0072MediumArrayHash TablePrefix Sum

Balanced Shift Stretch

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Trains the technique from

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 press logs one entry per hour for a whole run: states[i] is 1 when the press was cutting during hour i and 0 when it sat idle.

Maintenance wants the longest block of consecutive hours in which the press spent exactly as many hours cutting as it spent idle. Return the number of hours in that block.

Return 0 when no such block exists. The log can hold up to 10^5 hours, so checking every block separately is too slow.

Examples

Example 1

Input: states = [1, 1, 0, 1, 0]
Output: 4

Hours 1 through 4 hold two cutting hours and two idle hours. Adding hour 0 would tip the count to three cutting hours, so 4 is the longest.

Example 2

Input: states = [1, 1, 1, 1]
Output: 0

The press never idled, so no block can be balanced and the answer is the sentinel 0.

Example 3

Input: states = [0, 1, 0, 0, 1, 1]
Output: 6

Across the whole log there are three idle hours and three cutting hours, so the entire run is balanced.

Constraints

1 <= states.length <= 10^5
states[i] is either 0 or 1

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_balanced_stretch(states: list[int]) -> int:

Java

public int longestBalancedStretch(int[] states)

The intended approach

Time O(n)Space O(n)

Score a cutting hour as +1 and an idle hour as -1 and track the running score. A block is balanced exactly when the running score at its two ends is identical, so record the earliest hour at which each score value appeared and measure the distance whenever that score returns. Seeding the table with score 0 before the log starts covers blocks that begin at hour 0.

Pythonentry point: longest_balanced_stretch

def longest_balanced_stretch(states):
    first_seen = {0: -1}
    balance = 0
    best = 0
    for i, state in enumerate(states):
        balance += 1 if state == 1 else -1
        if balance in first_seen:
            span = i - first_seen[balance]
            if span > best:
                best = span
        else:
            first_seen[balance] = i
    return best

The obvious version

Try every start hour, extend one hour at a time while keeping a running tally, and record every block that lands on a tie, O(n^2).

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_balanced_stretch(states):
    n = len(states)
    best = 0
    for start in range(n):
        running = 0
        for end in range(start, n):
            running += 1 if states[end] == 1 else -1
            if running == 0 and end - start + 1 > best:
                best = end - start + 1
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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