Dada una array arr[] de tamaño N que contiene strings, la tarea es contar el número de strings que tienen una suma de valores ASCII de caracteres iguales a un número de Armstrong o un número primo.
Ejemplos:
Entrada: arr[] = {“hello”, “nace”}
Salida: El
número de strings Armstrong es: 1
El número de strings principales es: 0
Explicación: la suma de los valores ASCII de los caracteres de cada string es: {532, 407}, de los cuales 407 es un número de Armstrong, y ninguno de ellos es un número primo.
Por lo tanto, la string valorada por armstrong es «nace».Entrada: arr[] = {“geeksforgeeks”, “a”, “computer”, “science”, “portal”, “for”, “geeks”}
Salida: El
número de strings Armstrong es: 0
El número de strings principales es: 2
Explicación: La suma de los valores ASCII de los caracteres de cada string es: {1381, 97, 879, 730, 658, 327, 527}, de los cuales 1381 y 97 son números primos y ninguno de ellos es un número de Armstrong.
Por lo tanto, las strings con valores primos son «geeksforgeeks» y «a».
Enfoque: este problema se puede resolver calculando el valor ASCII de cada string. Siga los pasos a continuación para resolver este problema:
- Inicialice dos variables, countPrime y countArmstrong como 0 , para almacenar el recuento de strings con valores Prime y Armstrong.
- Itere sobre el rango de índices [0, N – 1] usando una variable, digamos i y realice los siguientes pasos:
- Almacene la suma de los valores ASCII de los caracteres de la string actual arr[i] en una variable, digamos val .
- Si el número val es un número de Armstrong , incremente countArmstrong en 1 .
- Si el valor del número es un número primo , incremente countPrime en 1 .
- Imprime los valores de countPrime y countArmstrong como resultado.
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 check if a
// number is prime number
bool isPrime(int num)
{
// Define a flag variable
bool flag = false;
if (num > 1)
{
// Check for factors of num
for(int i = 2; i < num; i++)
{
// If factor is found,
// set flag to True and
// break out of loop
if ((num % i) == 0)
{
flag = true;
break;
}
}
}
// Check if flag is True
if (flag)
return false;
else
return true;
}
// Function to calculate
// order of the number x
int order(int x)
{
int n = 0;
while (x != 0)
{
n = n + 1;
x = x / 10;
}
return n;
}
// Function to check whether the given
// number is Armstrong number or not
bool isArmstrong(int x)
{
int n = order(x);
int temp = x;
int sum1 = 0;
while (temp != 0)
{
int r = temp % 10;
sum1 = sum1 + pow(r, n);
temp = temp / 10;
}
// If the condition satisfies
return (sum1 == x);
}
// Function to count
// Armstrong valued strings
int count_armstrong(vector<string> li)
{
// Stores the count of
// Armstrong valued strings
int c = 0;
// Iterate over the list
for(string ele : li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che:ele)
val += che;
// Check if it an Armstrong number
if (isArmstrong(val))
c += 1;
}
return c;
}
// Function to count
// prime valued strings
int count_prime(vector<string> li)
{
// Store the count of
// prime valued strings
int c = 0;
// Iterate over the list
for(string ele:li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che : ele)
val += che;
// Check if it
// is a Prime Number
if (isPrime(val))
c += 1;
}
return c;
}
// Driver code
int main()
{
vector<string> arr = { "geeksforgeeks", "a", "computer",
"science", "portal", "for", "geeks"};
// Function Call
cout << "Number of Armstrong Strings are: "
<< count_armstrong(arr) << endl;
cout << "Number of Prime Strings are: "
<< count_prime(arr) << endl;
}
// This code is contributed by mohit kumar 29
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check if a
// number is prime number
static boolean isPrime(int num)
{
// Define a flag variable
boolean flag = false;
if (num > 1) {
// Check for factors of num
for (int i = 2; i < num; i++) {
// If factor is found,
// set flag to True and
// break out of loop
if ((num % i) == 0) {
flag = true;
break;
}
}
}
// Check if flag is True
if (flag)
return false;
else
return true;
}
// Function to calculate
// order of the number x
static int order(int x)
{
int n = 0;
while (x != 0) {
n = n + 1;
x = x / 10;
}
return n;
}
// Function to check whether the given
// number is Armstrong number or not
static boolean isArmstrong(int x)
{
int n = order(x);
int temp = x;
int sum1 = 0;
while (temp != 0) {
int r = temp % 10;
sum1 = sum1 + (int)(Math.pow(r, n));
temp = temp / 10;
}
// If the condition satisfies
return (sum1 == x);
}
// Function to count
// Armstrong valued strings
static int count_armstrong(String[] li)
{
// Stores the count of
// Armstrong valued strings
int c = 0;
// Iterate over the list
for(String ele : li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che : ele.toCharArray()) val += che;
// Check if it an Armstrong number
if (isArmstrong(val))
c += 1;
}
return c;
}
// Function to count
// prime valued strings
static int count_prime(String[] li)
{
// Store the count of
// prime valued strings
int c = 0;
// Iterate over the list
for(String ele : li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che : ele.toCharArray()) val += che;
// Check if it
// is a Prime Number
if (isPrime(val))
c += 1;
}
return c;
}
// Driver code
public static void main (String[] args) {
String[] arr
= { "geeksforgeeks", "a", "computer",
"science", "portal", "for",
"geeks" };
// Function Call
System.out.println(
"Number of Armstrong Strings are: "
+ count_armstrong(arr));
System.out.println("Number of Prime Strings are: "
+ count_prime(arr));
}
}
// This code is contributed by patel2127.
