Dado un número entero N , la tarea es verificar si el producto de cada conjunto consecutivo de dígitos es distinto o no.
Ejemplos:
Entrada: N = 234
Salida: Sí
Establecer Producto {2} 2 {2, 3} 2 * 3 = 6 {2, 3, 4} 2 * 3 * 4 = 24 {3} 3 {3, 4} 3 * 4 = 12 {4} 4 Todos los productos son distintos.
Entrada: N = 1234
Salida: No
Establezca {1, 2} y {2} ambos del mismo producto, es decir, 2.
Método: almacenar el producto de dígitos de cada subsecuencia contigua en un conjunto . Si el producto que se va a insertar ya está presente en el conjunto en algún momento, la respuesta es «No», de lo contrario, todos los productos son distintos al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if the product
// of every digit of a contiguous subsequence
// is distinct
bool productsDistinct(int N)
{
// To store the given number as a string
string s = "";
// Append all the digits
// starting from the end
while (N) {
s += (char)(N % 10 + '0');
N /= 10;
}
// Reverse the string to get
// the original number
reverse(s.begin(), s.end());
// Store size of the string
int sz = s.size();
// Set to store product of
// each contiguous subsequence
set<int> se;
// Find product of every
// contiguous subsequence
for (int i = 0; i < sz; i++) {
int product = 1;
for (int j = i; j < sz; j++) {
product *= (int)(s[j] - '0');
// If current product already
// exists in the set
if (se.find(product) != se.end())
return false;
else
se.insert(product);
}
}
return true;
}
// Driver code
int main()
{
int N = 2345;
if (productsDistinct(N))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if
// the product of every digit of a
// contiguous subsequence is distinct
static boolean productsDistinct(int N)
{
// To store the given number
// as a string
String s = "";
// Append all the digits
// starting from the end
while (N > 0)
{
s += (char)(N % 10 + '0');
N /= 10;
}
// Reverse the string to get
// the original number
s = reverse(s);
// Store size of the string
int sz = s.length();
// Set to store product of
// each contiguous subsequence
HashSet<Integer> se = new HashSet<Integer>();
// Find product of every
// contiguous subsequence
for (int i = 0; i < sz; i++)
{
int product = 1;
for (int j = i; j < sz; j++)
{
product *= (int)(s.charAt(j) - '0');
// If current product already
// exists in the set
if (se.contains(product))
return false;
else
se.add(product);
}
}
return true;
}
static String reverse(String input)
{
char[] a = input.toCharArray();
int l, r;
r = a.length - 1;
for (l = 0; l < r; l++, r--)
{
// Swap values of l and r
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver code
public static void main(String[] args)
{
int N = 2345;
if (productsDistinct(N))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed
// by PrinciRaj1992
Python3
# Python 3 implementation of the approach
# Function that returns true if the product
# of every digit of a contiguous subsequence
# is distinct
def productsDistinct(A):
A = str(A)
n = len(A)
hs = set()
for i in range(n):
prod = 1
for j in range(i, n):
prod = prod * int(A[j])
if(prod in hs):
return False
else:
hs.add(prod)
return True
# Driver code
if __name__ == '__main__':
N = 2345
if (productsDistinct(N)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that returns true if
// the product of every digit of a
// contiguous subsequence is distinct
static Boolean productsDistinct(int N)
{
// To store the given number
// as a string
String s = "";
// Append all the digits
// starting from the end
while (N > 0)
{
s += (char)(N % 10 + '0');
N /= 10;
}
// Reverse the string to get
// the original number
s = reverse(s);
// Store size of the string
int sz = s.Length;
// Set to store product of
// each contiguous subsequence
HashSet<int> se = new HashSet<int>();
// Find product of every
// contiguous subsequence
for (int i = 0; i < sz; i++)
{
int product = 1;
for (int j = i; j < sz; j++)
{
product *= (int)(s[j] - '0');
// If current product already
// exists in the set
if (se.Contains(product))
return false;
else
se.Add(product);
}
}
return true;
}
static String reverse(String input)
{
char[] a = input.ToCharArray();
int l, r;
r = a.Length - 1;
for (l = 0; l < r; l++, r--)
{
// Swap values of l and r
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver code
public static void Main(String[] args)
{
int N = 2345;
if (productsDistinct(N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if the product
// of every digit of a contiguous subsequence
// is distinct
function productsDistinct(N)
{
// To store the given number as a string
var s = "";
// Append all the digits
// starting from the end
while (N)
{
s += String.fromCharCode(
N % 10 + '0'.charCodeAt(0));
N = parseInt(N / 10);
}
// Reverse the string to get
// the original number
s = s.split('').reverse().join('');
// Store size of the string
var sz = s.length;
// Set to store product of
// each contiguous subsequence
var se = new Set();
// Find product of every
// contiguous subsequence
for(var i = 0; i < sz; i++)
{
var product = 1;
for(var j = i; j < sz; j++)
{
product *= (s[j].charCodeAt(0) -
'0'.charCodeAt(0));
// If current product already
// exists in the set
if (se.has(product))
return false;
else
se.add(product);
}
}
return true;
}
// Driver code
var N = 2345;
if (productsDistinct(N))
document.write("Yes");
else
document.write("No");
// This code is contributed by rrrtnx
</script>
Producción:
Yes
Complejidad del tiempo: O((log 10 N) 2 *log(log 10 N))
Espacio Auxiliar: O((log 10 N) 2 )
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA