0819MediumHash TableStringBit ManipulationPrefix Sum

Nearly Even Telemetry Stretches

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Trains the technique from

LeetCode 1915Number of Wonderful Substrings

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 probe reports ten kinds of event, written as the lowercase letters a through j. One session of reports arrives as the string stream, one character per event, in the order they happened.

A stretch is any non-empty contiguous run of the session. A stretch is nearly even when at most one of the ten kinds occurs an odd number of times inside it, so either every kind present occurs an even number of times, or exactly one kind occurs an odd number of times and all the others occur an even number of times.

Return how many nearly even stretches the session contains. Two stretches that cover different positions count separately even when they read the same.

Examples

Example 1

Input: stream = "efe"
Output: 4

Four stretches are nearly even: each of the three single events, where the one kind present occurs once, and the whole session `efe`, where `e` occurs twice and `f` once.

Example 2

Input: stream = "gghg"
Output: 7

Seven stretches are nearly even: the four single events, the stretch `gg` at indices 0 and 1 where `g` occurs twice, and the stretches `ggh` and `ghg`, each holding `g` twice and `h` once.

Example 3

Input: stream = "hiih"
Output: 8

Eight stretches are nearly even, among them the whole session, where `h` occurs twice and `i` occurs twice, so no kind at all occurs an odd number of times.

Constraints

1 <= stream.length <= 10^5
stream consists of the lowercase letters 'a' through 'j' only.
The count can exceed the range of a signed 32-bit integer; it stays below 10^11.

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 count_nearly_even(stream: str) -> int:

Java

public long countNearlyEven(String stream)

The intended approach

Time O(n * k) for a session of n events over k = 10 kinds.Space O(2^k) for the tally of parity summaries.

Only the parity of each kind's count matters, so the prefix read so far can be summarised by ten bits. A stretch is nearly even when the summary at its right end and the summary just before its left end agree everywhere, or disagree in exactly one bit. Walking the session once while keeping a tally of how often every summary has already been seen, and seeding that tally with the empty prefix, turns each position into eleven tally lookups.

Pythonentry point: count_nearly_even

def count_nearly_even(stream):
    # Keep the parity of each of the ten event kinds over the prefix read so far,
    # packed into ten bits, together with a tally of how many earlier prefixes
    # carried each pattern. A stretch is nearly even exactly when the pattern at
    # its right end equals the pattern just before its left end, or differs from it
    # in a single bit, so each position costs eleven tally lookups.
    # Time: O(n * k) for a stream of n characters over k = 10 kinds.
    # Space: O(2^k) for the tally of patterns.
    seen = [0] * 1024
    seen[0] = 1
    pattern = 0
    total = 0
    for ch in stream:
        pattern ^= 1 << (ord(ch) - 97)
        total += seen[pattern]
        for bit in range(10):
            total += seen[pattern ^ (1 << bit)]
        seen[pattern] += 1
    return total

The obvious version

Take every start, extend the stretch one event at a time keeping a plain count per kind, and for each stretch count outright how many kinds have an odd count. Quadratic in the session length and free of any parity argument.

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_count_nearly_even(stream):
    """Every stretch in turn, counting outright how many kinds occur an odd number
    of times inside it."""
    n = len(stream)
    total = 0
    for i in range(n):
        counts = {}
        for j in range(i, n):
            counts[stream[j]] = counts.get(stream[j], 0) + 1
            odd = 0
            for v in counts.values():
                if v % 2 == 1:
                    odd += 1
            if odd <= 1:
                total += 1
    return total

How this problem was checked

Reference solution executed against 20 test cases — 20 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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