Description#
You are given an integer array nums and an integer k.
Find the longest subsequence of nums that meets the following requirements:
- The subsequence is strictly increasing and
- The difference between adjacent elements in the subsequence is at most
k.
Return the length of the longest subsequence that meets the requirements.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [4,2,1,4,3,4,5,8,15], k = 3
Output: 5
Explanation:
The longest subsequence that meets the requirements is [1,3,4,5,8].
The subsequence has a length of 5, so we return 5.
Note that the subsequence [1,3,4,5,8,15] does not meet the requirements because 15 - 8 = 7 is larger than 3.
Example 2:
Input: nums = [7,4,5,1,8,12,4,7], k = 5
Output: 4
Explanation:
The longest subsequence that meets the requirements is [4,5,8,12].
The subsequence has a length of 4, so we return 4.
Example 3:
Input: nums = [1,5], k = 1
Output: 1
Explanation:
The longest subsequence that meets the requirements is [1].
The subsequence has a length of 1, so we return 1.
Constraints:
1 <= nums.length <= 1051 <= nums[i], k <= 105
Solutions#
Solution 1: Segment Tree#
We assume that $f[v]$ represents the length of the longest increasing subsequence ending with the number $v$.
We traverse each element $v$ in the array $nums$, with the state transition equation: $f[v] = \max(f[v], f[x])$, where the range of $x$ is $[v-k, v-1]$.
Therefore, we need a data structure to maintain the maximum value of the interval. It is not difficult to think of using a segment tree.
The segment tree divides the entire interval into multiple discontinuous subintervals, and the number of subintervals does not exceed $log(width)$. To update the value of an element, only $log(width)$ intervals need to be updated, and these intervals are all contained in a large interval that contains the element.
- Each node of the segment tree represents an interval;
- The segment tree has a unique root node, which represents the entire statistical range, such as $[1,N]$;
- Each leaf node of the segment tree represents an elementary interval of length $1$, $[x, x]$;
- For each internal node $[l,r]$, its left child is $[l,mid]$, and the right child is $[mid+1,r]$, where $mid = \left \lfloor \frac{l+r}{2} \right \rfloor$.
For this problem, the information maintained by the segment tree node is the maximum value within the interval range.
The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $nums$.
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| class Node:
def __init__(self):
self.l = 0
self.r = 0
self.v = 0
class SegmentTree:
def __init__(self, n):
self.tr = [Node() for _ in range(4 * n)]
self.build(1, 1, n)
def build(self, u, l, r):
self.tr[u].l = l
self.tr[u].r = r
if l == r:
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
def modify(self, u, x, v):
if self.tr[u].l == x and self.tr[u].r == x:
self.tr[u].v = v
return
mid = (self.tr[u].l + self.tr[u].r) >> 1
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)
def pushup(self, u):
self.tr[u].v = max(self.tr[u << 1].v, self.tr[u << 1 | 1].v)
def query(self, u, l, r):
if self.tr[u].l >= l and self.tr[u].r <= r:
return self.tr[u].v
mid = (self.tr[u].l + self.tr[u].r) >> 1
v = 0
if l <= mid:
v = self.query(u << 1, l, r)
if r > mid:
v = max(v, self.query(u << 1 | 1, l, r))
return v
class Solution:
def lengthOfLIS(self, nums: List[int], k: int) -> int:
tree = SegmentTree(max(nums))
ans = 1
for v in nums:
t = tree.query(1, v - k, v - 1) + 1
ans = max(ans, t)
tree.modify(1, v, t)
return ans
|
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| class Solution {
public int lengthOfLIS(int[] nums, int k) {
