Dada una array arr[] que consiste en N enteros positivos, la tarea para cada elemento de la array arr[i] es encontrar todos los dígitos de [0, 9] que no dividen ningún dígito presente en arr[i] .
Ejemplos:
Entrada: arr[] = {4162, 1152, 99842}
Salida:
4162 -> 5 7 8 9
1152 -> 3 4 6 7 8 9
99842 -> 5 6 7
Explicación:
Para arr[0] ( = 4162): Ninguno de los dígitos del elemento 4162 son divisibles por 5, 7, 8, 9.
Para arr[1]( = 1152): Ninguno de los dígitos del elemento 1152 son divisibles por 9, 8, 7, 6, 4, 3 Para arr [
2]( = 99842): Ninguno de los dígitos del elemento 99842 es divisible por 7, 6, 5.Entrada: arr[] = {2021}
Salida:
2021 -> 3 4 5 6 7 8 9
Enfoque: siga los pasos a continuación para resolver el problema:
- Recorra la array dada arr[] y realice los siguientes pasos:
- Itere sobre el rango [2, 9] usando la variable i , y si no existe ningún dígito en el elemento arr[i] que sea divisible por i , imprima el dígito i .
- De lo contrario, continúe para la siguiente iteración.
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 find digits for each array
// element that doesn't divide any digit
// of the that element
void indivisibleDigits(int arr[], int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
int num = 0;
cout << arr[i] << ": ";
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++) {
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
bool flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
cout << j << ' ';
}
}
cout << endl;
}
}
// Driver Code
int main()
{
int arr[] = { 4162, 1152, 99842 };
int N = sizeof(arr) / sizeof(arr[0]);
indivisibleDigits(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
static void indivisibleDigits(int[] arr, int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
System.out.print(arr[i] + ": ");
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++)
{
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
boolean flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
System.out.print(j + " ");
}
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 4162, 1152, 99842 };
int N = arr.length;
indivisibleDigits(arr, N);
}
}
// This code is contributed by sanjoy_62.
Python3
# Python3 program for the above approach # Function to find digits for each array # element that doesn't divide any digit # of the that element def indivisibleDigits(arr, N) : # Traverse the array arr[] for i in range(N): num = 0 print(arr[i], end = ' ') # Iterate over the range [2, 9] for j in range(2, 10): temp = arr[i] # Stores if there exists any digit # in arr[i] which is divisible by j flag = True while (temp > 0) : # If any digit of the number # is divisible by j if ((temp % 10) != 0 and (temp % 10) % j == 0) : flag = False break temp //= 10 # If the digit j doesn't # divide any digit of arr[i] if (flag) : print(j, end = ' ') print() # Driver Code arr = [ 4162, 1152, 99842 ] N = len(arr) indivisibleDigits(arr, N) # This code is contributed by susmitakundugoaldanga.
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
static void indivisibleDigits(int[] arr, int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
Console.Write(arr[i] + ": ");
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++)
{
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
bool flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
Console.Write(j + " ");
}
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
int[] arr = { 4162, 1152, 99842 };
int N = arr.Length;
indivisibleDigits(arr, N);
}
}
// This code is contributed by rishavmahato348.
Javascript
<script>
// javascript program for the above approach
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
function indivisibleDigits(arr , N) {
// Traverse the array arr
for (i = 0; i < N; i++) {
document.write(arr[i] + ": ");
// Iterate over the range [2, 9]
for (j = 2; j < 10; j++) {
var temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
var flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0 && (temp % 10) % j == 0) {
flag = false;
break;
}
temp = parseInt(temp/10);
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
document.write(j + " ");
}
}
document.write("<br/>");
}
}
// Driver Code
var arr = [ 4162, 1152, 99842 ];
var N = arr.length;
indivisibleDigits(arr, N);
// This code contributed by aashish1995
</script>
Producción:
4162: 5 7 8 9 1152: 3 4 6 7 8 9 99842: 5 6 7
Complejidad de tiempo: O(10*N*log 10 N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por subhammahato348 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA