Longest Increasing Subsequence

mediumcodingAlgorithmsDynamic ProgrammingArrays~30 min

Asked at Google, Meta, Stripe

Overview

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

A subsequence is derived from the array by deleting some or no elements without changing the relative order of the remaining elements.

Constraints

1 <= nums.length <= 2500
-10^4 <= nums[i] <= 10^4
Aim for O(n log n) but O(n²) DP is acceptable.

Examples

lengthOfLIS([10, 9, 2, 5, 3, 7, 101, 18]);
// => 4  (subsequence: [2, 3, 7, 101])

lengthOfLIS([0, 1, 0, 3, 2, 3]);
// => 4  (subsequence: [0, 1, 2, 3])

lengthOfLIS([7, 7, 7, 7]);
// => 1  (all equal — strictly increasing means length 1)

Solution

Reveal solution

function lengthOfLIS(nums) {
  const tails = [];

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

  return tails.length;
}

The patience sorting approach: maintain an array tails where tails[i] is the smallest tail element for an increasing subsequence of length i+1. Binary search for the insertion point of each element. O(n log n).

Resources

longest-increasing-subsequence.js

Longest Increasing Subsequence

mediumcodingAlgorithmsDynamic Programming

Overview

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

A subsequence is derived from the array by deleting some or no elements without changing the relative order of the remaining elements.

Constraints

1 <= nums.length <= 2500
-10^4 <= nums[i] <= 10^4
Aim for O(n log n) but O(n²) DP is acceptable.

Examples

lengthOfLIS([10, 9, 2, 5, 3, 7, 101, 18]);
// => 4  (subsequence: [2, 3, 7, 101])

lengthOfLIS([0, 1, 0, 3, 2, 3]);
// => 4  (subsequence: [0, 1, 2, 3])

lengthOfLIS([7, 7, 7, 7]);
// => 1  (all equal — strictly increasing means length 1)

Solution

Reveal solution

function lengthOfLIS(nums) {
  const tails = [];

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

  return tails.length;
}

The patience sorting approach: maintain an array tails where tails[i] is the smallest tail element for an increasing subsequence of length i+1. Binary search for the insertion point of each element. O(n log n).

Resources

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