Dados dos números positivos N y M , la tarea es contar el número de dígitos que están presentes tanto en N como en M .
Ejemplos:
Entrada: N = 748294, M = 34298156
Salida: 4
Explicación: Los dígitos que están presentes en ambos números son {4, 8, 2, 9}. Por lo tanto, el conteo requerido es 4.Entrada: N = 111222, M = 333444
Salida: 0
Explicación: No hay dígitos comunes presentes en los dos números dados.
Enfoque: El problema dado se puede resolver usando Hashing . Siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos contar como 0 , para almacenar la cantidad de dígitos que son comunes en ambos números.
- Inicialice dos arrays , digamos freq1[10] y freq2[10] como {0} , para almacenar el recuento de dígitos presentes en los números enteros N y M respectivamente.
- Itere sobre los dígitos del entero N e incremente el recuento de cada dígito en freq1[] en 1 .
- Itere sobre los dígitos del entero M e incremente el recuento de cada dígito en freq2[] en 1 .
- Itere sobre el rango [0, 9] e incremente el conteo en 1 si tanto freq1[i] como freq2[i] exceden 0 .
- Finalmente, luego de completar los pasos anteriores, imprima el conteo obtenido como la respuesta requerida.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count number of digits
// that are common in both N and M
int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int freq1[10] = { 0 };
// Stores the count of digits of M
int freq2[10] = { 0 };
// Iterate over the digits of N
while (N > 0) {
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0) {
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for (int i = 0; i < 10; i++) {
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0) {
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
// Input
int N = 748294;
int M = 34298156;
cout << CommonDigits(N, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count number of digits
// that are common in both N and M
static int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int freq1[] = new int[10];
// Stores the count of digits of M
int freq2[] = new int[10];
// Iterate over the digits of N
while (N > 0)
{
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0)
{
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for(int i = 0; i < 10; i++)
{
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0)
{
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 748294;
int M = 34298156;
System.out.print(CommonDigits(N, M));
}
}
// This code is contributed by gauravrajput1
Python3
# Python3 program for the above approach # Function to count number of digits # that are common in both N and M def CommonDigits(N, M): # Stores the count of common digits count = 0 # Stores the count of digits of N freq1 = [0] * 10 # Stores the count of digits of M freq2 = [0] * 10 # Iterate over the digits of N while (N > 0): # Increment the count of # last digit of N freq1[N % 10] += 1 # Update N N = N // 10 # Iterate over the digits of M while (M > 0): # Increment the count of # last digit of M freq2[M % 10] += 1 # Update M M = M // 10 # Iterate over the range [0, 9] for i in range(10): # If freq1[i] and freq2[i] both exceeds 0 if (freq1[i] > 0 and freq2[i] > 0): # Increment count by 1 count += 1 # Return the count return count # Driver Code if __name__ == '__main__': # Input N = 748294 M = 34298156 print (CommonDigits(N, M)) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Function to count number of digits
// that are common in both N and M
static int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int[] freq1 = new int[10];
// Stores the count of digits of M
int[] freq2 = new int[10];
// Iterate over the digits of N
while (N > 0)
{
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0)
{
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for(int i = 0; i < 10; i++)
{
// If freq1[i] and freq2[i]
// both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0)
{
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver code
static void Main()
{
// Input
int N = 748294;
int M = 34298156;
Console.WriteLine(CommonDigits(N, M));
}
}
// This code is contributed by sanjoy_62
Javascript
<script>
// javascript program for the above approach
// Function to count number of digits
// that are common in both N and M
function CommonDigits(N,M)
{
// Stores the count of common digits
var count = 0;
// Stores the count of digits of N
var freq1 = Array(10).fill(0);
// Stores the count of digits of M
var freq2 = Array(10).fill(0);
// Iterate over the digits of N
while (N > 0) {
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = Math.floor(N / 10);
}
// Iterate over the digits of M
while (M > 0) {
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = Math.floor(M / 10);
}
var i;
// Iterate over the range [0, 9]
for (i = 0; i < 10; i++) {
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0) {
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
// Input
var N = 748294;
var M = 34298156;
document.write(CommonDigits(N, M));
</script>
Producción:
4
Complejidad de tiempo: O(dígitos(N)+dígitos(M))
Espacio auxiliar: O(10)
Publicación traducida automáticamente
Artículo escrito por vikkycirus y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA