LeetCode 3909: Compare Sums of Bitonic Parts

Meta Data

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Difficulty: medium First Attempt: 2026-04-25

• Submissions: Accepted (first try)

Approach

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The array strictly rises then falls, so the peak is unique. Left is nums[0..peak] and right is nums[peak..n-1] (Array Template 1). The peak belongs to both, adding the same amount to each side, so it can be left out of the comparison.

My Solution

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class Solution:
    def compareBitonicSums(self, nums: List[int]) -> int:
        peak = nums.index(max(nums))
        left, right = sum(nums[:peak + 1]), sum(nums[peak:])
        if left > right:
            return 0
        if left < right:
            return 1
        return -1

I originally walked two pointers from both ends toward the peak, summing as they went (without the peak). With a strictly bitonic array the maximum is the peak, so max finds it too.

Complexity

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Time O(n), space O(n) (slices).

Mistakes to Avoid

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• Return values: 0 when the left is larger, 1 when the right is, -1 when equal.

Interview Explanation

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In a strictly bitonic array the maximum is the peak; I find it, sum both sides, and compare. The peak is on both sides, so it does not affect the result. Time O(n).