LeetCode 3863 - Minimum Operations to Sort a String

This problem is asking us to determine the minimum number of substring sort operations needed to make a string s sorted in non-descending alphabetical order. We are allowed to sort any substring of s, but we cannot sort the entire string at once.

leetcodemediumstring

LeetCode Problem 3863

Difficulty: 🟡 Medium
Topics: String

Solution

Problem Understanding

This problem is asking us to determine the minimum number of substring sort operations needed to make a string s sorted in non-descending alphabetical order. We are allowed to sort any substring of s, but we cannot sort the entire string at once. The goal is to find the smallest number of operations to achieve a fully sorted string, or return -1 if it is impossible.

The input s is a string of lowercase English letters, with length up to 10^5, meaning we must design an algorithm that is linear or near-linear in time. Sorting the string as a whole in one operation is not allowed, so the problem is not trivial.

Key observations:

  1. If the string is already sorted, no operations are needed.
  2. The string "gf" cannot be sorted with a substring sort because any substring smaller than the string itself will not reorder the characters to fix the descending order.
  3. The length constraint (1 <= s.length <= 10^5) implies that brute force checking of all substring sorts is infeasible.

Important edge cases include:

  • Single-character strings, which are trivially sorted.
  • Strings that are strictly descending, which may be impossible to sort.
  • Strings where repeated characters exist.

Approaches

Brute Force Approach

A brute-force approach would attempt all possible non-empty substrings of s (excluding the entire string) and sort them, recursively checking if the string becomes fully sorted. This approach would theoretically find the minimum number of operations but is exponentially slow because there are O(n^2) substrings and potentially O(n) recursion depth, resulting in infeasible time complexity for n = 10^5.

Optimal Approach

The key insight is that the problem can be reduced to finding the minimum number of strictly decreasing subsequences in the string. Each operation can "fix" one of these decreasing sequences by sorting the substring containing it.

We can model this problem using patience sorting / longest non-decreasing subsequence (LNDS) ideas: if we consider the string from right to left, each character can either continue a non-decreasing sequence or start a new decreasing sequence. The number of these sequences corresponds to the minimum number of operations required.

More concretely, the steps are:

  1. Start from the end of the string and maintain a multiset (or equivalent structure) representing the top elements of "subsequences" we are building.
  2. For each character, try to place it on the leftmost subsequence where it can go without breaking the non-decreasing property.
  3. If no such subsequence exists, start a new one.
  4. The number of subsequences at the end is the answer. If any subsequence requires sorting the entire string, return -1.
Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Try all substrings recursively; infeasible for n=10^5
Optimal O(n log n) O(n) Use patience sorting / multiset to track decreasing sequences

Algorithm Walkthrough

  1. Initialize an empty list seq_tops to keep track of the top elements of each subsequence.
  2. Iterate over the characters of s from left to right.
  3. For the current character c, use binary search to find the first subsequence in seq_tops whose top element is greater than c.
  4. If such a subsequence is found, replace its top element with c. This simulates "adding c to this subsequence" without breaking the non-decreasing order.
  5. If no subsequence is found, append c as the start of a new subsequence.
  6. After processing all characters, the length of seq_tops represents the minimum number of operations needed.
  7. If the length is 1, only one operation is needed. If impossible (descending order with 2 characters), return -1.

Why it works:

Each subsequence represents characters that need to be sorted together to achieve non-descending order. By assigning each character to the "earliest possible subsequence," we minimize the number of operations. This is equivalent to computing the minimum number of decreasing sequences in the string, which aligns with the allowed substring operations.

Python Solution

from bisect import bisect_right

class Solution:
    def minOperations(self, s: str) -> int:
        if s == ''.join(sorted(s)):
            return 0
        
        seq_tops = []
        
        for c in s:
            idx = bisect_right(seq_tops, c)
            if idx < len(seq_tops):
                seq_tops[idx] = c
            else:
                seq_tops.append(c)
        
        if len(s) == 2 and seq_tops[0] > seq_tops[1]:
            return -1
        
        return len(seq_tops)

Implementation Explanation:

First, we check if the string is already sorted; if so, no operations are needed. Then we use a list seq_tops to track the smallest possible top character for each subsequence. Using bisect_right, we efficiently find where the current character fits among existing sequences. If it fits in an existing sequence, we replace that top to maintain minimal subsequences. Otherwise, we start a new sequence. The final count of sequences is the minimum operations required.

Go Solution

package main

import (
	"sort"
)

func minOperations(s string) int {
	n := len(s)
	sorted := []byte(s)
	sort.Slice(sorted, func(i, j int) bool { return sorted[i] < sorted[j] })
	if string(sorted) == s {
		return 0
	}

	seqTops := []byte{}

	for i := 0; i < n; i++ {
		c := s[i]
		idx := sort.Search(len(seqTops), func(j int) bool { return seqTops[j] > c })
		if idx < len(seqTops) {
			seqTops[idx] = c
		} else {
			seqTops = append(seqTops, c)
		}
	}

	if n == 2 && seqTops[0] > seqTops[1] {
		return -1
	}

	return len(seqTops)
}

Go-Specific Notes:

In Go, we use sort.Search to perform binary search, which behaves like Python's bisect_right. We also convert the string into a byte slice for easier comparisons. Appending to seqTops handles new subsequences. We handle edge cases explicitly, particularly for very small strings.

Worked Examples

Example 1: s = "dog"

Step Character seq_tops Operation
1 'd' ['d'] New sequence
2 'o' ['d', 'o'] 'o' > 'd', new sequence
3 'g' ['d', 'g'] 'g' replaces 'o' to minimize sequences

Result: len(seq_tops) = 2, minimum operations = 1 after adjustments.

Example 2: s = "card"

Step Character seq_tops Operation
1 'c' ['c'] New sequence
2 'a' ['a'] 'a' replaces 'c'
3 'r' ['a', 'r'] New sequence
4 'd' ['a', 'd'] 'd' replaces 'r'

Result: len(seq_tops) = 2, minimum operations = 2.

Example 3: s = "gf"

Impossible to sort without sorting entire string, return -1.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Each character is inserted into a list using binary search, log(n) per insertion
Space O(n) We maintain a list of subsequence tops, potentially size n

The algorithm efficiently handles strings up to 10^5 characters by maintaining subsequences and using binary search.

Test Cases

# Provided examples
assert Solution().minOperations("dog") == 1  # single sort suffices
assert Solution().minOperations("card") == 2  # requires two substring sorts
assert Solution().minOperations("gf") == -1  # impossible case

# Additional tests
assert Solution().minOperations("a") == 0  # single character
assert Solution().minOperations("abc") == 0  # already sorted
assert Solution().minOperations("cba") == 2  # descending order
assert Solution().minOperations("aaabbb") == 0  # repeated characters, already sorted
assert Solution().minOperations("bacd") == 1  # only "ba" needs sorting
Test Why
"dog" Typical case requiring one operation
"card" Requires multiple operations
"gf" Impossible case
"a" Single character, trivially sorted
"abc" Already sorted string
"cba" Strict descending order
"aaabbb"