Maximum length of a sub-array with ugly numbers
Given an array arr[] of N elements (0 ? arr[i] ? 1000). The task is to find the maximum length of the sub-array that contains only ugly numbers. Ugly numbers are numbers whose only prime factors are 2, 3, or 5.
The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ….. shows the first few ugly numbers. By convention, 1 is included.
Examples:
Input: arr[] = {1, 2, 7, 9, 120, 810, 374}
Output: 3
The Longest possible sub-array of ugly number is {9, 120, 810}
Input: arr[] = {109, 480, 320, 142, 121, 1}
Output: 2
Approach:
- Take a unordered_set, and insert all the ugly numbers which are less than 1000 in the set.
- Traverse the array with two variables named current_max and max_so_far.
- Check for each element if it is present in the set.
- If an ugly number is found then increment current_max and compare it with max_so_far.
- If current_max > max_so_far then max_so_far = current_max.
- Every time a non-ugly element is found, reset current_max = 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to get the nth ugly number unsigned uglyNumber( int n) { // To store ugly numbers int ugly[n]; int i2 = 0, i3 = 0, i5 = 0; int next_multiple_of_2 = 2; int next_multiple_of_3 = 3; int next_multiple_of_5 = 5; int next_ugly_no = 1; ugly[0] = 1; for ( int i = 1; i < n; i++) { next_ugly_no = min(next_multiple_of_2, min(next_multiple_of_3, next_multiple_of_5)); ugly[i] = next_ugly_no; if (next_ugly_no == next_multiple_of_2) { i2 = i2 + 1; next_multiple_of_2 = ugly[i2] * 2; } if (next_ugly_no == next_multiple_of_3) { i3 = i3 + 1; next_multiple_of_3 = ugly[i3] * 3; } if (next_ugly_no == next_multiple_of_5) { i5 = i5 + 1; next_multiple_of_5 = ugly[i5] * 5; } } return next_ugly_no; } // Function to return the length of the // maximum sub-array of ugly numbers int maxUglySubarray( int arr[], int n) { unordered_set< int > s; int i = 1; // Insert ugly numbers in set // which are less than 1000 while (1) { int next_ugly_number = uglyNumber(i); if (next_ugly_number > 1000) break ; s.insert(next_ugly_number); i++; } int current_max = 0, max_so_far = 0; for ( int i = 0; i < n; i++) { // Check if element is non ugly if (s.find(arr[i]) == s.end()) current_max = 0; // If element is ugly, then update // current_max and max_so_far accordingly else { current_max++; max_so_far = max(current_max, max_so_far); } } return max_so_far; } // Driver code int main() { int arr[] = { 1, 0, 6, 7, 320, 800, 100, 648 }; int n = sizeof (arr) / sizeof (arr[0]); cout << maxUglySubarray(arr, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to get the nth ugly number static int uglyNumber( int n) { // To store ugly numbers int []ugly = new int [n]; int i2 = 0 , i3 = 0 , i5 = 0 ; int next_multiple_of_2 = 2 ; int next_multiple_of_3 = 3 ; int next_multiple_of_5 = 5 ; int next_ugly_no = 1 ; ugly[ 0 ] = 1 ; for ( int i = 1 ; i < n; i++) { next_ugly_no = Math.min(next_multiple_of_2, Math.min(next_multiple_of_3, next_multiple_of_5)); ugly[i] = next_ugly_no; if (next_ugly_no == next_multiple_of_2) { i2 = i2 + 1 ; next_multiple_of_2 = ugly[i2] * 2 ; } if (next_ugly_no == next_multiple_of_3) { i3 = i3 + 1 ; next_multiple_of_3 = ugly[i3] * 3 ; } if (next_ugly_no == next_multiple_of_5) { i5 = i5 + 1 ; next_multiple_of_5 = ugly[i5] * 5 ; } } return next_ugly_no; } // Function to return the length of the // maximum sub-array of ugly numbers static int maxUglySubarray( int arr[], int n) { HashSet<Integer> s = new HashSet<>(); int i = 1 ; // Insert ugly numbers in set // which are less than 1000 while ( true ) { int next_ugly_number = uglyNumber(i); if (next_ugly_number > 1000 ) break ; s.add(next_ugly_number); i++; } int current_max = 0 , max_so_far = 0 ; for (i = 0 ; i < n; i++) { // Check if element is non ugly if (!s.contains(arr[i])) current_max = 0 ; // If element is ugly, then update // current_max and max_so_far accordingly else { current_max++; max_so_far = Math.max(current_max, max_so_far); } } return max_so_far; } // Driver code public static void main(String[] args) { int arr[] = { 1 , 0 , 6 , 7 , 320 , 800 , 100 , 648 }; int n = arr.length; System.out.println(maxUglySubarray(arr, n)); } } // This code is contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach # Function to get the nth ugly number def uglyNumber(n): # To store ugly numbers ugly = [ None for i in range (n)] i2 = 0 i3 = 0 i5 = 0 next_multiple_of_2 = 2 