Ugly Numbers

• Difficulty Level : Medium
• Last Updated : 18 Jan, 2022

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 11 ugly numbers. By convention, 1 is included.
Given a number n, the task is to find n’th Ugly number.

Examples:

Input  : n = 7
Output : 8

Input  : n = 10
Output : 12

Input  : n = 15
Output : 24

Input  : n = 150
Output : 5832

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Simple)
Loop for all positive integers till the ugly number count is smaller than n, if an integer is ugly than increment ugly number count.
To check if a number is ugly, divide the number by greatest divisible powers of 2, 3 and 5, if the number becomes 1 then it is an ugly number otherwise not.

For example, let us see how to check for 300 is ugly or not. Greatest divisible power of 2 is 4, after dividing 300 by 4 we get 75. Greatest divisible power of 3 is 3, after dividing 75 by 3 we get 25. Greatest divisible power of 5 is 25, after dividing 25 by 25 we get 1. Since we get 1 finally, 300 is ugly number.

Below is the implementation of the above approach:

C++

 // C++ program to find nth ugly number#include using namespace std; // This function divides a by greatest// divisible power of bint maxDivide(int a, int b){    while (a % b == 0)        a = a / b;             return a;} // Function to check if a number is ugly or notint isUgly(int no){    no = maxDivide(no, 2);    no = maxDivide(no, 3);    no = maxDivide(no, 5);     return (no == 1) ? 1 : 0;} // Function to get the nth ugly numberint getNthUglyNo(int n){    int i = 1;         // Ugly number count    int count = 1;     // Check for all integers until ugly    // count becomes n    while (n > count)    {        i++;        if (isUgly(i))            count++;    }    return i;} // Driver Codeint main(){         // Function call    unsigned no = getNthUglyNo(150);    cout << "150th ugly no. is " << no;    return 0;} // This code is contributed by shivanisinghss2110

C

 // C program to find nth ugly number#include #include  // This function divides a by greatest divisible//  power of bint maxDivide(int a, int b){    while (a % b == 0)        a = a / b;    return a;} // Function to check if a number is ugly or notint isUgly(int no){    no = maxDivide(no, 2);    no = maxDivide(no, 3);    no = maxDivide(no, 5);     return (no == 1) ? 1 : 0;} // Function to get the nth ugly numberint getNthUglyNo(int n){        int i = 1;         // ugly number count    int count = 1;     // Check for all integers until ugly count    //  becomes n    while (n > count) {        i++;        if (isUgly(i))            count++;    }    return i;} // Driver Codeint main(){    // Function call    unsigned no = getNthUglyNo(150);    printf("150th ugly no. is %d ", no);    getchar();    return 0;}

Java

 // Java program to find nth ugly numberclass GFG {     /*This function divides a by greatest    divisible power of b*/    static int maxDivide(int a, int b)    {        while (a % b == 0)            a = a / b;        return a;    }     /* Function to check if a number    is ugly or not */    static int isUgly(int no)    {        no = maxDivide(no, 2);        no = maxDivide(no, 3);        no = maxDivide(no, 5);         return (no == 1) ? 1 : 0;    }     /* Function to get the nth ugly    number*/    static int getNthUglyNo(int n)    {        int i = 1;         // ugly number count        int count = 1;         // check for all integers        // until count becomes n        while (n > count) {            i++;            if (isUgly(i) == 1)                count++;        }        return i;    }     /* Driver Code*/    public static void main(String args[])    {        int no = getNthUglyNo(150);               // Function call        System.out.println("150th ugly "                           + "no. is " + no);    }} // This code has been contributed by// Amit Khandelwal (Amit Khandelwal 1)

Python3

 # Python3 code to find nth ugly number # This function divides a by greatest# divisible power of b  def maxDivide(a, b):    while a % b == 0:        a = a / b    return a # Function to check if a number# is ugly or notdef isUgly(no):    no = maxDivide(no, 2)    no = maxDivide(no, 3)    no = maxDivide(no, 5)    return 1 if no == 1 else 0 # Function to get the nth ugly numberdef getNthUglyNo(n):    i = 1         # ugly number count    count = 1      # Check for all integers until    # ugly count becomes n    while n > count:        i += 1        if isUgly(i):            count += 1    return i  # Driver codeno = getNthUglyNo(150)print("150th ugly no. is ", no) # This code is contributed by "Sharad_Bhardwaj".

