0906EasyArrayHash TableSegment Tree

Squared Variety Over Every Window

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Trains the technique from

LeetCode 2913Subarrays Distinct Element Sum of Squares I

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.

Readings are given as nums. For a contiguous stretch of them, its variety is how many distinct readings it holds.

Return the total of the squares of the varieties over every non-empty contiguous stretch.

Examples

Example 1

Input: nums = [4, 9, 4, 7, 9]
Output: 70

Every one of the fifteen stretches contributes the square of its own variety, and repeats inside a stretch add nothing, which is what keeps this total below what all-distinct readings would give.

Example 2

Input: nums = [3, 8]
Output: 6

The two single readings each have variety 1, and the whole pair has variety 2, giving 1 plus 1 plus 4.

Example 3

Input: nums = [3, 3]
Output: 3

The pair holds one distinct reading, so all three stretches have variety 1.

Constraints

1 <= nums.length <= 100
1 <= nums[i] <= 100

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 sum_counts(nums: list[int]) -> int:

Java

public int sumCounts(int[] nums)

The intended approach

Time O(n^2) for n readings, one step per stretch.Space O(V) for the tally over the reading range.

Stretches sharing a left end grow one reading at a time, and each step either meets a reading already inside, leaving the variety alone, or meets a new one, raising it by one. So fix the left end and walk the right end outward, carrying a tally of which readings are present and the current variety; add the square of that variety at each step. The tally is reset once per left end, and no stretch is ever rebuilt from scratch.

Pythonentry point: sum_counts

def sum_counts(nums):
    n = len(nums)
    total = 0
    for start in range(n):
        present = {}
        variety = 0
        for end in range(start, n):
            v = nums[end]
            if present.get(v, 0) == 0:
                variety += 1
            present[v] = present.get(v, 0) + 1
            total += variety * variety
    return total

The obvious version

Take every stretch and work out its variety from scratch by walking it and collecting the readings into a set. Cubic in the reading count and it carries nothing from one stretch to the next.

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_sum_counts(nums):
    n = len(nums)
    total = 0
    for start in range(n):
        for end in range(start, n):
            seen = set()
            for at in range(start, end + 1):
                seen.add(nums[at])
            total += len(seen) * 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

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