Given an array nums[] of size N, the task is to check whether the given array can be converted into a permutation of 1 to N after performing given operations any number of times (may be 0). An operation is defined as: Pick any element of the array say ‘x’, and replace it with ‘x/2’.
Note: In a permutation of 1 to N all the numbers in range [1, N] are present in any order and each number is present only once.
Examples:
Input: N = 4, nums[] = {1, 8, 25, 2}
Output: true
Explanation: Sequence of operations followed are listed below:
Replace 8 with 8/2 = 4, nums[] = {1, 4, 25, 2}
Replace 25 with 25/2 = 12, nums[] = {1, 4, 12, 2}
Replace 12 with 12/2 = 6, nums[] = {1, 4, 6, 2}
Replace 6 with 6/2 = 3, nums[] = {1, 4, 3, 2} (Here element from 1 to 4 is present exactly once)Input: N = 4, nums[] = {24, 7, 16, 7}
Output: false
Approach: The solution to the problem is based on sorting. Create an array freq of type boolean and size (N+1) to keep track of the numbers of the desired permutation. Follow the below steps:
- Sort the given array.
- Traverse from the back of the array and at each step:
- If val is less than N and freq[val] is false then mark it as true.
- If val is greater than N or freq[val] is true then divide it by 2 while (val > 0 && freq[val] == true)
- At last, check if the desired permutation is achieved or not.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to check is it possible// to make array or notbool possibleOrNot(vector<int>& nums, int N){ // Sorting array sort(nums.begin(), nums.end()); // Initializing freq[] array which // keeps track of elements from 1 to N vector<bool> freq(N + 1, 0); // Iterating from backwards for (int i = N - 1; i >= 0; i--) { int val = nums[i]; // Dividing val by 2 while // it is greater than N // or freq[val] is true while (val > N || (freq[val] && val >= 1)) { val /= 2; } // Updating freq array if (val != 0) freq[val] = true; } // Checking if every element from // 1 to N is present or not for (int i = 1; i < freq.size(); i++) if (!freq[i]) { return false; } return true;}// Driver Codeint main(){ int N = 4; vector<int> nums = { 1, 8, 25, 2 }; bool ans = possibleOrNot(nums, N); cout << (ans); return 0;} // This code is contributed by rakeshsahni |
Java
// Java code to implement the above approachimport java.util.*;class GFG { // Driver Code public static void main(String[] args) { int N = 4; int[] nums = { 1, 8, 25, 2 }; boolean ans = possibleOrNot(nums, N); System.out.println(ans); } // Function to check is it possible // to make array or not public static boolean possibleOrNot(int[] nums, int N) { // Sorting array Arrays.sort(nums); // Initializing freq[] array which // keeps track of elements from 1 to N boolean[] freq = new boolean[N + 1]; // Iterating from backwards for (int i = N - 1; i >= 0; i--) { int val = nums[i]; // Dividing val by 2 while // it is greater than N // or freq[val] is true while (val > N || (freq[val] && val >= 1)) { val /= 2; } // Updating freq array if (val != 0) freq[val] = true; } // Checking if every element from // 1 to N is present or not for (int i = 1; i < freq.length; i++) if (!freq[i]) { return false; } return true; }} |
Python3
# python3 code to implement the above approach# Function to check is it possible# to make array or notdef possibleOrNot(nums, N): # Sorting array nums.sort() # Initializing freq[] array which # keeps track of elements from 1 to N freq = [0 for _ in range(N + 1)] # Iterating from backwards for i in range(N - 1, -1, -1): val = nums[i] # Dividing val by 2 while # it is greater than N # or freq[val] is true while (val > N or (freq[val] and val >= 1)): val //= 2 # Updating freq array if (val != 0): freq[val] = True # Checking if every element from # 1 to N is present or not for i in range(1, len(freq)): if (not freq[i]): return False return True# Driver Codeif __name__ == "__main__": N = 4 nums = [1, 8, 25, 2] ans = possibleOrNot(nums, N) print(ans) # This code is contributed by rakeshsahni |
C#
using System;public class GFG{ // Function to check is it possible // to make array or not static bool possibleOrNot(int[] nums, int N) { // Sorting array Array.Sort(nums); // Initializing freq[] array which // keeps track of elements from 1 to N bool[] freq = new bool[N + 1]; // Iterating from backwards for (int i = N - 1; i >= 0; i--) { int val = nums[i]; // Dividing val by 2 while // it is greater than N // or freq[val] is true while (val > N || (freq[val] && val >= 1)) { val /= 2; } // Updating freq array if (val != 0) freq[val] = true; } // Checking if every element from // 1 to N is present or not for (int i = 1; i < freq.Length; i++) if (!freq[i]) { return false; } return true; } // Driver Code static public void Main (){ int N = 4; int[] nums = { 1, 8, 25, 2 }; bool ans = possibleOrNot(nums, N); if(ans == true) Console.WriteLine(1); else Console.WriteLine(0); }}// This code is contributed by hrithikgrg03188. |
Javascript
<script>// Javascript code to implement the above approach// Function to check is it possible// to make array or notfunction possibleOrNot(nums, N){ // Sorting array nums.sort((a, b) => a - b); // Initializing freq[] array which // keeps track of elements from 1 to N let freq = new Array(N + 1).fill(0) // Iterating from backwards for (let i = N - 1; i >= 0; i--) { let val = nums[i]; // Dividing val by 2 while // it is greater than N // or freq[val] is true while (val > N || (freq[val] && val >= 1)) { val = Math.floor(val / 2); } // Updating freq array if (val != 0) freq[val] = true; } // Checking if every element from // 1 to N is present or not for (let i = 1; i < freq.length; i++) if (!freq[i]) { return false; } return true;}// Driver Codelet N = 4;let nums = [1, 8, 25, 2]let ans = possibleOrNot(nums, N);document.write(ans);// This code is contributed by saurabh_jaiswal.</script> |
Output
true
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
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