C#

 // C# program to find nth ugly numberusing System; class GFG {     /*This function divides a by    greatest divisible power of b*/    static int maxDivide(int a, int b)    {        while (a % b == 0)            a = a / b;        return a;    }     /* Function to check if a number    is ugly or not */    static int isUgly(int no)    {        no = maxDivide(no, 2);        no = maxDivide(no, 3);        no = maxDivide(no, 5);         return (no == 1) ? 1 : 0;    }     /* Function to get the nth ugly    number*/    static int getNthUglyNo(int n)    {        int i = 1;         // ugly number count        int count = 1;         // Check for all integers        // until count becomes n        while (n > count) {            i++;            if (isUgly(i) == 1)                count++;        }        return i;    }     // Driver code    public static void Main()    {        int no = getNthUglyNo(150);         // Function call        Console.WriteLine("150th ugly"                          + " no. is " + no);    }} // This code is contributed by Sam007.

PHP

 \$count){    \$i++;        if (isUgly(\$i))    \$count++;}return \$i;}     // Driver Code    \$no = getNthUglyNo(150);    echo "150th ugly no. is ". \$no; // This code is contributed by Sam007?>

Javascript


Output
150th ugly no. is 5832

This method is not time efficient as it checks for all integers until ugly number count becomes n, but space complexity of this method is O(1)

Method 2 (Use Dynamic Programming)
Here is a time efficient solution with O(n) extra space. The ugly-number sequence is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
because every number can only be divided by 2, 3, 5, one way to look at the sequence is to split the sequence to three groups as below:
(1) 1×2, 2×2, 3×2, 4×2, 5×2, …
(2) 1×3, 2×3, 3×3, 4×3, 5×3, …
(3) 1×5, 2×5, 3×5, 4×5, 5×5, …
We can find that every subsequence is the ugly-sequence itself (1, 2, 3, 4, 5, …) multiply 2, 3, 5. Then we use similar merge method as merge sort, to get every ugly number from the three subsequences. Every step we choose the smallest one, and move one step after.

1 Declare an array for ugly numbers:  ugly[n]
2 Initialize first ugly no:  ugly = 1
3 Initialize three array index variables i2, i3, i5 to point to
1st element of the ugly array:
i2 = i3 = i5 =0;
4 Initialize 3 choices for the next ugly no:
next_multiple_of_2 = ugly[i2]*2;
next_multiple_of_3 = ugly[i3]*3
next_multiple_of_5 = ugly[i5]*5;
5 Now go in a loop to fill all ugly numbers till 150:
For (i = 1; i < 150; i++ )
{
/* These small steps are not optimized for good
readability. Will optimize them in C program */
next_ugly_no  = Min(next_multiple_of_2,
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;
}

}/* end of for loop */
6.return next_ugly_no

Example:
Let us see how it works

initialize
ugly[] =  | 1 |
i2 =  i3 = i5 = 0;

First iteration
ugly = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
= Min(2, 3, 5)
= 2
ugly[] =  | 1 | 2 |
i2 = 1,  i3 = i5 = 0  (i2 got incremented )

Second iteration
ugly = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
= Min(4, 3, 5)
= 3
ugly[] =  | 1 | 2 | 3 |
i2 = 1,  i3 =  1, i5 = 0  (i3 got incremented )

Third iteration
ugly = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
= Min(4, 6, 5)
= 4
ugly[] =  | 1 | 2 | 3 |  4 |
i2 = 2,  i3 =  1, i5 = 0  (i2 got incremented )

Fourth iteration
ugly = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
= Min(6, 6, 5)
= 5
ugly[] =  | 1 | 2 | 3 |  4 | 5 |
i2 = 2,  i3 =  1, i5 = 1  (i5 got incremented )

Fifth iteration
ugly = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
= Min(6, 6, 10)
= 6
ugly[] =  | 1 | 2 | 3 |  4 | 5 | 6 |
i2 = 3,  i3 =  2, i5 = 1  (i2 and i3 got incremented )

