Left and Right Sum Differences
Real-world version: Silo Row ImbalanceThe statement here is our own: the same inputs, outputs and technique.
A grain terminal lines up its silos in one row. Silo i currently holds nums[i] tonnes.
For each silo the operator wants its imbalance: the absolute difference between the tonnage stored in the silos strictly before it in the row and the tonnage stored in the silos strictly after it. The silo itself counts on neither side, and an empty side contributes zero.
Return an array of the imbalances, one entry per silo, in row order.
Examples
Example 1
- Input
- nums = [6,2,9,4]
- Output
- [15,7,4,17]
For silo 0 nothing lies before it and 2 + 9 + 4 = 15 lies after it. For silo 1 the sides hold 6 and 13. For silo 2 they hold 8 and 4. For silo 3 they hold 17 and nothing.
Example 2
- Input
- nums = [12,5]
- Output
- [5,12]
Silo 0 has nothing before it and 5 tonnes after it. Silo 1 has 12 tonnes before it and nothing after it.
Example 3
- Input
- nums = [3,3,3]
- Output
- [6,0,6]
The middle silo has 3 tonnes on each side, so its imbalance is 0, while each end silo faces 6 tonnes on its inner side and nothing on its outer side.
Constraints
- 1 <= nums.length <= 1000
- 1 <= nums[i] <= 10^5
The signature
The editor starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.
def left_right_difference(nums: list[int]) -> list[int]:public int[] leftRightDifference(int[] nums)Checking your plan…
How this problem was checked
- Passed: Reference solution run against 8 test cases: 8 passed
- Passed: 3 worked examples confirmed by running the code
- Passed: 0 eight-word sequences shared with the parent problem's text
- Passed: Agrees with an independent brute-force solution on random inputs
- Passed: Test set shown to catch deliberately broken solutions
- Passed: Reviewed by a separate quality pass
- Passed: Function arity matches the parent problem's