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 Program` `int` `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 program` `if` `__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.` |

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 contribute.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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