0009MediumArrayPrefix Sum

Amplifier Bypass Gains

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Trains the technique from

LeetCode 238Product of Array Except Self

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 studio rack wires several amplifier stages in series. Stage i scales whatever signal reaches it by the integer factor gains[i]: a negative factor flips the signal's phase, and a factor of 0 means that stage is muted and kills the signal outright. The rack's overall factor is what you get by applying every stage in turn.

The engineer is auditioning patch cables, and for each stage wants to know the rack's overall factor when that single stage is jumpered out of the chain and all the other stages stay exactly as they are.

Given gains, return an array bypassed of the same length, where bypassed[i] is the factor the rack would have with stage i removed.

Your routine has to finish in linear time, and it may not use division anywhere: muted stages make dividing the rack's overall factor by a single stage's factor useless.

Examples

Example 1

Input: gains = [3, 1, 4, 2]
Output: [8, 24, 6, 12]

Jumpering out the first stage leaves 1, 4 and 2, which combine to 8. Removing the second leaves 3 * 4 * 2 = 24, removing the third leaves 3 * 1 * 2 = 6, and removing the fourth leaves 3 * 1 * 4 = 12.

Example 2

Input: gains = [-2, 0, 5, 3]
Output: [0, -30, 0, 0]

Only removing the muted stage revives the rack, giving -2 * 5 * 3 = -30; every other chain still contains the muted stage and reads 0.

Example 3

Input: gains = [4, 0, 0, -6]
Output: [0, 0, 0, 0]

Two stages are muted, so whichever single stage is jumpered out a muted one is still in the chain and the rack stays silent.

Constraints

2 <= gains.length <= 10^5
-30 <= gains[i] <= 30
Inputs are chosen so that every entry of the returned array fits in a signed 32-bit integer.
Division may not be used.

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 bypass_gains(gains: list[int]) -> list[int]:

Java

public int[] bypassGains(int[] gains)

The intended approach

Time O(n)Space O(1) extra beyond the returned array

The chain without stage `i` is just everything upstream of `i` combined with everything downstream of `i`, so no division is needed anywhere. Sweep left to right carrying a running factor of the stages seen so far and park it at each position, then sweep right to left carrying a running factor of the stages behind you and multiply it into the value already parked there. Zeros are handled for free because they are never divided out, only skipped over by whichever sweep they fall outside of.

Pythonentry point: bypass_gains

def bypass_gains(gains):
    n = len(gains)
    bypassed = [1] * n
    upstream = 1
    for i in range(n):
        bypassed[i] = upstream
        upstream *= gains[i]
    downstream = 1
    for i in range(n - 1, -1, -1):
        bypassed[i] *= downstream
        downstream *= gains[i]
    return bypassed

The obvious version

Recompute each product from scratch, O(n^2), no division.

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_bypass(gains):
    out = []
    for i in range(len(gains)):
        p = 1
        for j, g in enumerate(gains):
            if j != i:
                p *= g
        out.append(p)
    return out

How this problem was checked

Reference solution executed against 16 test cases — 16 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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