# 673. Number of Longest Increasing Subsequence

## Description

Given an integer array nums, return the number of longest increasing subsequences.

Notice that the sequence has to be strictly increasing.

Example 1:

Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].


Example 2:

Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.


Constraints:

• 1 <= nums.length <= 2000
• -106 <= nums[i] <= 106

## Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the length of the longest increasing subsequence ending with $nums[i]$, and $cnt[i]$ as the number of longest increasing subsequences ending with $nums[i]$. Initially, $f[i]=1$, $cnt[i]=1$. Also, we define $mx$ as the length of the longest increasing subsequence, and $ans$ as the number of longest increasing subsequences.

For each $nums[i]$, we enumerate all elements $nums[j]$ in $nums[0:i-1]$. If $nums[j] < nums[i]$, then $nums[i]$ can be appended after $nums[j]$ to form a longer increasing subsequence. If $f[i] < f[j] + 1$, it means the length of the longest increasing subsequence ending with $nums[i]$ has increased, so we need to update $f[i]=f[j]+1$ and $cnt[i]=cnt[j]$. If $f[i]=f[j]+1$, it means we have found a longest increasing subsequence with the same length as before, so we need to increase $cnt[i]$ by $cnt[j]$. Then, if $mx < f[i]$, it means the length of the longest increasing subsequence has increased, so we need to update $mx=f[i]$ and $ans=cnt[i]$. If $mx=f[i]$, it means we have found a longest increasing subsequence with the same length as before, so we need to increase $ans$ by $cnt[i]$.

Finally, we return $ans$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Solution 2: Binary Indexed Tree

We can use a binary indexed tree to maintain the length and count of the longest increasing subsequence in the prefix interval. We remove duplicates from the array $nums$ and sort it to get the array $arr$. Then we enumerate each element $x$ in $nums$, find the position $i$ of $x$ in the array $arr$ by binary search, then query the length and count of the longest increasing subsequence in $[1,i-1]$, denoted as $v$ and $cnt$, then update the length and count of the longest increasing subsequence in $[i]$ to $v+1$ and $\max(cnt,1)$. Finally, we query the length and count of the longest increasing subsequence in $[1,m]$, where $m$ is the length of the array $arr$, which is the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class BinaryIndexedTree {
private int n;
private int[] c;
private int[] d;

public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
d = new int[n + 1];
}

public void update(int x, int v, int cnt) {
while (x <= n) {
if (c[x] < v) {
c[x] = v;
d[x] = cnt;
} else if (c[x] == v) {
d[x] += cnt;
}
x += x & -x;
}
}

public int[] query(int x) {
int v = 0, cnt = 0;
while (x > 0) {
if (c[x] > v) {
v = c[x];
cnt = d[x];
} else if (c[x] == v) {
cnt += d[x];
}
x -= x & -x;
}
return new int[] {v, cnt};
}
}

public class Solution {
public int findNumberOfLIS(int[] nums) {
// int[] arr = Arrays.stream(nums).distinct().sorted().toArray();
int[] arr = nums.clone();
Arrays.sort(arr);
int m = arr.length;
BinaryIndexedTree tree = new BinaryIndexedTree(m);
for (int x : nums) {
int i = Arrays.binarySearch(arr, x) + 1;
int[] t = tree.query(i - 1);
int v = t[0];
int cnt = t[1];
tree.update(i, v + 1, Math.max(cnt, 1));
}
return tree.query(m)[1];
}
}

• class BinaryIndexedTree {
private:
int n;
vector<int> c;
vector<int> d;

public:
BinaryIndexedTree(int n)
: n(n)
, c(n + 1, 0)
, d(n + 1, 0) {}

void update(int x, int v, int cnt) {
while (x <= n) {
if (c[x] < v) {
c[x] = v;
d[x] = cnt;
} else if (c[x] == v) {
d[x] += cnt;
}
x += x & -x;
}
}

pair<int, int> query(int x) {
int v = 0, cnt = 0;
while (x > 0) {
if (c[x] > v) {
v = c[x];
cnt = d[x];
} else if (c[x] == v) {
cnt += d[x];
}
x -= x & -x;
}
return {v, cnt};
}
};

class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
vector<int> arr = nums;
sort(arr.begin(), arr.end());
arr.erase(unique(arr.begin(), arr.end()), arr.end());
int m = arr.size();
BinaryIndexedTree tree(m);
for (int x : nums) {
auto it = lower_bound(arr.begin(), arr.end(), x);
int i = distance(arr.begin(), it) + 1;
auto [v, cnt] = tree.query(i - 1);
tree.update(i, v + 1, max(cnt, 1));
}
return tree.query(m).second;
}
};

