LeetCode Meditations: Longest Increasing Subsequence

The description for this problem simply states:

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

For example:

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.

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.

Or:

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

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

Or:

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

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

Similar to the previous problem in this series, we can look at a bottom-up dynamic programming approach here as well.

For each value in the nums array, the length of the largest subsequence we can have starting from index $i$ is either:

1 (that value itself)

1 + the number of the largest subsequence we can have starting from index $i + 1$ .

However, we can't include the second option if nums[i + 1] is less than nums[i].

First, we can start out by creating a dp array to hold the length of the subsequences we can have from each index of nums onwards. That is, dp[0] will have the length of the largest subsequence that we can have from nums[0] onwards, dp[1] has the length of the largest subsequence that we can have from nums[1] onwards, and so on:

let dp = Array.from({ length: nums.length }, () => 1);

let dp = Array.from({ length: nums.length }, () => 1);

Then, we can start iterating from the last index of nums backwards (as it is the simplest position where there is only one way to form a subsequence onwards, just taking the value itself):

for (let i = nums.length - 1; i >= 0; i--) {
 /* ... */
}

for (let i = nums.length - 1; i >= 0; i--) {
 /* ... */
}

For each option, we can iterate from the next index to see if we can include the largest subsequence that can be formed from that index onwards, if so, we can get the maximum value between dp[i] and 1 + dp[j]:

for (let i = nums.length - 1; i >= 0; i--) {
  for (let j = i + 1; j < nums.length; j++) {
    if (nums[i] < nums[j]) {
      dp[i] = Math.max(dp[i], 1 + dp[j]);
    }
  }
}

for (let i = nums.length - 1; i >= 0; i--) {
  for (let j = i + 1; j < nums.length; j++) {
    if (nums[i] < nums[j]) {
      dp[i] = Math.max(dp[i], 1 + dp[j]);
    }
  }
}

Finally, we can return the largest value in dp:

function lengthOfLIS(nums: number[]): number {
  /* ... */
  return Math.max(...dp); 
}

function lengthOfLIS(nums: number[]): number {
  /* ... */
  return Math.max(...dp); 
}

And, the final solution looks like this:

function lengthOfLIS(nums: number[]): number {
  let dp = Array.from({ length: nums.length }, () => 1);

  for (let i = nums.length - 1; i >= 0; i--) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] < nums[j]) {
        dp[i] = Math.max(dp[i], 1 + dp[j]);
      }
    }
  }

  return Math.max(...dp);
}

function lengthOfLIS(nums: number[]): number {
  let dp = Array.from({ length: nums.length }, () => 1);

  for (let i = nums.length - 1; i >= 0; i--) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] < nums[j]) {
        dp[i] = Math.max(dp[i], 1 + dp[j]);
      }
    }
  }

  return Math.max(...dp);
}

Time and space complexity

The time complexity is $O(n ^ 2)$ as we iterate through each item in nums for each item in nums. The space complexity is $O(n)$ as we keep a dp array and its size will increase as the length of nums increases.

This was the last dynamic programming problem in this series. Next up, we will start a new chapter on intervals. Until then, happy coding.

Eda Eren

LeetCode Meditations: Longest Increasing Subsequence

Time and space complexity