Introduction

In this blog post, we will explore how to solve the LeetCode problem "1830" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a string s (0-indexed)​​​​​​. You are asked to perform the following operation on s​​​​​​ until you get a sorted string: Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1]. 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. Swap the two characters at indices i - 1​​​​ and j​​​​​. 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". Constraints: 1 <= s.length <= 3000 s​​​​​​ consists only of lowercase English letters.

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • The problem essentially asks for the number of permutations required to sort a given string using a specific algorithm. This is related to finding the lexicographical index of the given string among all its permutations.
  • We can iterate through the string from left to right. For each character, count the number of characters to its right that are smaller than it. This count helps determine how many permutations are lexicographically smaller.
  • To avoid redundant calculations, we can use a frequency count of the remaining characters to efficiently count smaller characters and update the counts after each step.

• Complexity:

  • Runtime Complexity: O(n^2), where n is the length of the string.
  • Storage Complexity: O(n), where n is the length of the string (for frequency counts).

Code

    def solve():
    s = input()
    n = len(s)
    ans = 0
    mod = 10**9 + 7

    for i in range(n):
        counts = {}
        for j in range(i + 1, n):
            if s[j] < s[i]:
                counts[s[j]] = counts.get(s[j], 0) + 1

        smaller_count = sum(counts.values())

        factorial = 1
        remaining_length = n - i - 1
        for k in range(1, remaining_length + 1):
            factorial = (factorial * k) % mod

        denominator = 1
        for char in set(s[i+1:]):
          count = s[i+1:].count(char)
          fact = 1
          for k in range(1, count + 1):
            fact = (fact * k) % mod
          denominator = (denominator * fact) % mod


        def mod_inverse(a, m):
            m0 = m
            y = 0
            x = 1

            if (m == 1):
                return 0

            while (a > 1):
                q = a // m
                t = m
                m = a % m
                a = t
                t = y
                y = x - q * y
                x = t

            if (x < 0):
                x = x + m0

            return x

        inverse_denominator = mod_inverse(denominator, mod)

        ans = (ans + (smaller_count * factorial * inverse_denominator) % mod) % mod

    print(ans)

solve()