Given an array arr containing N positive integers, the task is to check if the given array arr represents a permutation or not.
A sequence of N integers is called a permutation if it contains all integers from 1 to N exactly once.
Examples:
Input: arr[] = {1, 2, 5, 3, 2}
Output: No
Explanation:
The given array contains 2 twice, and 4 is missing for the array to represent a permutation of length 5.
Input: arr[] = {1, 2, 5, 3, 4}
Output: Yes
Explanation:
The given array contains all integers from 1 to 5 exactly once. Hence, it represents a permutation of length 5.
Naive Approach: in O(N2) Time
This approach is mentioned here
Another Approach: in O(N) Time and O(N) Space
This approach is mentioned here.
Efficient Approach: Using HashTable
- Create a HashTable of N size to store the frequency count of each number from 1 to N
- Traverse through the given array and store the frequency of each number in the HashTable.
- Then traverse the HashTable and check if all the numbers from 1 to N have a frequency of 1 or not.
- Print “Yes” if the above condition is True, Else “No”.
Below is the implementation of the above approach:
CPP
// C++ program to decide if an array// represents a permutation or not#include <bits/stdc++.h>using namespace std;// Function to check if an// array represents a permutation or notstring permutation(int arr[], int N){ int hash[N + 1] = { 0 }; // Counting the frequency for (int i = 0; i < N; i++) { hash[arr[i]]++; } // Check if each frequency is 1 only for (int i = 1; i <= N; i++) { if (hash[i] != 1) return "No"; } return "Yes";}// Driver codeint main(){ int arr[] = { 1, 1, 5, 5, 3 }; int n = sizeof(arr) / sizeof(int); cout << permutation(arr, n) << endl; return 0;} |
Java
// Java program to decide if an array// represents a permutation or notclass GFG{ // Function to check if an// array represents a permutation or notstatic String permutation(int arr[], int N){ int []hash = new int[N + 1]; // Counting the frequency for (int i = 0; i < N; i++) { hash[arr[i]]++; } // Check if each frequency is 1 only for (int i = 1; i <= N; i++) { if (hash[i] != 1) return "No"; } return "Yes";} // Driver codepublic static void main(String[] args){ int arr[] = { 1, 1, 5, 5, 3 }; int n = arr.length; System.out.print(permutation(arr, n) +"\n");}}// This code is contributed by Princi Singh |
Python3
# Python3 program to decide if an array# represents a permutation or not# Function to check if an# array represents a permutation or notdef permutation(arr, N) : hash = [0]*(N + 1); # Counting the frequency for i in range(N) : hash[arr[i]] += 1; # Check if each frequency is 1 only for i in range(1, N + 1) : if (hash[i] != 1) : return "No"; return "Yes";# Driver codeif __name__ == "__main__" : arr = [ 1, 1, 5, 5, 3 ]; n = len(arr); print(permutation(arr, n)); # This code is contributed by Yash_R |
C#
// C# program to decide if an array// represents a permutation or notusing System;class GFG{ // Function to check if an // array represents a permutation or not static string permutation(int []arr, int N) { int []hash = new int[N + 1]; // Counting the frequency for (int i = 0; i < N; i++) { hash[arr[i]]++; } // Check if each frequency is 1 only for (int i = 1; i <= N; i++) { if (hash[i] != 1) return "No"; } return "Yes"; } // Driver code public static void Main(string[] args) { int []arr = { 1, 1, 5, 5, 3 }; int n = arr.Length; Console.Write(permutation(arr, n) +"\n"); }}// This code is contributed by Yash_R |
Javascript
<script>// JavaScript program to decide if an array// represents a permutation or not// Function to check if an// array represents a permutation or notfunction permutation(arr, N){ var hash = Array(N+1).fill(0); // Counting the frequency for (var i = 0; i < N; i++) { hash[arr[i]]++; } // Check if each frequency is 1 only for (var i = 1; i <= N; i++) { if (hash[i] != 1) return "No"; } return "Yes";}// Driver codevar arr = [1, 1, 5, 5, 3];var n = arr.length;document.write( permutation(arr, n));</script> |
Output:
No
Time Complexity: O(N)
Auxiliary Space Complexity: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
