15. Longest Increasing Subsequence
mediumAsked at ConfluentReturn the length of the longest strictly increasing subsequence — Confluent uses it to probe DP/binary-search hybrid thinking, which connects to ordered-offset analysis over a Kafka log.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an integer array nums, return the length of the longest strictly increasing subsequence. The subsequence need not be contiguous.
Constraints
1 <= nums.length <= 2500-10^4 <= nums[i] <= 10^4
Examples
Example 1
nums=[10,9,2,5,3,7,101,18]4Example 2
nums=[0,1,0,3,2,3]4Approaches
1. DP O(n^2)
dp[i] = 1 + max(dp[j]) for j < i with nums[j] < nums[i]; answer is max(dp).
- Time
- O(n^2)
- Space
- O(n)
const dp=new Array(n).fill(1);
for (let i=0;i<n;i++)
for (let j=0;j<i;j++)
if (nums[j]<nums[i]) dp[i]=Math.max(dp[i],dp[j]+1);
return Math.max(...dp);Tradeoff:
2. Patience binary search
Maintain piles whose tops form an increasing array. For each number, binary-search for the leftmost top >= x and replace; the array length is the LIS length.
- Time
- O(n log n)
- Space
- O(n)
function lengthOfLIS(nums) {
const tails = [];
for (const x of nums) {
let lo = 0, hi = tails.length;
while (lo < hi) {
const m = (lo + hi) >> 1;
if (tails[m] < x) lo = m + 1; else hi = m;
}
tails[lo] = x;
if (lo === tails.length) tails.push(x);
}
return tails.length;
}Tradeoff:
Confluent-specific tips
Confluent will pivot to streaming — explain how each partition can maintain its own LIS array and emit it on rebalance so the downstream merger can recompute the global LIS without re-reading from the log.
Solve it now
Free. No sign-up. Python and JavaScript run instantly in your browser.
Practice these live with InterviewChamp.AI
Drill Longest Increasing Subsequence and other Confluent interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.
Practice these live with InterviewChamp.AI →