0554MediumHash TableStringSliding Window

Longest Belt Run Within a Colour Budget

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Trains the technique from

LeetCode 340Longest Substring with At Most K Distinct Characters

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Parts travel down a belt in a fixed order, and each part carries a single-character colour code. The string s lists those codes from the first part to the last.

A packing cell can handle a stretch of consecutive parts only while that stretch uses at most k different colour codes. Return the number of parts in the longest such stretch. When k is 0 no stretch qualifies, so the answer is 0.

Examples

Example 1

Input: s = "abaccc", k = 2
Output: 4

The four parts from position 2 to position 5 carry the codes a, c, c, c, which is two different codes and so within the budget of two.

Example 2

Input: s = "xyxyzz", k = 1
Output: 2

With a budget of one code, a stretch may only hold repeats of a single code. The two parts at positions 4 and 5 both carry z, giving a run of two.

Example 3

Input: s = "abcdef", k = 0
Output: 0

A budget of zero codes admits no part at all, so the answer is 0.

Constraints

1 <= s.length <= 5 * 10^4
0 <= k <= 50
s contains only lowercase English letters and digits.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def length_of_longest_substring_k_distinct(s: str, k: int) -> int:

Java

public int lengthOfLongestSubstringKDistinct(String s, int k)

The intended approach

Time O(n) in the number of parts; each edge advances at most n times.Space O(k) for the tally, which never names more than k + 1 codes.

Sweep a window along the belt. The right edge advances one part per step and the part's code is added to a tally of counts. If the tally then names more codes than `k` allows, advance the left edge, decrementing the departing part's count and removing the code from the tally the moment its count hits zero, and keep advancing until the tally is small enough again. The tally size is therefore always the number of distinct codes actually inside the window. Record the widest window seen. Each edge only ever moves forward, so the whole sweep is linear. A budget of zero is answered directly.

Pythonentry point: length_of_longest_substring_k_distinct

def length_of_longest_substring_k_distinct(s, k):
    # O(n) time, O(k) space. Hold a window over the belt and a tally of how many
    # parts of each code it contains. Extend it one part at a time; whenever the
    # tally holds more codes than allowed, drop parts off the left end until it
    # does not, deleting a code from the tally as soon as its count reaches zero
    # so the tally size is the true number of distinct codes in the window.
    if k == 0:
        return 0
    tally = {}
    left = 0
    best = 0
    for right, code in enumerate(s):
        tally[code] = tally.get(code, 0) + 1
        while len(tally) > k:
            gone = s[left]
            tally[gone] -= 1
            if tally[gone] == 0:
                del tally[gone]
            left += 1
        span = right - left + 1
        if span > best:
            best = span
    return best

The obvious version

Start a stretch at every part in turn and extend it one part at a time, collecting the codes seen into a set. Stop that stretch as soon as the set grows past `k`, and keep the longest stretch that stayed within budget.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_length_of_longest_substring_k_distinct(s, k):
    n = len(s)
    best = 0
    for start in range(n):
        seen = set()
        for end in range(start, n):
            seen.add(s[end])
            if len(seen) > k:
                break
            if end - start + 1 > best:
                best = end - start + 1
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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