next_multiple_of_3 = 3 next_multiple_of_5 = 5 next_ugly_no = 1 ugly[ 0 ] = 1 for i in range ( 1 , n, 1 ): next_ugly_no = min (next_multiple_of_2, min (next_multiple_of_3, next_multiple_of_5)) ugly[i] = next_ugly_no if (next_ugly_no = = next_multiple_of_2): i2 = i2 + 1 next_multiple_of_2 = ugly[i2] * 2 if (next_ugly_no = = next_multiple_of_3): i3 = i3 + 1 next_multiple_of_3 = ugly[i3] * 3 if (next_ugly_no = = next_multiple_of_5): i5 = i5 + 1 next_multiple_of_5 = ugly[i5] * 5 return next_ugly_no # Function to return the length of the # maximum sub-array of ugly numbers def maxUglySubarray(arr, n): s = set () i = 1 # Insert ugly numbers in set # which are less than 1000 while ( 1 ): next_ugly_number = uglyNumber(i) if (next_ugly_number > = 1000 ): break s.add(next_ugly_number) i + = 1 current_max = 0 max_so_far = 0 for i in range (n): # Check if element is non ugly if (arr[i] not in s): current_max = 0 # If element is ugly, then update # current_max and max_so_far accordingly else : current_max + = 1 max_so_far = max (current_max, max_so_far) return max_so_far # Driver code if __name__ = = '__main__' : arr = [ 1 , 0 , 6 , 7 , 320 , 800 , 100 , 648 ] n = len (arr) print (maxUglySubarray(arr, n)) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { // Function to get the nth ugly number static int uglyNumber( int n) { // To store ugly numbers int []ugly = new int [n]; int i2 = 0, i3 = 0, i5 = 0; int next_multiple_of_2 = 2; int next_multiple_of_3 = 3; int next_multiple_of_5 = 5; int next_ugly_no = 1; ugly[0] = 1; for ( int i = 1; i < n; i++) { next_ugly_no = Math.Min(next_multiple_of_2, Math.Min(next_multiple_of_3, next_multiple_of_5)); ugly[i] = next_ugly_no; if (next_ugly_no == next_multiple_of_2) { i2 = i2 + 1; next_multiple_of_2 = ugly[i2] * 2; } if (next_ugly_no == next_multiple_of_3) { i3 = i3 + 1; next_multiple_of_3 = ugly[i3] * 3; } if (next_ugly_no == next_multiple_of_5) { i5 = i5 + 1; next_multiple_of_5 = ugly[i5] * 5; } } return next_ugly_no; } // Function to return the length of the // maximum sub-array of ugly numbers static int maxUglySubarray( int []arr, int n) { HashSet< int > s = new HashSet< int >(); int i = 1; // Insert ugly numbers in set // which are less than 1000 while ( true ) { int next_ugly_number = uglyNumber(i); if (next_ugly_number > 1000) break ; s.Add(next_ugly_number); i++; } int current_max = 0, max_so_far = 0; for (i = 0; i < n; i++) { // Check if element is non ugly if (!s.Contains(arr[i])) current_max = 0; // If element is ugly, then update // current_max and max_so_far accordingly else { current_max++; max_so_far = Math.Max(current_max, max_so_far); } } return max_so_far; } // Driver code public static void Main(String[] args) { int []arr = { 1, 0, 6, 7, 320, 800, 100, 648 }; int n = arr.Length; Console.WriteLine(maxUglySubarray(arr, n)); } } // This code is contributed by Princi Singh |
Javascript
<script> // javascript implementation of the approach // Function to get the nth ugly number function uglyNumber( n) { // To store ugly numbers var ugly = []; var i2 = 0, i3 = 0, i5 = 0; var next_multiple_of_2 = 2; var next_multiple_of_3 = 3; var next_multiple_of_5 = 5; var next_ugly_no = 1; ugly[0] = 1; for ( var i = 1; i < n; i++) { next_ugly_no = Math.min(next_multiple_of_2, Math.min(next_multiple_of_3, next_multiple_of_5)); ugly[i] = next_ugly_no; if (next_ugly_no == next_multiple_of_2) { i2 = i2 + 1; next_multiple_of_2 = ugly[i2] * 2; } if (next_ugly_no == next_multiple_of_3) { i3 = i3 + 1; next_multiple_of_3 = ugly[i3] * 3; } if (next_ugly_no == next_multiple_of_5) { i5 = i5 + 1; next_multiple_of_5 = ugly[i5] * 5; } } return next_ugly_no; } // Function to return the length of the // maximum sub-array of ugly numbers function maxUglySubarray(arr, n) { var s = [] var i = 1; // Insert ugly numbers in set // which are less than 1000 while ( true ) { var next_ugly_number = uglyNumber(i); if (next_ugly_number > 1000) break ; s.push(next_ugly_number); i++; } var current_max = 0, max_so_far = 0; for (i = 0; i < n; i++) { // Check if element is non ugly if (!s.includes(arr[i])) current_max = 0; // If element is ugly, then update // current_max and max_so_far accordingly else { current_max++; max_so_far = Math.max(current_max, max_so_far); } } return max_so_far; } // Driver code var arr = [ 1, 0, 6, 7, 320, 800, 100, 648 ]; var n = arr.length; document.write(maxUglySubarray(arr, n)); </script> |
Output:
4