0762EasyStringGreedyCounting

Most Even Conveyor Segments

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Trains the technique from

LeetCode 1221Split a String in Balanced Strings

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 conveyor gate writes one character per event into log: 'I' when a crate comes in and 'O' when a crate goes out.

Call a run of consecutive events even when it records exactly as many 'I' events as 'O' events. Cut the whole log into consecutive non-empty runs so that every run is even, using every event exactly once and keeping the events in order.

Return the largest number of runs such a cut can produce. The log as a whole is even, so at least one such cut always exists.

Examples

Example 1

Input: log = "IOIOIO"
Output: 3

Cutting after every second event gives the runs "IO", "IO" and "IO", each recording one arrival and one departure, so three runs is achievable.

Example 2

Input: log = "IIOOIO"
Output: 2

The runs "IIOO" and "IO" use every event in order and each is even, giving two runs.

Example 3

Input: log = "OIOIIO"
Output: 3

The runs "OI", "OI" and "IO" are all even, so three runs is achievable. A run may start with a departure.

Constraints

2 <= log.length <= 1000
log[i] is either 'I' or 'O'.
log records the same number of 'I' events as 'O' events.

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 most_even_segments(log: str) -> int:

Java

public int mostEvenSegments(String log)

The intended approach

Time O(n) for a log of n events.Space O(1).

Sweep the log once holding a running balance that rises by one on `'I'` and falls by one on `'O'`. A run starting where the balance was last zero is even exactly when the balance is zero again at its end, so every zero crossing closes a run, and closing a run at the earliest possible zero never blocks a later one. Counting the zero crossings is therefore the answer, and the last event is always one of them.

Pythonentry point: most_even_segments

def most_even_segments(log):
    # Time O(n), space O(1).
    balance = 0
    segments = 0
    for ch in log:
        balance += 1 if ch == "I" else -1
        if balance == 0:
            segments += 1
    return segments

The obvious version

Work backwards over start positions with a table: the best cut of the log from position `i` onwards tries every end position `j` such that `i..j` is even, and takes one plus the best cut of the rest. Quadratic in the log length.

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_most_even_segments(log):
    n = len(log)
    best = [None] * (n + 1)
    best[n] = 0
    for i in range(n - 1, -1, -1):
        balance = 0
        top = None
        for j in range(i, n):
            balance += 1 if log[j] == "I" else -1
            if balance == 0 and best[j + 1] is not None:
                cand = best[j + 1] + 1
                if top is None or cand > top:
                    top = cand
        best[i] = top
    return best[0]

How this problem was checked

Reference solution executed against 12 test cases — 12 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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