NOVICE43 - Problem 3
When I first learned backtracking I made a program to find all the permutations of the English alphabets in lexicographically increasing. Filled with elation I showed the program to Rohil. Rohil being someone who likes to do stuff off the league was not impressed and gave me the following variation of the problem help me to solve the problem:
You have to find the number of permutations of length N(1<=N<=11) such that at whenever an alphabet (say 'c' ) appears in the permutation all the alphabets smaller than 'c' should have appeared before it at least once. An alphabet is smaller than another if it appears before the other in the English alphabet. ‘a’ being the smallest and ‘z’ being the largest. For example when N=2 then aa,ab are the only valid permutations and ba,bb is invalid since in ba all the alphabets smaller than b have not appeared at least once before it. See example for further clarification.
Line 1: T(no. of test cases)
Line 2: n1
Line 3: n2
Line 1: no. of such permutations of length n1
Input: 2 2 3 Output: 2 5
Explanation for N=3, the possible permutations are: abc aba abb aab aaa
AC in one go.!!
Basically, there are 11 possible values of n, which means you can do the backtrack. Or you can precalculate the answer and get AC in O(1).
Answer for n=11 will be 6785**Last edit: 2017-12-15 12:40:28
@Mahesh_Chandra_Sharma why its showing tle...with n^2, or O(11*11), infact it is showing tle also when i am storing the value of ans for all n=11 and then just using that array to output could u check my submissions
what are the permuations for N=4 ? can someone list them out please?
Easy one.. ;)
The answer for N=11 is 6785**