Push the empty slots to the end
A picking list uses zero for a line that was cancelled. The screen keeps the live lines in order and pushes the blanks to the bottom.
- Every non-zero value keeps its position relative to the others.
- All the zeros end up together at the end.
- The list comes back the same length it went in.
moveBlanksLast(lines: list<int>) → list<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 moveBlanksLast(lines []int) []int {
}Worked examples
| Call | Result |
|---|---|
moveBlanksLast([]int{0, 1, 0, 3, 12}) | []int{1, 3, 12, 0, 0} |
moveBlanksLast([]int{1, 2, 3}) | []int{1, 2, 3} |
moveBlanksLast([]int{0, 0}) | []int{0, 0} |
moveBlanksLast([]int{}) | []int{} |
Hint
Keep a write index. Walk the list once copying every non-zero value to that index and advancing it; then fill what is left with zeros.
Reference solution in Go
func moveBlanksLast(lines []int) []int {
out := make([]int, len(lines))
write := 0
for _, value := range lines {
if value != 0 {
out[write] = value
write++
}
}
return out
}