Given an array of n elements. The task is to check whether a permutation of given array exists, such that each element indicate number of element present before or after it. Print “Yes” if exists, else print “No”.
Examples :
Input : arr[] = {1, 3, 3, 2}
Output : Yes
{3, 1, 2, 3} is a permutation in which each element
indicate number of element present before or after it.
There is one more permutation {3, 2, 1, 3}
Input : arr[] = {4, 1, 2, 3, 0}
Output : Yes
There are two permutations {0, 1, 2, 3, 4} or
{4, 3, 2, 1, 0}
The idea is to use hashing. Observe, for each index i in the array, arr[i] can have value i or n – i. We traverse the given array and find the frequency of each element present in the array. Now, for each index i, check availability of value i and n-i and accordingly decrement the frequency. Note that an item with value i can either go to index i or n-i-1. Similarly, an item with value n-i-1 can go to either index i or n-i-1.
Below is the implementation of this approach:
C++
// C++ program to check if array permutation// exists such that each element indicates// either number of elements before or after// it.#include <bits/stdc++.h>using namespace std;// Check if array permutation exist such that// each element indicate either number of// elements before or after it.bool check(int arr[], int n){ map<int, int> freq; // Finding the frequency of each number. for (int i = 0; i < n; i++) freq[arr[i]]++; for (int i = 0; i < n; i++) { // Try to find number of element before // the current index. if (freq[i]) freq[i]--; // Try to find number of element after // the current index. else if (freq[n-i-1]) freq[n-i-1]--; // If no such number find, return false. else return false; } return true;}// Driven Programint main(){ int arr[] = {1, 3, 3, 2}; int n = sizeof(arr)/sizeof(arr[0]); check(arr, n)? (cout << "Yes" << endl) : (cout << "No" << endl); return 0;} |
Java
// Java program to check if array permutation// exists such that each element indicates// either number of elements before or after// it.import java.io.*;class GFG{ // Check if array permutation exist such that // each element indicate either number of // elements before or after it. public static boolean check(int arr[]) { int n = arr.length; int[] freq = new int[n]; // Finding the frequency of each number. for (int i = 0; i < n; i++) freq[arr[i]]++; for (int i = 0; i < n; i++) { // Try to find number of element before // the current index. if (freq[i]!= 0) freq[i]--; // Try to find number of element after // the current index. else if (freq[n-i-1]!= 0) freq[n-i-1]--; // If no such number find, return false. else return false; } return true; } //Driver program public static void main (String[] args) { int arr[] = {1, 3, 3, 2}; boolean bool = check(arr); if(bool) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python program to check if array permutation# exists such that each element indicates# either number of elements before or after# it.# Check if array permutation exist such that# each element indicate either number of# elements before or after it.def check(arr): n = len(arr); freq = [0] * n; # Finding the frequency of each number. for i in range(n): freq[arr[i]] += 1; for i in range(n): # Try to find number of element before # the current index. if (freq[i] != 0): freq[i] -= 1; # Try to find number of element after # the current index. elif (freq[n - i - 1] != 0): freq[n - i - 1] -= 1; # If no such number find, return false. else: return False; return True; # Driver programif __name__ == '__main__': arr = [1, 3, 3, 2]; bool = check(arr); if(bool): print("Yes"); else: print("No");# This code is contributed by 29AjayKumar |
C#
// C# program to check if array permutation// exists such that each element indicates// either number of elements before or after it.using System;class GFG{ // Check if array permutation exist such that // each element indicate either number of // elements before or after it. public static bool check(int []arr) { int n = arr.Length; int []freq = new int[n]; // Finding the frequency of each number. for (int i = 0; i < n; i++) freq[arr[i]]++; for (int i = 0; i < n; i++) { // Try to find number of element // before the current index. if (freq[i]!= 0) freq[i]--; // Try to find number of element // after the current index. else if (freq[n-i-1] != 0) freq[n-i-1]--; // If no such number find, return false. else return false; } return true; } //Driver program public static void Main () { int []arr = {1, 3, 3, 2}; bool boo = check(arr); if(boo) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by nitin mittal. |
Javascript
<script> // JavaScript program to check if array permutation // exists such that each element indicates // either number of elements before or after it. // Check if array permutation exist such that // each element indicate either number of // elements before or after it. function check(arr) { var n = arr.length; var freq = new Array(n).fill(0); // Finding the frequency of each number. for (var i = 0; i < n; i++) freq[arr[i]]++; for (var i = 0; i < n; i++) { // Try to find number of element // before the current index. if (freq[i] !== 0) freq[i]--; // Try to find number of element // after the current index. else if (freq[n - i - 1] !== 0) freq[n - i - 1]--; // If no such number find, return false. else return false; } return true; } // Driver program var arr = [1, 3, 3, 2]; var boo = check(arr); if (boo) document.write("Yes"); else document.write("No"); // This code is contributed by rdtank. </script> |
Output:
Yes
Time Complexity: O(n).
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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