1830. Minimum Number of Operations to Make String Sorted

# Time:  O(26 * n) = O(n)
# Space: O(max_n) = O(max_n)

inv = [0, 1]


class Solution(object):
    def makeStringSorted(self, s):  # count of prev_permutation
        """
        :type s: str
        :rtype: int
        """
        def inverse(n, m):
            i = len(inv)
            while len(inv) <= n:  # lazy initialization
                inv.append(inv[m%i]*(m-m//i) % m)  # https://cp-algorithms.com/algebra/module-inverse.html
                i += 1
            return inv[n]
    
        MOD = 10**9+7
        count, result, comb_total = [0]*26, 0, 1
        for i in reversed(xrange(len(s))):
            num = ord(s[i])-ord('a') 
            count[num] += 1
            comb_total = (comb_total*(len(s)-i))*inverse(count[num], MOD)
            result = (result + (comb_total*sum(count[:num]))*inverse(len(s)-i, MOD)) % MOD
        return result

Solution from kamyu104/LeetCode-Solutions · MIT