Will continue same way till I < 150

C++

 // C++ program to find n'th Ugly number#include using namespace std; // Function to get the nth ugly numberunsigned getNthUglyNo(unsigned n){    // To store ugly numbers    unsigned ugly[n];    unsigned i2 = 0, i3 = 0, i5 = 0;    unsigned next_multiple_of_2 = 2;    unsigned next_multiple_of_3 = 3;    unsigned next_multiple_of_5 = 5;    unsigned next_ugly_no = 1;     ugly = 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;        }    }        // End of for loop (i=1; i

Java

 // Java program to find nth ugly numberimport java.lang.Math; class UglyNumber{    // Function to get the nth ugly number    int getNthUglyNo(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 = 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;            }        }                 // End of for loop (i=1; i

Python

 # Python program to find n'th Ugly number # Function to get the nth ugly number  def getNthUglyNo(n):     ugly =  * n  # To store ugly numbers     # 1 is the first ugly number    ugly = 1     # i2, i3, i5 will indicate indices for    # 2,3,5 respectively    i2 = i3 = i5 = 0     # Set initial multiple value    next_multiple_of_2 = 2    next_multiple_of_3 = 3    next_multiple_of_5 = 5     # Start loop to find value from    # ugly to ugly[n]    for l in range(1, n):         # Choose the min value of all        # available multiples        ugly[l] = min(next_multiple_of_2,                      next_multiple_of_3,                      next_multiple_of_5)         # Increment the value of index accordingly        if ugly[l] == next_multiple_of_2:            i2 += 1            next_multiple_of_2 = ugly[i2] * 2         if ugly[l] == next_multiple_of_3:            i3 += 1            next_multiple_of_3 = ugly[i3] * 3         if ugly[l] == next_multiple_of_5:            i5 += 1            next_multiple_of_5 = ugly[i5] * 5     # Return ugly[n] value    return ugly[-1] # Driver Codedef main():     n = 150     print getNthUglyNo(n)  if __name__ == '__main__':    main() # This code is contributed by Neelam Yadav

C#

 // C# program to count inversions in an arrayusing System;using System.Collections.Generic; class GFG {     // Function to get the nth ugly number    static int getNthUglyNo(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 = 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;    }     // Driver code    public static void Main()    {        int n = 150;               // Function call        Console.WriteLine(getNthUglyNo(n));    }} // This code is contributed by Sam007



Javascript


Output
5832

Time Complexity: O(n)
Auxiliary Space: O(n)
Super Ugly Number (Number whose prime factors are in the given set)

Method 3 (Using SET Data Structure in C++ and TreeSet in JAVA)

In this method, SET data structure to store the minimum of the 3 generated ugly numbers which will the ith Ugly Number stored at the first position of the set. SET Data Structure as a set stores all the element in ascending order

Below is the implementation of the above approach:

C++

 // C++ Implementation of the above approach#include using namespace std; int nthUglyNumber(int n){    // Base cases...    if (n == 1 or n == 2 or n == 3 or n == 4 or n == 5)        return n;     set s;    s.insert(1);    n--;     while (n) {        auto it = s.begin();         // Get the beginning element of the set        long long int x = *it;         // Deleting the ith element        s.erase(it);         // Inserting all the other options        s.insert(x * 2);        s.insert(x * 3);        s.insert(x * 5);        n--;    }     // The top of the set represents the nth ugly number    return *s.begin();} // Driver Codeint main(){    int n = 150;     // Function call    cout << nthUglyNumber(n);} // Contributed by:- Soumak Poddar

Java

 /*package whatever //do not write package name here */ import java.io.*;import java.util.*; class GFG {     static long nthUglyNumber(int n)    {         TreeSet t = new TreeSet<>();        // Ugly number sequence starts with 1        t.add(1L);        int i = 1;        // when i==n we have the nth ugly number        while (i < n) {            // remove the ith ugly number and add back its            // multiples of 2,3 and 5 back to the set            long temp = t.pollFirst();            t.add(temp * 2);            t.add(temp * 3);            t.add(temp * 5);            i++;            // the first element of set is always the ith            // ugly number        }         return t.pollFirst();    }     public static void main(String[] args)    {        int n = 150;        System.out.println(nthUglyNumber(n));    }}// Contributed by:- Saswata Halder

