Problem statement

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: [2,3,7,101] (or [2,5,7,101]).

Example 2

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

Constraints

• 1 <= nums.length <= 30
• -100 <= nums[i] <= 100

Follow-up: the O(n log n) 'patience sorting' solution keeps an array of smallest tail values.

Hints for Longest Increasing Subsequence

1. Hint 1

   O(n²): dp[i] = 1 + max(dp[j]) over j < i with nums[j] < nums[i].

2. Hint 2

   O(n log n): maintain tails[k] = smallest tail of an increasing subsequence of length k+1; binary-search each element in.

How to solve Longest Increasing Subsequence

Let best[i] be the length of the longest increasing subsequence ending exactly at index i. Any such subsequence's second-to-last element sits at some earlier index j with a smaller value, so best[i] = 1 + max(best[j]) over those j. The answer is the largest entry in the table.

Approach

1. Set every best[i] to 1 — each element alone is a subsequence.
2. For each i, scan all earlier j.
3. If nums[j] < nums[i], the subsequence ending at j can be extended: take best[j] + 1 if it improves best[i].
4. Return the maximum value in best.

Why it works

Anchoring the state at the ending index is what removes the ambiguity — subsequences are not contiguous, so "the best up to i" is not enough information to extend correctly, but "the best ending at i" is. Every increasing subsequence has a unique last element, so scanning all endpoints covers each exactly once. An O(n log n) variant keeps a list of the smallest possible tail value for each length and binary-searches it, which matters at larger inputs.

Complexity

• Time — O(n^2)
• Space — O(n)

Pitfalls

• The comparison is strict (<); using <= counts equal values and turns this into non-decreasing subsequences.
• The answer is the maximum over the whole table, not best[n-1] — the longest run need not end at the last element.
• Initialise to 1, not 0; every element is a subsequence of length one.
• An all-equal array answers 1, which is a good check of the strictness.

Reference solution

Python

from typing import List
import bisect

def lengthOfLIS(nums: List[int]) -> int:
    tails = []
    for x in nums:
        i = bisect.bisect_left(tails, x)
        if i == len(tails):
            tails.append(x)
        else:
            tails[i] = x
    return len(tails)

Java

public static int lengthOfLIS(int[] nums) {
    int n = nums.length;
    int[] best = new int[n];
    Arrays.fill(best, 1);
    int answer = 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[j] < nums[i] && best[j] + 1 > best[i]) best[i] = best[j] + 1;
        }
        if (best[i] > answer) answer = best[i];
    }
    return answer;
}

C++

int lengthOfLIS(vector<int>& nums) {
    int n = (int) nums.size();
    vector<int> best(n, 1);
    int answer = 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[j] < nums[i] && best[j] + 1 > best[i]) best[i] = best[j] + 1;
        }
        if (best[i] > answer) answer = best[i];
    }
    return answer;
}

JavaScript

var lengthOfLIS = function(nums) {
    const tails = [];
    for (const x of nums) {
        let lo = 0, hi = tails.length;
        while (lo < hi) {
            const mid = (lo + hi) >> 1;
            if (tails[mid] < x) lo = mid + 1;
            else hi = mid;
        }
        tails[lo] = x;
    }
    return tails.length;
};

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