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Hello!,

I tried to use the code for deleting k digits but it seems it fails in some test cases.

Given num = 891, k =1.
Output 89

I think the correct solution is 81 instead. I've a similar code to get the min/max number by removing k digits, this will return the correct solution.

Please let me know if i'm missing something here.

Thanks

finish the problem in this repo: graph from links, Eulerian path, change return program. credit card validator, dijkstra’s algorithm , count words in a string, factor finder, coin flip simulation, closest pair problem

spreadsheet, Cover board with tile, Divide integers, Pizza picking problem, Boggle puzzle, Kth element in two sorted arrays, Celebrity problem, Recover the queue, Number of closed rectangles in a matrix, Check whether a tree exists, Trim binary search tree, Flip a binary tree, Climb ladder, Flip bits, Pair with largest common words, Reverse index search

I came across 2 problem in your approach:

1. If you are doing it O(n) then you could do a simple scan and just minimum value and then return its index. This could have avoided your complex code.
2. Time complexity is O(log n):
   http://stackoverflow.com/questions/2796413/binary-search-to-find-the-rotation-point-in-a-rotated-sorted-list

I don't know how big you expect m to be, but if m is an int, n is long long and k is int - there's no need for simulating a big number via a string (also, there are better ways to create big numbers).

The solution code can be

long long get_m_to_nth_mod_10_to_kth(int m, long long n, int k)
{
    long long exponent = std::pow(10L, k);
    if (n == 1)
        return m % exponent;
    if (n%2) //odd
        return (get_m_to_nth_mod_10_to_kth(m,n-1,k) * m) % exponent;
    else
        return  static_cast<long long>(std::pow(get_m_to_nth_mod_10_to_kth(m,n/2,k), 2L)) % exponent;
    
}