0483MediumHash TableStringCounting

Total Spread Across Every Stretch of the Log

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Trains the technique from

LeetCode 1781Sum of Beauty of All 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 machine shop writes its shift log as the string log, one lowercase letter per job, in the order the jobs ran.

A stretch is any run of consecutive letters of the log; a stretch is picked out by where it starts and where it ends, so two stretches at different places count separately even when they read the same. The spread of a stretch is how many times its busiest letter occurs in it, less how many times its quietest letter occurs in it, counting only letters that occur in that stretch at all.

Return the spreads of every stretch of log added together.

Examples

Example 1

Input: log = "ppqqp"
Output: 4

Most stretches here even out to nothing. The stretch `"ppq"` contributes 1 and the whole log contributes 1, and the remaining contributions are similarly small.

Example 2

Input: log = "xyz"
Output: 0

Every letter of this log is different, so no stretch has a busier letter than its quietest one.

Example 3

Input: log = "mmmnnm"
Output: 9

The long run of `"m"` at the front leaves several stretches lopsided, and their contributions add up to the figure returned.

Constraints

1 <= log.length <= 500
log consists of 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 total_spread(log: str) -> int:

Java

public int totalSpread(String log)

The intended approach

Time O(n^2 * 26)Space O(26), a fixed 26-slot tally

Hold the left end of the stretch still and push the right end along one letter at a time, keeping a tally of twenty-six counts that is grown rather than rebuilt. Adding a letter can only push the busiest count up, so track that in a single variable; the quietest count among the letters that are present is read by looking over the twenty-six slots. Add the difference for every stretch as it is formed, and start a fresh tally for each new left end.

Pythonentry point: total_spread

def total_spread(log):
    # Fix the left end and extend the right end, keeping a 26-slot tally.
    # The busiest count can only rise as the stretch grows, so it needs no rescan;
    # the quietest present count is read off the 26 slots each step.
    # time O(n^2 * 26), space O(26)
    n = len(log)
    total = 0
    for start in range(n):
        tally = [0] * 26
        busiest = 0
        for end in range(start, n):
            slot = ord(log[end]) - 97
            tally[slot] += 1
            if tally[slot] > busiest:
                busiest = tally[slot]
            quietest = busiest
            for count in tally:
                if count and count < quietest:
                    quietest = count
            total += busiest - quietest
    return total

The obvious version

Take every start and end, slice out that stretch, count its letters from scratch in a dictionary, and add the difference between the largest and smallest count found.

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_total_spread(log):
    n = len(log)
    total = 0
    for i in range(n):
        for j in range(i, n):
            seen = {}
            for ch in log[i:j + 1]:
                seen[ch] = seen.get(ch, 0) + 1
            total += max(seen.values()) - min(seen.values())
    return total

How this problem was checked

Reference solution executed against 8 test cases — 8 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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