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Maximum length of a sub-array with ugly numbers

  • Difficulty Level : Medium
  • Last Updated : 14 May, 2021

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:
The Longest possible sub-array of ugly number sis {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, than 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, than 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, than 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, than 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, than 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

 

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