0900EasyArrayHash Table

Two Neighbouring Pairs of Equal Load

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Trains the technique from

LeetCode 2395Find Subarrays With Equal Sum

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 holds readings nums.

Consider the pairs of neighbouring readings: the pair at position i is nums[i] together with nums[i + 1]. Two such pairs are different when they start at different positions, even if they hold the same readings.

Return true when two different neighbouring pairs have the same total.

Examples

Example 1

Input: nums = [14, 9, 5, 18, 6, 17]
Output: true

The neighbouring pairs total 23, 14, 23, 24 and 23, so the pair starting at position 0 and the one starting at position 2 match.

Example 2

Input: nums = [10, 20, 30, 40, 50]
Output: false

The totals are 30, 50, 70 and 90, all different.

Example 3

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

Both pairs hold the same readings and so total the same, and they start at different positions, which is what makes them different pairs.

Constraints

2 <= nums.length <= 1000
-10^9 <= nums[i] <= 10^9

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 find_subarrays(nums: list[int]) -> bool:

Java

public boolean findSubarrays(int[] nums)

The intended approach

Time O(n) for n readings, one pass.Space O(n) for the set of totals seen.

There is one pair per starting position, so there are one fewer pairs than readings, and the question is only whether two of their totals coincide. Walk the conveyor working out each pair's total and keep the totals seen so far in a set; the first total already in the set answers yes. Reaching the end with no repeat answers no. Nothing has to be compared against everything, since membership in the set stands in for all the earlier comparisons at once.

Pythonentry point: find_subarrays

def find_subarrays(nums):
    seen = set()
    for i in range(len(nums) - 1):
        total = nums[i] + nums[i + 1]
        if total in seen:
            return True
        seen.add(total)
    return False

The obvious version

Work out every neighbouring pair's total and compare each against every other. Quadratic in the reading count and it keeps nothing between comparisons.

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_find_subarrays(nums):
    n = len(nums)
    for i in range(n - 1):
        for j in range(i + 1, n - 1):
            if nums[i] + nums[i + 1] == nums[j] + nums[j + 1]:
                return True
    return False

How this problem was checked

Reference solution executed against 11 test cases — 11 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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