The imbalance across a split
A load balancer divides a batch at one index and wants the difference between the two halves.
- Sum the values strictly left of `at`, and the values from `at` onward inclusive.
- The answer is the left sum minus the right sum.
- A cut at or before the start takes all the weight as the right side: negate the whole sum.
- A cut at or past the end leaves everything on the left: just the whole sum.
splitImbalance(values: list<int>, at: int) → int
Go needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
func splitImbalance(values []int, at int) int {
}Worked examples
| Call | Result |
|---|---|
splitImbalance([]int{1, 2, 3, 4}, 2) | -4 |
splitImbalance([]int{5, 3, 8}, 1) | -6 |
splitImbalance([]int{1, 2, 3}, 0) | -6 |
splitImbalance([]int{1, 2, 3}, 3) | 6 |
Hint
Handle the out-of-range slices first, then sum each side and subtract.
Reference solution in Go
func splitImbalance(values []int, at int) int {
total := 0
for _, v := range values {
total += v
}
if at <= 0 {
return -total
}
if at >= len(values) {
return total
}
left := 0
for i := 0; i < at; i++ {
left += values[i]
}
right := 0
for i := at; i < len(values); i++ {
right += values[i]
}
return left - right
}