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// There's definitely faster ways to do this but solution works well. | |
func maxSubsequence(nums []int, k int) []int { | |
h := NewIntHeap(0, 0, true) | |
heap.Init(h) | |
for _, num := range nums { | |
heap.Push(h, num) // push every number onto descending heap | |
} | |
maxSubUnordered := map[int]int{} | |
// pop the k-1 elements to find the minimum | |
i := 0 | |
for i < k { | |
i++ | |
maxSubUnordered[heap.Pop(h).(int)]++ | |
} | |
maxSub := make([]int, k)[0:0] | |
for _, num := range nums { | |
if count := maxSubUnordered[num]; count > 0 { | |
maxSubUnordered[num]-- | |
maxSub = append(maxSub, num) | |
} | |
} | |
return maxSub | |
} | |
func NewIntHeap(initialSize int, maxSize int, desc bool) *IntHeap { | |
return &IntHeap{ | |
data: make([]int, initialSize)[0:0], | |
desc: desc, | |
maxSize: maxSize, | |
} | |
} | |
type IntHeap struct { | |
data []int | |
desc bool // true = descending, false = ascending | |
initialSize int | |
maxSize int | |
} | |
func (h *IntHeap) resetData() { | |
h.data = make([]int, h.initialSize)[0:0] | |
fmt.Println("Reset backing data", h) | |
heap.Init(h) | |
} | |
func (h IntHeap) Len() int { | |
return len(h.data) | |
} | |
func (h IntHeap) Less(i, j int) bool { | |
if h.desc { | |
return h.data[i] > h.data[j] | |
} | |
return h.data[i] < h.data[j] | |
} | |
func (h IntHeap) Swap(i, j int) { | |
h.data[i], h.data[j] = h.data[j], h.data[i] | |
} | |
func (h *IntHeap) Push(x interface{}) { | |
h.data = append(h.data, x.(int)) | |
} | |
func (h *IntHeap) Prune() { | |
if h.maxSize == 0 || h.Len() < h.maxSize { | |
return | |
} | |
// pop the desired values | |
r := make([]int, h.maxSize) | |
for i := 0; i < h.maxSize; i++ { | |
r[i] = heap.Pop(h).(int) | |
} | |
// reset heap data | |
h.resetData() | |
// push back | |
for _, v := range r { | |
heap.Push(h, v) | |
} | |
} | |
func (h *IntHeap) Pop() interface{} { | |
old := *h | |
n := len(old.data) | |
x := old.data[n-1] | |
h.data = old.data[0:n-1] | |
return x | |
} |
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