Hello coders, today we are going to solve Ambiguous Permutations CodeChef Solution whose Problem Code is PERMUT2.

Task
Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.
Input Specification
The input contains several test cases.
The first line of each test case contains an integer n (1 ≤ n ≤ 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.
Output Specification
For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.
Sample Input
4
1 4 3 2
5
2 3 4 5 1
1
1
0
Sample Output
ambiguous
not ambiguous
ambiguous
Solution – Ambiguous Permutations | CodeChef Solution
C++
#include <iostream> #include <stdlib.h> using namespace std; int main() { // your code goes here while(1) { long int n, i, flag = 0; cin >> n; if(!n) break; long int arr[n]; for (i = 0; i < n; i++) cin >> arr[i]; for (i = 0; i < n; i++) if (i + 1 != arr[arr[i] - 1]) { flag = -1; break; } if (flag) cout << "not ambiguous\n"; else cout << "ambiguous\n"; } return 0; }
Python
#Solution Provided by CodingBroz n = int(input()) while(n != 0): li = [int(i) for i in input().split()] li1 = list('0' * len(li)) for i in range(len(li)): li1[li[i] - 1] = i + 1 if(li == li1): print("ambiguous") else: print("not ambiguous") n = int(input())
Java
/* package codechef; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Codechef { public static void main (String[] args) throws java.lang.Exception { // your code goes here Scanner sc = new Scanner(System.in); while(true) { int N = sc.nextInt(); int a[] = new int[N]; int b[] = new int[N]; int count = 0; for(int i = 0; i<N; i++) { a[i] = sc.nextInt(); b[a[i]-1] = i+1; } if(N == 0) { break; } for(int i = 0; i<N; i++) { if(a[i] == b[i]) { count++; } } if(count == N) { System.out.println("ambiguous"); } else { System.out.println("not ambiguous"); } } } }
Disclaimer: The above Problem (Ambiguous Permutations) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.