#3398

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Smallest Substring With Identical Characters I

Minimize the length of the longest substring with identical characters after at most numOps changes in a binary string.

Problem Statement

You are given a binary string s of length n and an integer numOps. Your goal is to minimize the length of the longest contiguous substring of s such that all the characters in the substring are identical.

You are allowed to change up to numOps characters in s to either '0' or '1'. After making these changes, return the minimized length of the longest substring with identical characters.

Examples

Example 1

Input: s = "000001", numOps = 1

Output: 2

By changing s[2] to '1' , s becomes "001001" . The longest substrings with identical characters are s[0..1] and s[3..4] .

Example 2

Input: s = "0000", numOps = 2

Output: 1

By changing s[0] and s[2] to '1' , s becomes "1010" .

Example 3

Input: s = "0101", numOps = 0

Output: 1

Example details omitted.

Constraints

1 <= n == s.length <= 1000
s consists only of '0' and '1'.
0 <= numOps <= n

Solution Approach

Binary Search Over Answer Space

Perform binary search to find the smallest possible value for the longest substring length. The valid range for the substring length is between 1 and the length of the string.

Greedy Sliding Window

For each candidate length from the binary search, use a sliding window approach to check if it's possible to form substrings of that length with at most numOps changes.

Optimization with Prefix Sums

Use prefix sums to efficiently count the number of changes required to make all characters in a window identical. This can speed up the greedy check for each binary search iteration.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n), where n is the length of the string, due to the binary search over possible substring lengths and the linear scan required for each candidate length. The space complexity is O(n) due to the prefix sum array and other auxiliary data structures used for checking the sliding window.

What Interviewers Usually Probe

Look for a clear understanding of binary search and sliding window techniques.
Evaluate the ability to optimize the approach with prefix sums or similar techniques.
Check if the candidate can adapt their solution based on the constraints, particularly with larger strings.

Common Pitfalls or Variants

Common pitfalls

Failing to implement binary search correctly over the answer space.
Not optimizing the sliding window check for each candidate substring length.
Mismanaging the number of allowed operations, leading to incorrect checks or results.

Follow-up variants

Increase the string size n to test scalability of the solution.
Add constraints on numOps to limit the number of changes even further.
Consider different initial configurations of the binary string, such as alternating or uniform strings.

FAQ

How can binary search be applied to this problem?

Binary search is applied over the possible lengths of the longest identical substring. The goal is to minimize this length by checking feasibility using sliding windows.

What is the optimal time complexity for this problem?

The optimal time complexity is O(n log n), achieved by using binary search over the answer space and a sliding window to validate each candidate length.

What does the sliding window technique do here?

The sliding window technique is used to check if we can form a substring of a given length with at most numOps changes. It slides through the string while counting the number of required operations.

What are the edge cases for this problem?

Edge cases include strings that are already optimal, strings with no possible changes (numOps = 0), and strings where numOps is equal to the length of the string.

How does GhostInterview help with this problem?

GhostInterview provides structured hints for solving this problem using binary search and sliding windows, ensuring a deep understanding of the solution pattern.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minLength(self, s: str, numOps: int) -> int:
        def check(m: int) -> bool:
            cnt = 0
            if m == 1:
                t = "01"
                cnt = sum(c == t[i & 1] for i, c in enumerate(s))
                cnt = min(cnt, n - cnt)
            else:
                k = 0
                for i, c in enumerate(s):
                    k += 1
                    if i == len(s) - 1 or c != s[i + 1]:
                        cnt += k // (m + 1)
                        k = 0
            return cnt <= numOps

        n = len(s)
        return bisect_left(range(n), True, lo=1, key=check)

Continue Practicing

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