0896HardHash TableStringDynamic Programming

Total Variety Across Every Window

Tracked in this browser only

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 reel of labels reads s, a string of lowercase letters. The variety of a contiguous stretch of the reel is how many distinct letters it holds.

Return the total variety over every non-empty contiguous stretch of the reel.

Examples

Example 1

Input: s = "mississippi"
Output: 148

Every one of the sixty-six stretches contributes its own count of distinct letters, and those counts add up to this total.

Example 2

Input: s = "ab"
Output: 4

The stretches are "a" with variety 1, "b" with variety 1, and "ab" with variety 2, giving 4.

Example 3

Input: s = "aa"
Output: 3

Each single label has variety 1 and the whole reel still has variety 1, since the repeat adds no new letter, giving 3.

Constraints

1 <= s.length <= 10^5
s 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 appeal_sum(s: str) -> int:

Java

public long appealSum(String s)

The intended approach

Time O(n) for a reel of length n, one pass.Space O(1): a table of twenty-six last-seen positions.

Adding the varieties one stretch at a time is far too slow, so count each letter's contribution instead. A stretch counts a letter once however often it appears, which invites double counting, so pin that single count to one chosen copy: the **first** copy of the letter inside the stretch. Then position `i` is credited by exactly the stretches that reach `i` and in which `i` holds the first copy of its letter, which means the left end is at or before `i` but after the previous copy of that same letter, and the right end is anywhere from `i` onward. The two ends are chosen independently, so the credit for position `i` is one count times the other: `i` minus the previous position of the same letter, times the number of positions from `i` to the end. Remembering where each letter was last seen turns the whole thing into one pass, and every stretch's variety is counted exactly once because each letter it holds has exactly one first copy in it.

Pythonentry point: appeal_sum

def appeal_sum(s):
    n = len(s)
    last = [-1] * 26
    total = 0
    for i in range(n):
        c = ord(s[i]) - 97
        total += (i - last[c]) * (n - i)
        last[c] = i
    return total

The obvious version

Walk every start, extend the stretch one label at a time while collecting the letters met in a set, and add the set's size each time. Quadratic in the reel length and it counts varieties directly rather than crediting positions.

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_appeal_sum(s):
    n = len(s)
    total = 0
    for start in range(n):
        seen = set()
        for end in range(start, n):
            seen.add(s[end])
            total += len(seen)
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in