Ambiguous Permutations Codechef Solution

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

Problem

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

Ambiguous Permutations CodeChef Solution

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()) 

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int n,a[100001],i;
    while(1)
    {
        scanf("%d",&n);
        if(n==0)break;
        for(i=1;i<=n;i++)
             scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
        {
            if(a[a[i]]!=i){printf("not ambiguous \n");break;}
            if(i==n)printf("ambiguous \n");
        }
    }
    return 0;
}

Disclaimer: The above Problem (Ambiguous Permutations) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.