## Permutation PRMA SOLUTIONS BRBG2020 Code Chef

Rohit collects coins: he has exactly one coin for every year from 1 to n. Naturally, Rohit keeps all the coins in his collection in the order in which they were released. Once Rohit’s younger brother made a change — he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that, the segment’s endpoints did not coincide. For example, if n = 8, then initially Rohit’s coins were kept in the order 1 2 3 4 5 6 7 8. If Rohit’s younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Rohit suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 … n using exactly one segment reversal. If it is possible, find the segment itself.

Input:

The first line contains an integer N which is the number of coins in Rohit’s collection.

The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.

Output:

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l, r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 … n the given one.

Constraints

1≤N≤1000

1≤A[N]≤109

Sample Input:

8

1 6 5 4 3 2 7 8

Sample Output:

2 6

**SOLUTION:**

#CODE IN PYTHON3.6n=int(input())a=list(map(int,input().split()))l,r=-1,-1for i in range(n):if a[i]!=i+1:l=ibreakfor i in range(n-1,-1,-1):if a[i]!=i+1:r=ibreakj=r+1for i in range(l,r+1):if a[i]==j:j-=1continueelse:print(0,0)exit(0)print(l+1,r+1)

**Contest Link: https://www.codechef.com/BRBG2020/problems/PRMA**

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