原文:https://www . geesforgeks . org/count-array-elements-having-sum-of-digits-equal-k/
给定一个大小为 N 的数组 arr[] ，任务是计算其位数之和等于 K 的数组元素的数量。
示例:
输入: arr[] = {23，54，87，29，92，62}，K = 11 输出: 2 解释: 29 = 2+9 = 11 92 = 9+2 = 11
输入: arr[]= {11，04，57，99，98，32}，K = 18 输出: 1
方法:按照以下步骤解决问题:
- 初始化一个变量，比如 N ，来存储数组的大小。
- 初始化一个变量，比如说计数，来存储具有等于 K 的数字总和的元素。
- 声明一个函数， sumOfDigits() 到计算一个数字的位数之和。
- 遍历数组 arr[] ，对于每个数组元素，检查位数之和是否等于 K 。如果发现为真，则将计数增加 1 。
- 打印的值，将计为所需答案。
下面是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
# Prthe 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.
java 描述语言
<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
时间复杂度: O(N * logN) 辅助空间: O(1)
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