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300. Longest Increasing Subsequence

Description

Given an integer array nums, return the length of the longest strictly increasing subsequence.

 

Example 1:

Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2:

Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]
Output: 1

 

Constraints:

  • 1 <= nums.length <= 2500
  • -104 <= nums[i] <= 104

 

Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

Solutions

Solution 1

Thinking

A subsequence must keep relative order and increase strictly. Enumerating all subsequences explodes with \(n\); \(n \le 2500\) still allows an \(O(n^2)\) transfer by ending index.

The LIS ending at \(nums[i]\) depends only on smaller values to its left. Define \(f[i]\) as that length, and for each \(j < i\) with \(nums[j] < nums[i]\) take \(\max(f[j]+1)\). A singleton has length \(1\), and the answer is \(\max f\).

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class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    f[i] = max(f[i], f[j] + 1)
        return max(f)

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class Solution {
    public int lengthOfLIS(int[] nums) {
        int n = nums.length;
        int[] f = new int[n];
        Arrays.fill(f, 1);
        int ans = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
            ans = Math.max(ans, f[i]);
        }
        return ans;
    }
}

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class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> f(n, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    f[i] = max(f[i], f[j] + 1);
                }
            }
        }
        return *max_element(f.begin(), f.end());
    }
};

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func lengthOfLIS(nums []int) int {
    n := len(nums)
    f := make([]int, n)
    for i := range f {
        f[i] = 1
    }
    ans := 1
    for i := 1; i < n; i++ {
        for j := 0; j < i; j++ {
            if nums[j] < nums[i] {
                f[i] = max(f[i], f[j]+1)
                ans = max(ans, f[i])
            }
        }
    }
    return ans
}

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function lengthOfLIS(nums: number[]): number {
    const n = nums.length;
    const f: number[] = new Array(n).fill(1);
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (nums[j] < nums[i]) {
                f[i] = Math.max(f[i], f[j] + 1);
            }
        }
    }
    return Math.max(...f);
}

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impl Solution {
    pub fn length_of_lis(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut f = vec![1; n];
        for i in 1..n {
            for j in 0..i {
                if nums[j] < nums[i] {
                    f[i] = f[i].max(f[j] + 1);
                }
            }
        }
        *f.iter().max().unwrap()
    }
}

Solution 2

Thinking

Method 1 scans the left side linearly for the best \(f[j]\) below \(x\), which costs \(O(n^2)\). After discretizing values, that query is a prefix maximum on ranks, which a Fenwick tree maintains in \(O(\log n)\).

Scan each \(x\) in original order: query the prefix max strictly below \(x\), then write \(x\). Only left-hand information is used, so the recurrence matches Method 1 in \(O(n \log n)\) time.

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

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = 0
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        s = sorted(set(nums))
        m = len(s)
        tree = BinaryIndexedTree(m)
        for x in nums:
            x = bisect_left(s, x) + 1
            t = tree.query(x - 1) + 1
            tree.update(x, t)
        return tree.query(m)

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class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] s = nums.clone();
        Arrays.sort(s);
        int m = 0;
        int n = s.length;
        for (int i = 0; i < n; ++i) {
            if (i == 0 || s[i] != s[i - 1]) {
                s[m++] = s[i];
            }
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m);
        for (int x : nums) {
            x = search(s, x, m);
            int t = tree.query(x - 1) + 1;
            tree.update(x, t);
        }
        return tree.query(m);
    }

    private int search(int[] nums, int x, int r) {
        int l = 0;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l + 1;
    }
}

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

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

    public void update(int x, int v) {
        while (x <= n) {
            c[x] = Math.max(c[x], v);
            x += x & -x;
        }
    }

    public int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = Math.max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

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class BinaryIndexedTree {
public:
    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) {}

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

    int query(int x) {
        int mx = 0;
        while (x) {
            mx = max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }

private:
    int n;
    vector<int> c;
};

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> s = nums;
        sort(s.begin(), s.end());
        s.erase(unique(s.begin(), s.end()), s.end());
        BinaryIndexedTree tree(s.size());
        for (int x : nums) {
            x = lower_bound(s.begin(), s.end(), x) - s.begin() + 1;
            int t = tree.query(x - 1) + 1;
            tree.update(x, t);
        }
        return tree.query(s.size());
    }
};

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type BinaryIndexedTree struct {
    n int
    c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
    return &BinaryIndexedTree{n, make([]int, n+1)}
}

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

func (bit *BinaryIndexedTree) query(x int) int {
    mx := 0
    for x > 0 {
        mx = max(mx, bit.c[x])
        x -= x & -x
    }
    return mx
}

func lengthOfLIS(nums []int) int {
    n := len(nums)
    s := make([]int, n)
    copy(s, nums)
    sort.Ints(s)
    m := 0
    for i, x := range s {
        if i == 0 || x != s[i-1] {
            s[m] = x
            m++
        }
    }
    tree := newBinaryIndexedTree(m)
    for _, x := range nums {
        x = sort.SearchInts(s[:m], x) + 1
        t := tree.query(x-1) + 1
        tree.update(x, t)
    }
    return tree.query(m)
}

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

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

    update(x: number, v: number) {
        while (x <= this.n) {
            this.c[x] = Math.max(this.c[x], v);
            x += x & -x;
        }
    }

    query(x: number): number {
        let mx = 0;
        while (x) {
            mx = Math.max(mx, this.c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

function lengthOfLIS(nums: number[]): number {
    const s = [...new Set(nums)].sort((a, b) => a - b);
    const m = s.length;
    const tree = new BinaryIndexedTree(m);
    for (let x of nums) {
        x = search(s, x);
        const t = tree.query(x - 1) + 1;
        tree.update(x, t);
    }
    return tree.query(m);
}

function search(nums: number[], x: number): number {
    let l = 0,
        r = nums.length - 1;
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= x) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l + 1;
}

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