0718MediumStringBinary SearchSliding WindowPrefix Sum

Longest Stretch Retinted Within Budget

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Trains the technique from

LeetCode 1208Get Equal Substrings Within Budget

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.

A ribbon press holds two rolls of the same length. Every position on a roll carries one shade, written as a lowercase letter, so panel and pattern are strings of equal length and panel[i], pattern[i] are the shades sitting at position i on the two rolls.

Give each letter its place in the alphabet, with "a" worth 0 and "z" worth 25. Retinting one position of panel so that it carries the shade pattern has at that position costs the absolute difference of the two values in units of ink, so a position that already carries the right shade costs nothing.

You hold budget units of ink. Pick one contiguous stretch of positions and retint every position inside it so the stretch matches pattern there, spending at most budget units in total across the stretch. Positions outside the stretch are left alone.

Return the number of positions in the longest stretch you can finish. An empty stretch is always allowed and has length 0.

Examples

Example 1

Input: panel = "dqgh", pattern = "cqfi", budget = 2
Output: 3

Retinting positions 0, 1 and 2 costs 1 + 0 + 1 = 2 units of ink, which the budget covers, and that stretch holds 3 positions.

Example 2

Input: panel = "wq", pattern = "aa", budget = 0
Output: 0

Position 0 needs 22 units and position 1 needs 16, so with no ink at all only the empty stretch is affordable.

Example 3

Input: panel = "cbbb", pattern = "aaaa", budget = 3
Output: 3

The four positions need 2, 1, 1 and 1 units. Positions 1, 2 and 3 together need 3 units, which the budget covers, and that stretch holds 3 positions.

Constraints

1 <= panel.length <= 10^5
pattern.length == panel.length
0 <= budget <= 10^6
panel and pattern hold lowercase English letters only.

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 longest_retinted_stretch(panel: str, pattern: str, budget: int) -> int:

Java

public int longestRetintedStretch(String panel, String pattern, int budget)

The intended approach

Time O(n) for a roll of n positions: each edge of the window advances at most n times.Space O(1) beyond the two input rolls.

The ink a stretch needs is the sum of the per-position distances inside it, and every distance is non-negative, so the cost of a stretch grows when it widens and falls when it narrows. Walk a right edge across the roll, adding the new position's distance, and whenever the running total passes the budget release positions from the left edge, giving back their distances, until the total is affordable again. Every position is added once and released at most once, and the widest affordable window seen along the way is the answer.

Pythonentry point: longest_retinted_stretch

def longest_retinted_stretch(panel, pattern, budget):
    # Slide a window over the positions and keep the ink the whole window needs.
    # Widening by one position adds that position's ink; while the window is over
    # budget the leftmost position leaves and gives its ink back.
    # Time O(n) for a roll of n positions.  Space O(1).
    spent = 0
    left = 0
    best = 0
    for right in range(len(panel)):
        spent += abs(ord(panel[right]) - ord(pattern[right]))
        while spent > budget:
            spent -= abs(ord(panel[left]) - ord(pattern[left]))
            left += 1
        width = right - left + 1
        if width > best:
            best = width
    return best

The obvious version

Precompute the ink each single position needs, then start a stretch at every position and extend it one position at a time, adding ink as it goes and abandoning that start as soon as the running total passes the budget. Keeps the widest affordable stretch found anywhere.

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_longest_retinted_stretch(panel, pattern, budget):
    n = len(panel)
    ink = [abs(ord(panel[i]) - ord(pattern[i])) for i in range(n)]
    best = 0
    for i in range(n):
        total = 0
        for j in range(i, n):
            total += ink[j]
            if total > budget:
                break
            if j - i + 1 > best:
                best = j - i + 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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