Leetcode #1830: Minimum Number of Operations to Make String Sorted
In this guide, we solve Leetcode #1830 Minimum Number of Operations to Make String Sorted in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Math, String, Combinatorics
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
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".
Python Solution
n = 3010
mod = 10**9 + 7
f = [1] + [0] * n
g = [1] + [0] * n
for i in range(1, n):
f[i] = f[i - 1] * i % mod
g[i] = pow(f[i], mod - 2, mod)
class Solution:
def makeStringSorted(self, s: str) -> int:
cnt = Counter(s)
ans, n = 0, len(s)
for i, c in enumerate(s):
m = sum(v for a, v in cnt.items() if a < c)
t = f[n - i - 1] * m
for v in cnt.values():
t = t * g[v] % mod
ans = (ans + t) % mod
cnt[c] -= 1
if cnt[c] == 0:
cnt.pop(c)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.