• class BinaryIndexedTree:
__slots__ = ["n", "c", "d"]

def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
self.d = [0] * (n + 1)

def update(self, x, v, cnt):
while x <= self.n:
if self.c[x] < v:
self.c[x] = v
self.d[x] = cnt
elif self.c[x] == v:
self.d[x] += cnt
x += x & -x

def query(self, x):
v = cnt = 0
while x:
if self.c[x] > v:
v = self.c[x]
cnt = self.d[x]
elif self.c[x] == v:
cnt += self.d[x]
x -= x & -x
return v, cnt

class Solution:
def findNumberOfLIS(self, nums: List[int]) -> int:
arr = sorted(set(nums))
m = len(arr)
tree = BinaryIndexedTree(m)
for x in nums:
i = bisect_left(arr, x) + 1
v, cnt = tree.query(i - 1)
tree.update(i, v + 1, max(cnt, 1))
return tree.query(m)[1]


• type BinaryIndexedTree struct {
n int
c []int
d []int
}

func newBinaryIndexedTree(n int) BinaryIndexedTree {
return BinaryIndexedTree{
n: n,
c: make([]int, n+1),
d: make([]int, n+1),
}
}

func (bit *BinaryIndexedTree) update(x, v, cnt int) {
for x <= bit.n {
if bit.c[x] < v {
bit.c[x] = v
bit.d[x] = cnt
} else if bit.c[x] == v {
bit.d[x] += cnt
}
x += x & -x
}
}

func (bit *BinaryIndexedTree) query(x int) (int, int) {
v, cnt := 0, 0
for x > 0 {
if bit.c[x] > v {
v = bit.c[x]
cnt = bit.d[x]
} else if bit.c[x] == v {
cnt += bit.d[x]
}
x -= x & -x
}
return v, cnt
}

func findNumberOfLIS(nums []int) int {
arr := make([]int, len(nums))
copy(arr, nums)
sort.Ints(arr)
m := len(arr)
tree := newBinaryIndexedTree(m)
for _, x := range nums {
i := sort.SearchInts(arr, x) + 1
v, cnt := tree.query(i - 1)
tree.update(i, v+1, max(cnt, 1))
}
_, ans := tree.query(m)
return ans
}

• class BinaryIndexedTree {
private n: number;
private c: number[];
private d: number[];

constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
this.d = Array(n + 1).fill(0);
}

update(x: number, v: number, cnt: number): void {
while (x <= this.n) {
if (this.c[x] < v) {
this.c[x] = v;
this.d[x] = cnt;
} else if (this.c[x] === v) {
this.d[x] += cnt;
}
x += x & -x;
}
}

query(x: number): [number, number] {
let v = 0;
let cnt = 0;
while (x > 0) {
if (this.c[x] > v) {
v = this.c[x];
cnt = this.d[x];
} else if (this.c[x] === v) {
cnt += this.d[x];
}
x -= x & -x;
}
return [v, cnt];
}
}

function findNumberOfLIS(nums: number[]): number {
const arr: number[] = [...new Set(nums)].sort((a, b) => a - b);
const m: number = arr.length;
const tree: BinaryIndexedTree = new BinaryIndexedTree(m);
const search = (x: number): number => {
let l = 0,
r = arr.length;
while (l < r) {
const mid = (l + r) >> 1;
if (arr[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
};
for (const x of nums) {
const i: number = search(x);
const [v, cnt]: [number, number] = tree.query(i - 1);
tree.update(i, v + 1, Math.max(cnt, 1));
}
return tree.query(m)[1];
}


• struct BinaryIndexedTree {
n: usize,
c: Vec<i32>,
d: Vec<i32>,
}

impl BinaryIndexedTree {
fn new(n: usize) -> BinaryIndexedTree {
BinaryIndexedTree {
n,
c: vec![0; n + 1],
d: vec![0; n + 1],
}
}

fn update(&mut self, x: usize, v: i32, cnt: i32) {
let mut x = x as usize;
while x <= self.n {
if self.c[x] < v {
self.c[x] = v;
self.d[x] = cnt;
} else if self.c[x] == v {
self.d[x] += cnt;
}
x += x & x.wrapping_neg();
}
}

fn query(&self, mut x: usize) -> (i32, i32) {
let (mut v, mut cnt) = (0, 0);
while x > 0 {
if self.c[x] > v {
v = self.c[x];
cnt = self.d[x];
} else if self.c[x] == v {
cnt += self.d[x];
}
x -= x & x.wrapping_neg();
}
(v, cnt)
}
}

impl Solution {
pub fn find_number_of_lis(nums: Vec<i32>) -> i32 {
let mut arr: Vec<i32> = nums.iter().cloned().collect();
arr.sort();
let m = arr.len();
let mut tree = BinaryIndexedTree::new(m);
for x in nums.iter() {
if let Ok(i) = arr.binary_search(x) {
let (v, cnt) = tree.query(i);
tree.update(i + 1, v + 1, cnt.max(1));
}
}
let (_, ans) = tree.query(m);
ans
}
}