Algorithm
Problem link-
PERMUT2 - Ambiguous Permutations
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
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
import java.util.*;
import java.lang.*;
class Main
{
public static void main(String args[]){
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t != 0){
int[] input = new int[t];
for(int i= 0; i < t; i++){
input[i] = s.nextInt();
}
int i;
for( i = 1; i <= t/2; i++){
if(i != input[input[i-1]-1]){
System.out.println("not ambiguous");
break;
}
}
if(i>t/2)
System.out.println("ambiguous");
t = s.nextInt();
}
}
}Input
1 4 3 2
5
2 3 4 5 1
1
1
0
Output
not ambiguous
ambiguous
#2 Code Example with C++ Programming
Code -
C++ Programming
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int n;
while(1)
{
scanf("%d",&n);
if(n==0)
return 0;
int i;
int a[n+1];
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
{
if(a[a[i]]!=i)
{
break;
}
}
if(i!=n+1)
printf("not ambiguous\n");
else
printf("ambiguous\n");
}
return 0;
}start coding...
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Input
1 4 3 2
5
2 3 4 5 1
1
1
0
Output
not ambiguous
ambiguous
Demonstration
SPOJ Solution-Ambiguous Permutations-Solution in C, C++, Java, Python