Python3

 # Python Implementation of the above approachdef nthUglyNumber(n):         # Base cases...    if (n == 1 or n == 2 or n == 3 or n == 4 or n == 5):        return n    s =     n-=1     while (n):        it = s                 # Get the beginning element of the set        x = it                 # Deleting the ith element        s = s[1:]        s = set(s)                 # Inserting all the other options        s.add(x * 2)        s.add(x * 3)        s.add(x * 5)        s = list(s)        s.sort()        n -= 1    # The top of the set represents the nth ugly number    return s # Driver Coden = 150 # Function callprint( nthUglyNumber(n)) # This code is contributed by Shubham Singh
Output
5832

Time Complexity:- O(N log N)
Auxiliary Space:- O(N)

Method 4(Using Binary Search)

1. This method is suitable if you have a max value for n. The no will be of form x=pow(2,p)*pow(3,q)*pow(5,r).
2. Search from low=1 to high =21474836647. We are expecting nth ugly no to be in this range.
3. So we do a binary search. Now suppose we are at mid now we are going to find the total number of ugly numbers less than mid and put our conditions accordingly.

Below is the rough CPP code:

C++

 // CPP program for the above approach#include using namespace std; // Print nth Ugly numberint nthUglyNumber(int n){   int pow = { 1 };   // stored powers of 2 from  // pow(2,0) to pow(2,30)  for (int i = 1; i <= 30; ++i)    pow[i] = pow[i - 1] * 2;   // Initialized low and high  int l = 1, r = 2147483647;   int ans = -1;   // Applying Binary Search  while (l <= r) {     // Found mid    int mid = l + ((r - l) / 2);     // cnt stores total numbers of ugly    // number less than mid    int cnt = 0;     // Iterate from 1 to mid    for (long long i = 1; i <= mid; i *= 5)     {      // Possible powers of i less than mid is i      for (long long j = 1; j * i <= mid; j *= 3)       {        // possible powers of 3 and 5 such that        // their product is less than mid         // using the power array of 2 (pow) we are        // trying to find the max power of 2 such        // that i*J*power of 2 is less than mid         cnt += upper_bound(pow, pow + 31,                           mid / (i * j)) - pow;      }    }     // If total numbers of ugly number    // less than equal    // to mid is less than n we update l    if (cnt < n)      l = mid + 1;     // If total numbers of ugly number    // less than qual to    // mid is greater than n we update    // r and ans simultaneously.    else      r = mid - 1, ans = mid;  }   return ans;} // Driver Codeint main(){         int n = 150;       // Function Call    cout << nthUglyNumber(n);    return 0;}

Java

 // JAVA program for the above approachimport java.util.*; class GFG{  static int upperBound(int[] a, int low,                        int high, int element)  {    while(low < high)    {      int middle = low + (high - low)/2;      if(a[middle] > element)        high = middle;      else        low = middle + 1;    }    return low;  }   // Print nth Ugly number  static int nthUglyNumber(int n)  {     int pow[] = new int;    Arrays.fill(pow, 1);     // stored powers of 2 from    // Math.pow(2,0) to Math.pow(2,30)    for (int i = 1; i <= 30; ++i)      pow[i] = pow[i - 1] * 2;     // Initialized low and high    int l = 1, r = 2147483647;     int ans = -1;     // Applying Binary Search    while (l <= r) {       // Found mid      int mid = l + ((r - l) / 2);       // cnt stores total numbers of ugly      // number less than mid      int cnt = 0;       // Iterate from 1 to mid      for (long i = 1; i <= mid; i *= 5)       {        // Possible powers of i less than mid is i        for (long j = 1; j * i <= mid; j *= 3)         {          // possible powers of 3 and 5 such that          // their product is less than mid           // using the power array of 2 (pow) we are          // trying to find the max power of 2 such          // that i*J*power of 2 is less than mid           cnt += upperBound(pow,0, 31,                            (int)(mid / (i * j)));        }      }       // If total numbers of ugly number      // less than equal      // to mid is less than n we update l      if (cnt < n)        l = mid + 1;       // If total numbers of ugly number      // less than qual to      // mid is greater than n we update      // r and ans simultaneously.      else {        r = mid - 1; ans = mid;      }    }     return ans;  }   // Driver Code  public static void main(String[] args)  {     int n = 150;     // Function Call    System.out.print(nthUglyNumber(n));  }} // This code is contributed by Rajput-Ji