int mx = nums[0];
for (int v : nums) {
mx = Math.max(mx, v);
}
SegmentTree tree = new SegmentTree(mx);
int ans = 0;
for (int v : nums) {
int t = tree.query(1, v - k, v - 1) + 1;
ans = Math.max(ans, t);
tree.modify(1, v, t);
}
return ans;
}
}
class Node {
int l;
int r;
int v;
}
class SegmentTree {
private Node[] tr;
public SegmentTree(int n) {
tr = new Node[4 * n];
for (int i = 0; i < tr.length; ++i) {
tr[i] = new Node();
}
build(1, 1, n);
}
public void build(int u, int l, int r) {
tr[u].l = l;
tr[u].r = r;
if (l == r) {
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
public void modify(int u, int x, int v) {
if (tr[u].l == x && tr[u].r == x) {
tr[u].v = v;
return;
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (x <= mid) {
modify(u << 1, x, v);
} else {
modify(u << 1 | 1, x, v);
}
pushup(u);
}
public void pushup(int u) {
tr[u].v = Math.max(tr[u << 1].v, tr[u << 1 | 1].v);
}
public int query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) {
return tr[u].v;
}
int mid = (tr[u].l + tr[u].r) >> 1;
int v = 0;
if (l <= mid) {
v = query(u << 1, l, r);
}
if (r > mid) {
v = Math.max(v, query(u << 1 | 1, l, r));
}
return v;
}
}
|
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| class Node {
public:
int l;
int r;
int v;
};
class SegmentTree {
public:
vector<Node*> tr;
SegmentTree(int n) {
tr.resize(4 * n);
for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
build(1, 1, n);
}
void build(int u, int l, int r) {
tr[u]->l = l;
tr[u]->r = r;
if (l == r) return;
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int x, int v) {
if (tr[u]->l == x && tr[u]->r == x) {
tr[u]->v = v;
return;
}
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (x <= mid)
modify(u << 1, x, v);
else
modify(u << 1 | 1, x, v);
pushup(u);
}
void pushup(int u) {
tr[u]->v = max(tr[u << 1]->v, tr[u << 1 | 1]->v);
}
int query(int u, int l, int r) {
if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
int mid = (tr[u]->l + tr[u]->r) >> 1;
int v = 0;
if (l <= mid) v = query(u << 1, l, r);
if (r > mid) v = max(v, query(u << 1 | 1, l, r));
return v;
}
};
class Solution {
public:
int lengthOfLIS(vector<int>& nums, int k) {
SegmentTree* tree = new SegmentTree(*max_element(nums.begin(), nums.end()));
int ans = 1;
for (int v : nums) {
int t = tree->query(1, v - k, v - 1) + 1;
ans = max(ans, t);
tree->modify(1, v, t);
}
return ans;
}
};
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| func lengthOfLIS(nums []int, k int) int {
mx := slices.Max(nums)
tree := newSegmentTree(mx)
ans := 1
for _, v := range nums {
t := tree.query(1, v-k, v-1) + 1
ans = max(ans, t)
tree.modify(1, v, t)
}
return ans
}
type node struct {
l int
r int
v int
}
type segmentTree struct {
tr []*node
}
func newSegmentTree(n int) *segmentTree {
tr := make([]*node, n<<2)
for i := range tr {
tr[i] = &node{}
}
t := &segmentTree{tr}
t.build(1, 1, n)
return t
}
func (t *segmentTree) build(u, l, r int) {
t.tr[u].l, t.tr[u].r = l, r
if l == r {
return
}
mid := (l + r) >> 1
t.build(u<<1, l, mid)
t.build(u<<1|1, mid+1, r)
t.pushup(u)
}
func (t *segmentTree) modify(u, x, v int) {
if t.tr[u].l == x && t.tr[u].r == x {
t.tr[u].v = v
return
}
mid := (t.tr[u].l + t.tr[u].r) >> 1
if x <= mid {
t.modify(u<<1, x, v)
} else {
t.modify(u<<1|1, x, v)
}
t.pushup(u)
}
func (t *segmentTree) query(u, l, r int) int {
if t.tr[u].l >= l && t.tr[u].r <= r {
return t.tr[u].v
}
mid := (t.tr[u].l + t.tr[u].r) >> 1
v := 0
if l <= mid {
v = t.query(u<<1, l, r)
}
if r > mid {
v = max(v, t.query(u<<1|1, l, r))
}
return v
}
func (t *segmentTree) pushup(u int) {
t.tr[u].v = max(t.tr[u<<1].v, t.tr[u<<1|1].v)
}
|
