Given an integer array arr[], the task is to sort only those elements which are ugly numbers at their relative positions in the array (positions of other elements must not be affected).
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 first few ugly numbers. By convention, 1 is included.
Examples:
Input: arr[] = {13, 9, 11, 3, 2}
Output: 13 2 11 3 9
9, 3 and 2 are the only ugly numbers in the given array.
Input: arr[] = {1, 2, 3, 7, 12, 10}
Output: 1 2 3 7 10 12
Approach:
- Start traversing the array and for every element arr[i], if arr[i] is an ugly number then store it in an ArrayList and update arr[i] = -1
- After all the ugly numbers have been stored in the ArrayList, sort the updated ArrayList.
- Again traverse the array and for every element,
- If arr[i] = -1 then print the first element from the ArrayList that hasn’t been printed before.
- Else, print arr[i].
Below is the implementation of the above approach:
C++
// CPP implementation of the approach#include<bits/stdc++.h>using namespace std;// Function that returns true if n is an ugly numberbool isUgly(int n){ // While divisible by 2, keep dividing while (n % 2 == 0) n = n / 2; // While divisible by 3, keep dividing while (n % 3 == 0) n = n / 3; // While divisible by 5, keep dividing while (n % 5 == 0) n = n / 5; // n must be 1 if it was ugly if (n == 1) return true; return false;}// Function to sort ugly numbers// in their relative positionsvoid sortUglyNumbers(int arr[], int n){ // To store the ugly numbers from the array vector<int> list; int i; for (i = 0; i < n; i++) { // If current element is an ugly number if (isUgly(arr[i])) { // Add it to the ArrayList // and set arr[i] to -1 list.push_back(arr[i]); arr[i] = -1; } } // Sort the ugly numbers sort(list.begin(),list.end()); int j = 0; for (i = 0; i < n; i++) { // Position of an ugly number if (arr[i] == -1) cout << list[j++] << " "; else cout << arr[i] << " "; }}// Driver codeint main(){ int arr[] = { 1, 2, 3, 7, 12, 10 }; int n = sizeof(arr)/sizeof(arr[0]); sortUglyNumbers(arr, n);}// This code is contributed by// Surendra_Gangwar |
Java
// Java implementation of the approachimport java.util.ArrayList;import java.util.Collections;class GFG { // Function that returns true if n is an ugly number static boolean isUgly(int n) { // While divisible by 2, keep dividing while (n % 2 == 0) n = n / 2; // While divisible by 3, keep dividing while (n % 3 == 0) n = n / 3; // While divisible by 5, keep dividing while (n % 5 == 0) n = n / 5; // n must be 1 if it was ugly if (n == 1) return true; return false; } // Function to sort ugly numbers // in their relative positions static void sortUglyNumbers(int arr[], int n) { // To store the ugly numbers from the array ArrayList<Integer> list = new ArrayList<>(); int i; for (i = 0; i < n; i++) { // If current element is an ugly number if (isUgly(arr[i])) { // Add it to the ArrayList // and set arr[i] to -1 list.add(arr[i]); arr[i] = -1; } } // Sort the ugly numbers Collections.sort(list); int j = 0; for (i = 0; i < n; i++) { // Position of an ugly number if (arr[i] == -1) System.out.print(list.get(j++) + " "); else System.out.print(arr[i] + " "); } } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 3, 7, 12, 10 }; int n = arr.length; sortUglyNumbers(arr, n); }} |
Python3
# Python3 implementation of the approach # Function that returns true if n is an ugly number def isUgly(n): # While divisible by 2, keep dividing while n % 2 == 0: n = n // 2 # While divisible by 3, keep dividing while n % 3 == 0: n = n // 3 # While divisible by 5, keep dividing while n % 5 == 0: n = n // 5 # n must be 1 if it was ugly if n == 1: return True return False# Function to sort ugly numbers # in their relative positions def sortUglyNumbers(arr, n): # To store the ugly numbers from the array list = [] for i in range(0, n): # If current element is an ugly number if isUgly(arr[i]): # Add it to the ArrayList # and set arr[i] to -1 list.append(arr[i]) arr[i] = -1 # Sort the ugly numbers list.sort() j = 0 for i in range(0, n): # Position of an ugly number if arr[i] == -1: print(list[j], end = " ") j += 1 else: print(arr[i], end = " ") # Driver code if __name__ == "__main__": arr = [1, 2, 3, 7, 12, 10] n = len(arr) sortUglyNumbers(arr, n) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG { // Function that returns true // if n is an ugly number static bool isUgly(int n) { // While divisible by 2, keep dividing while (n % 2 == 0) n = n / 2; // While divisible by 3, keep dividing while (n % 3 == 0) n = n / 3; // While divisible by 5, keep dividing while (n % 5 == 0) n = n / 5; // n must be 1 if it was ugly if (n == 1) return true; return false; } // Function to sort ugly numbers // in their relative positions static void sortUglyNumbers(int []arr, int n) { // To store the ugly numbers from the array List<int> list = new List<int>(); int i; for (i = 0; i < n; i++) { // If current element is an ugly number if (isUgly(arr[i])) { // Add it to the ArrayList // and set arr[i] to -1 list.Add(arr[i]); arr[i] = -1; } } // Sort the ugly numbers list.Sort(); int j = 0; for (i = 0; i < n; i++) { // Position of an ugly number if (arr[i] == -1) Console.Write(list[j++] + " "); else Console.Write(arr[i] + " "); } } // Driver code public static void Main(String[] args) { int []arr = { 1, 2, 3, 7, 12, 10 }; int n = arr.Length; sortUglyNumbers(arr, n); }}// This code contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if n is an ugly numberfunction isUgly(n){ // While divisible by 2, keep dividing while (n % 2 == 0) n = n / 2; // While divisible by 3, keep dividing while (n % 3 == 0) n = n / 3; // While divisible by 5, keep dividing while (n % 5 == 0) n = n / 5; // n must be 1 if it was ugly if (n == 1) return true; return false;}// Function to sort ugly numbers// in their relative positionsfunction sortUglyNumbers(arr, n){ // To store the ugly numbers from the array var list = []; var i; for (i = 0; i < n; i++) { // If current element is an ugly number if (isUgly(arr[i])) { // Add it to the ArrayList // and set arr[i] to -1 list.push(arr[i]); arr[i] = -1; } } // Sort the ugly numbers list.sort((a,b)=>a-b); var j = 0; for (i = 0; i < n; i++) { // Position of an ugly number if (arr[i] == -1) document.write( list[j++] + " "); else document.write( arr[i] + " "); }}// Driver codevar arr = [1, 2, 3, 7, 12, 10 ];var n = arr.length;sortUglyNumbers(arr, n);</script> |
Output:
1 2 3 7 10 12
Time Complexity: O(nlog(n))
Auxiliary Space: O(n)
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