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)
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