Python3

 # Python Program to implement# the above approach def upper_bound(a, low, high, element) :    while(low < high) :      middle = low + (high - low)//2      if(a[middle] > element) :        high = middle      else :        low = middle + 1         return low # Prnth Ugly numberdef nthUglyNumber(n) :    pow =  * 40    # stored powers of 2 from  # pow(2,0) to pow(2,30)  for i in range(1, 31):    pow[i] = pow[i - 1] * 2    # Initialized low and high  l = 1  r = 2147483647    ans = -1    # Applying Binary Search  while (l <= r) :      # Found mid    mid = l + ((r - l) // 2)      # cnt stores total numbers of ugly    # number less than mid    cnt = 0      # Iterate from 1 to mid    i = 1    while(i <= mid) :          # Possible powers of i less than mid is i        j = 1        while(j * i <= mid) :            # possible powers of 3 and 5 such that            # their product is less than mid              # using the power array of 2 (pow) we are            # trying to find the max power of 2 such            # that i*J*power of 2 is less than mid              cnt += upper_bound(pow, 0,  31, mid // (i * j))            j *= 3                 i *= 5                  # If total numbers of ugly number    # less than equal    # to mid is less than n we update l    if (cnt < n):      l = mid + 1      # If total numbers of ugly number    # less than qual to    # mid is greater than n we update    # r and ans simultaneously.    else :      r = mid - 1      ans = mid  return ans  # Driver Coden = 150    # Function Callprint(nthUglyNumber(n)) # This code is contributed by code_hunt.

C#

 // C# program for the above approachusing System; public class GFG {  static int upperBound(int[] a, int low,                        int high, int element)  {    while (low < high) {      int middle = low + (high - low) / 2;      if (a[middle] > element)        high = middle;      else        low = middle + 1;    }    return low;  }   // Print nth Ugly number  static int nthUglyNumber(int n) {     int []pow = new int;    for (int i = 0; i <40; ++i)      pow[i] = 1;         // stored powers of 2 from    // Math.Pow(2,0) to Math.Pow(2,30)    for (int i = 1; i <= 30; ++i)      pow[i] = pow[i - 1] * 2;     // Initialized low and high    int l = 1, r = 2147483647;     int ans = -1;     // Applying Binary Search    while (l <= r) {       // Found mid      int mid = l + ((r - l) / 2);       // cnt stores total numbers of ugly      // number less than mid      int cnt = 0;       // Iterate from 1 to mid      for (long i = 1; i <= mid; i *= 5)       {        // Possible powers of i less than mid is i        for (long j = 1; j * i <= mid; j *= 3)         {          // possible powers of 3 and 5 such that          // their product is less than mid           // using the power array of 2 (pow) we are          // trying to find the max power of 2 such          // that i*J*power of 2 is less than mid           cnt += upperBound(pow, 0, 31,                            (int) (mid / (i * j)));        }      }       // If total numbers of ugly number      // less than equal      // to mid is less than n we update l      if (cnt < n)        l = mid + 1;       // If total numbers of ugly number      // less than qual to      // mid is greater than n we update      // r and ans simultaneously.      else {        r = mid - 1;        ans = mid;      }    }     return ans;  }   // Driver Code  public static void Main(String[] args) {     int n = 150;     // Function Call    Console.Write(nthUglyNumber(n));  }}  // This code is contributed by gauravrajput1

Javascript


Output
5832

Time Complexity: O(log N)

Auxiliary Space: O(1)

Please write comments if you find any bug in the above program or other ways to solve the same problem.

My Personal Notes arrow_drop_up