0346EasyArrayPrefix Sum

Silo Row Imbalance

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Trains the technique from

LeetCode 2574Left and Right Sum Differences

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 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 is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def left_right_difference(nums: list[int]) -> list[int]:

Java

public int[] leftRightDifference(int[] nums)

The intended approach

Time O(n)Space O(n) for the returned array, O(1) beyond it

Add up the whole row once. Then walk the row keeping the tonnage already passed; the tonnage still ahead is the grand total minus what was passed and minus the current silo, so each imbalance costs constant work.

Pythonentry point: left_right_difference

def left_right_difference(nums):
    # One pass to learn the grand total, a second pass that carries the running
    # total to the left of the current silo; the total to its right is whatever is
    # left once the current silo and everything before it is taken away.
    # Time O(n).  Space O(n) for the returned answer, O(1) beyond it.
    grand = 0
    for v in nums:
        grand += v
    out = []
    left = 0
    for v in nums:
        right = grand - left - v
        gap = left - right
        if gap < 0:
            gap = -gap
        out.append(gap)
        left += v
    return out

The obvious version

For every silo, add up the tonnage on its left with one loop and the tonnage on its right with another. Quadratic.

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_left_right_gap(nums):
    n = len(nums)
    out = []
    for i in range(n):
        left = 0
        for j in range(0, i):
            left += nums[j]
        right = 0
        for j in range(i + 1, n):
            right += nums[j]
        out.append(abs(left - right))
    return out

How this problem was checked

Reference solution executed against 8 test cases — 8 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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