Python3
# Python program for the above approach
# Function to check if a
# number is prime number
def isPrime(num):
# Define a flag variable
flag = False
if num > 1:
# Check for factors of num
for i in range(2, num):
# If factor is found,
# set flag to True and
# break out of loop
if (num % i) == 0:
flag = True
break
# Check if flag is True
if flag:
return False
else:
return True
# Function to calculate
# order of the number x
def order(x):
n = 0
while (x != 0):
n = n + 1
x = x // 10
return n
# Function to check whether the given
# number is Armstrong number or not
def isArmstrong(x):
n = order(x)
temp = x
sum1 = 0
while (temp != 0):
r = temp % 10
sum1 = sum1 + r**n
temp = temp // 10
# If the condition satisfies
return (sum1 == x)
# Function to count
# Armstrong valued strings
def count_armstrong(li):
# Stores the count of
# Armstrong valued strings
c = 0
# Iterate over the list
for ele in li:
# Store the value
# of the string
val = 0
# Find value of the string
for che in ele:
val += ord(che)
# Check if it an Armstrong number
if isArmstrong(val):
c += 1
return c
# Function to count
# prime valued strings
def count_prime(li):
# Store the count of
# prime valued strings
c = 0
# Iterate over the list
for ele in li:
# Store the value
# of the string
val = 0
# Find value of the string
for che in ele:
val += ord(che)
# Check if it
# is a Prime Number
if isPrime(val):
c += 1
return c
# Driver code
arr = ["geeksforgeeks", "a", "computer",
"science", "portal", "for", "geeks"]
# Function Call
print("Number of Armstrong Strings are:", count_armstrong(arr))
print("Number of Prime Strings are:", count_prime(arr))
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if a
// number is prime number
static bool isPrime(int num)
{
// Define a flag variable
bool flag = false;
if (num > 1) {
// Check for factors of num
for (int i = 2; i < num; i++) {
// If factor is found,
// set flag to True and
// break out of loop
if ((num % i) == 0) {
flag = true;
break;
}
}
}
// Check if flag is True
if (flag)
return false;
else
return true;
}
// Function to calculate
// order of the number x
static int order(int x)
{
int n = 0;
while (x != 0) {
n = n + 1;
x = x / 10;
}
return n;
}
// Function to check whether the given
// number is Armstrong number or not
static bool isArmstrong(int x)
{
int n = order(x);
int temp = x;
int sum1 = 0;
while (temp != 0) {
int r = temp % 10;
sum1 = sum1 + (int)(Math.Pow(r, n));
temp = temp / 10;
}
// If the condition satisfies
return (sum1 == x);
}
// Function to count
// Armstrong valued strings
static int count_armstrong(string[] li)
{
// Stores the count of
// Armstrong valued strings
int c = 0;
// Iterate over the list
foreach(string ele in li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
foreach(char che in ele) val += che;
// Check if it an Armstrong number
if (isArmstrong(val))
c += 1;
}
return c;
}
// Function to count
// prime valued strings
static int count_prime(string[] li)
{
// Store the count of
// prime valued strings
int c = 0;
// Iterate over the list
foreach(string ele in li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
foreach(char che in ele) val += che;
// Check if it
// is a Prime Number
if (isPrime(val))
c += 1;
}
return c;
}
// Driver code
public static void Main()
{
string[] arr
= { "geeksforgeeks", "a", "computer",
"science", "portal", "for",
"geeks" };
// Function Call
Console.WriteLine(
"Number of Armstrong Strings are: "
+ count_armstrong(arr));
Console.WriteLine("Number of Prime Strings are: "
+ count_prime(arr));
}
}
// This code is contributed by ukasp.
Javascript
<script>
// JavaScript program for the above approach
// Function to check if a
// number is prime number
function isPrime(num) {
// Define a flag variable
let flag = false;
if (num > 1) {
// Check for factors of num
for (let i = 2; i < num; i++) {
// If factor is found,
// set flag to True and
// break out of loop
if ((num % i) == 0) {
flag = true;
break;
}
}
}
// Check if flag is True
if (flag)
return false;
else
return true;
}
// Function to calculate
// order of the number x
function order(x) {
let n = 0;
while (x != 0) {
n = n + 1;
x = x / 10;
}
return n;
}
// Function to check whether the given
// number is Armstrong number or not
function isArmstrong(x) {
let n = order(x);
let temp = x;
let sum1 = 0;
while (temp != 0) {
let r = temp % 10;
sum1 = sum1 + Math.pow(r, n);
temp = temp / 10;
}
// If the condition satisfies
return (sum1 == x);
}
// Function to count
// Armstrong valued strings
function count_armstrong(li) {
// Stores the count of
// Armstrong valued strings
let c = 0;
// Iterate over the list
for (let ele of li) {
// Store the value
// of the string
let val = 0;
// Find value of the string
for (let che of ele)
val += che.charCodeAt(0);
// Check if it an Armstrong number
if (isArmstrong(val))
c += 1;
}
return c;
}
// Function to count
// prime valued strings
function count_prime(li) {
// Store the count of
// prime valued strings
let c = 0;
// Iterate over the list
for (let ele of li) {
// Store the value
// of the string
let val = 0;
// Find value of the string
for (let che of ele)
val += che.charCodeAt(0);
// Check if it
// is a Prime Number
if (isPrime(val))
c += 1;
}
return c;
}
// Driver code
let arr = ["geeksforgeeks", "a", "computer",
"science", "portal", "for", "geeks"];
// Function Call
document.write("Number of Armstrong Strings are: "
+ count_armstrong(arr) + "<br>");
document.write("Number of Prime Strings are: "
+ count_prime(arr) + "<br>");
// This code is contributed by gfgking
</script>
Producción
Number of Armstrong Strings are: 0 Number of Prime Strings are: 2
Complejidad de tiempo: O(N*M), donde M es la longitud de la string más larga de la array arr[]
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por santhoshcharan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA