LeetCode 1830. Minimum Number of Operations to Make String Sorted

You are given a string s (0-indexed)​​​​​​. You are asked to perform the following operation on s​​​​​​ until you get a sorted string:

1. Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1].

2. Find the largest index j such that i <= j < s.length and s[k] < s[i - 1] for all the possible values of k in the range [i, j] inclusive.

3. Swap the two characters at indices i - 1​​​​ and j​​​​​.

4. Reverse the suffix starting at index i​​​​​​.

Return the number of operations needed to make the string sorted. Since the answer can be too large, return it modulo 109 + 7.

Example 1:

Input: s = "cba"
Output: 5
Explanation: The simulation goes as follows:
Operation 1: i=2, j=2. Swap s[1] and s[2] to get s="cab", then reverse the suffix starting at 2. Now, s="cab".
Operation 2: i=1, j=2. Swap s[0] and s[2] to get s="bac", then reverse the suffix starting at 1. Now, s="bca".
Operation 3: i=2, j=2. Swap s[1] and s[2] to get s="bac", then reverse the suffix starting at 2. Now, s="bac".
Operation 4: i=1, j=1. Swap s[0] and s[1] to get s="abc", then reverse the suffix starting at 1. Now, s="acb".
Operation 5: i=2, j=2. Swap s[1] and s[2] to get s="abc", then reverse the suffix starting at 2. Now, s="abc".

Example 2:

Input: s = "aabaa"
Output: 2
Explanation: The simulation goes as follows:
Operation 1: i=3, j=4. Swap s[2] and s[4] to get s="aaaab", then reverse the substring starting at 3. Now, s="aaaba".
Operation 2: i=4, j=4. Swap s[3] and s[4] to get s="aaaab", then reverse the substring starting at 4. Now, s="aaaab".

Example 3:

Input: s = "cdbea"
Output: 63

Example 4:

Input: s = "leetcodeleetcodeleetcode"
Output: 982157772

Constraints:

• 1 <= s.length <= 3000

• s​​​​​​ consists only of lowercase English letters.

Solution

English Version in Youtube

中文版解答Youtube Link

中文版解答Bilibili Link

class Solution {
    int mod = 1e9 + 7;
    
    long pow(long a, long p, long mod) {
        long ans = 1;
        while (p > 0) {
            if (p & 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
            p = p >> 1;
        }
        return ans;
    }
public:
    
    int makeStringSorted(string s) {
        map<int, int> freq; 
        for (char ch: s) {
            freq[ch - 'a']++;
        }
        
        int n = s.length();
        vector<long> fact(s.size() + 1, 1);
        for (int i = 1; i <= s.size(); i++) {
            fact[i] = (fact[i - 1] * i) % mod;
        }
        
        long ans = 0;
        for (char ch : s) {
            long freq_sum = 0;
            long duplicates = 1;
            for (int i = 0; i < 26; i++) {
                if (i < (ch - 'a')) {
                    freq_sum += freq[i];
                }
                duplicates = (duplicates * fact[freq[i]]) % mod;
            }
            ans += (freq_sum * fact[n - 1] % mod) * pow(duplicates, mod - 2, mod);
            ans %= mod;
            n--;
            freq[ch - 'a']--;
        }
        return ans;
    }
};