Given an integer N, the task is to check if the product of every consecutive set of digits is distinct or not.
Examples:
Input: N = 234
Output: Yes
Set Product {2} 2 {2, 3} 2 * 3 = 6 {2, 3, 4} 2 * 3 * 4 = 24 {3} 3 {3, 4} 3 * 4 = 12 {4} 4 All the products are distinct.
Input: N = 1234
Output: No
Set {1, 2} and {2} both the same product i.e. 2.
Approach: Store the product of digits of every contiguous subsequence in a set. If the product to be inserted is already present in the set at any point then the answer is “No” else all the product are distinct in the end.
Below is the implementation of the above approach:
// 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 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 |
# 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# 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 |
<script> // Javascript implementation of the approach // Function that returns true if the product // of every digit of a contiguous subsequence // is distinct // 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
const s = N.toString();
// Store size of the string
const sz = s.length;
// Set to store product of
// each contiguous subsequence
const products = new Set();
// Find product of every
// contiguous subsequence
for (let i = 0; i < sz; i++) {
let product = 1;
for (let j = i; j < sz; j++) {
product *= Number(s[j]);
// If current product already exists in the set
if (products.has(product)) return false ;
else products.add(product);
}
}
return true ;
} // Driver code const N = 2345; if (productsDistinct(N))
document.write( "Yes" );
else document.write( "No" );
</script> |
Yes
Time Complexity: O((log10N)2*log(log10N))
Auxiliary Space: O((log10N)2)