LeetCode 3398 - Smallest Substring With Identical Characters I

The problem gives us a binary string s consisting of only '0' and '1' characters and an integer numOps representing the maximum number of bit flips we can perform.

leetcodehardarraybinary-searchenumeration

LeetCode Problem 3398

Difficulty: 🔴 Hard
Topics: Array, Binary Search, Enumeration

Solution

Problem Understanding

The problem gives us a binary string s consisting of only '0' and '1' characters and an integer numOps representing the maximum number of bit flips we can perform. The objective is to minimize the length of the longest contiguous substring where all characters are identical after performing at most numOps flips.

In simpler terms, we are trying to "break" long blocks of consecutive '0's or '1's by flipping at most numOps bits to reduce the size of the largest homogeneous segment. The input string can be up to length 1000, so any algorithm significantly worse than O(n²) might become slow. The string always contains valid binary characters, and numOps ranges from 0 to the length of the string, so we must handle cases where no operations are allowed or where we can flip every bit.

Important edge cases include:

  • numOps = 0, where no flips are allowed, and the longest homogeneous segment is determined purely by the input.
  • Strings where all characters are the same initially, e.g., "0000".
  • numOps is large enough to potentially break all long blocks, e.g., numOps >= n/2.

Approaches

Brute Force Approach

The brute-force solution would try every possible combination of numOps flips, compute the resulting string, and determine the length of the longest homogeneous substring. Then we would take the minimum across all combinations. While correct, this approach is impractical because the number of ways to choose indices to flip is combinatorial (C(n, numOps)), which becomes infeasible for n up to 1000.

Optimal Approach

The key insight is that we do not need to try all combinations explicitly. Instead, we can use binary search on the answer combined with a sliding window or prefix-sum approach. Let L be the maximum length of a homogeneous substring after flips. The problem reduces to checking: can we ensure all blocks are at most length L with at most numOps flips?

To check efficiently:

  1. Identify all contiguous blocks of identical characters.
  2. For each block longer than L, compute how many flips are needed to break it into sub-blocks of size at most L.
  3. Sum these required flips for all blocks. If the total is ≤ numOps, then L is feasible.

We then perform binary search on L from 1 to n to find the minimum feasible length. This reduces complexity from combinatorial to O(n log n).

Approach Time Complexity Space Complexity Notes
Brute Force O(C(n, numOps) * n) O(n) Try all flip combinations; infeasible for large n
Optimal O(n log n) O(n) Binary search on maximum block length + prefix sums/sliding window to check feasibility

Algorithm Walkthrough

  1. Compute the lengths of consecutive blocks of identical characters and store them in a list.
  2. Define a function canAchieve(L) that checks if it is possible to break all blocks so that no block exceeds length L using at most numOps flips. For each block longer than L, compute the number of flips required as ceil(block_length / (L + 1)) - 1.
  3. Perform binary search on L between 1 and n. If canAchieve(mid) is true, try a smaller L; otherwise, try a larger L.
  4. Return the minimum feasible L after binary search completes.

Why it works: The algorithm relies on the observation that breaking blocks into smaller segments of size at most L requires flipping specific bits at intervals, and we can compute the total flips deterministically. Binary search ensures that we find the smallest L efficiently.

Python Solution

class Solution:
    def minLength(self, s: str, numOps: int) -> int:
        n = len(s)
        if n == 0:
            return 0

        # Compute blocks of identical characters
        blocks = []
        i = 0
        while i < n:
            j = i
            while j < n and s[j] == s[i]:
                j += 1
            blocks.append(j - i)
            i = j

        # Helper function to check if max block length L is achievable
        def canAchieve(L: int) -> bool:
            flips_needed = 0
            for block_len in blocks:
                if block_len > L:
                    flips_needed += (block_len - 1) // (L + 1)
            return flips_needed <= numOps

        # Binary search on the answer
        left, right = 1, n
        ans = n
        while left <= right:
            mid = (left + right) // 2
            if canAchieve(mid):
                ans = mid
                right = mid - 1
            else:
                left = mid + 1
        return ans

Implementation Explanation: We first transform the string into a list of consecutive block lengths. The canAchieve function efficiently computes flips needed per block using integer division. Binary search is applied to find the minimum feasible maximum block length.

Go Solution

func minLength(s string, numOps int) int {
    n := len(s)
    if n == 0 {
        return 0
    }

    // Compute blocks of identical characters
    blocks := []int{}
    for i := 0; i < n; {
        j := i
        for j < n && s[j] == s[i] {
            j++
        }
        blocks = append(blocks, j - i)
        i = j
    }

    // Helper function to check feasibility of max block length L
    canAchieve := func(L int) bool {
        flips := 0
        for _, block := range blocks {
            if block > L {
                flips += (block - 1) / (L + 1)
            }
        }
        return flips <= numOps
    }

    left, right, ans := 1, n, n
    for left <= right {
        mid := (left + right) / 2
        if canAchieve(mid) {
            ans = mid
            right = mid - 1
        } else {
            left = mid + 1
        }
    }
    return ans
}

Go-Specific Notes: Slices are used to store block lengths. Integer division naturally computes the number of flips needed per block. No pointer handling or special cases for nil slices are needed as the slice is always initialized.

Worked Examples

Example 1: s = "000001", numOps = 1

Step Blocks L tested Flips needed Result
Compute blocks [5,1] 3 (5-1)//(3+1)=1 feasible
Binary search mid=2 2 (5-1)//(2+1)=1 feasible
Answer 2 - - 2

Example 2: s = "0000", numOps = 2

Step Blocks L tested Flips needed Result
Compute blocks [4] 2 (4-1)//(2+1)=1 feasible
Binary search mid=1 1 (4-1)//(1+1)=1 feasible
Answer 1 - - 1

Example 3: s = "0101", numOps = 0

Step Blocks L tested Flips needed Result
Compute blocks [1,1,1,1] 1 0 feasible
Answer 1 - - 1

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) O(n) to compute blocks and O(log n) iterations of binary search
Space O(n) Store block lengths in a list of size up to n

The complexity is efficient for n ≤ 1000 and scales well due to logarithmic binary search combined with linear block computation.

Test Cases

# Provided examples
assert Solution().minLength("000001", 1) == 2  # flips 1 bit to reduce max block
assert Solution().minLength("0000", 2) == 1    # enough flips to make alternating pattern
assert Solution().minLength("0101", 0) == 1    # no flips allowed, longest block is 1

# Additional edge cases
assert Solution().minLength("1111", 0) == 4    # no flips, string uniform
assert Solution().minLength("1111", 2) == 1    # flips sufficient to break into single bits
assert Solution().minLength("101010", 0) == 1  # already alternating
assert Solution().minLength("1", 0) == 1       # single character
assert Solution().minLength("1", 1) == 1       # single character, flip does not reduce
assert Solution().minLength("0001000", 1) == 3 # flip the middle '1