Longest Increasing Subsequence in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with 1s for each element.

DSA Typescript

const dp = new Array(nums.length).fill([1]);

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A0

B1

Cnums.length

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp with 0 instead of 1

Using nums.length as fill value

Each element starts with a subsequence length of 1 (the element itself).

2fill in blank

medium

Complete the code to check if nums[j] is less than nums[i] to extend the subsequence.

DSA Typescript

if (nums[j] [1] nums[i]) {

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A<

B>=

C===

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <

Using >= which allows equal values

We extend the subsequence only if nums[j] is smaller than nums[i].

3fill in blank

hard

Fix the error in updating dp[i] to the maximum length found so far.

DSA Typescript

dp[i] = Math.[1](dp[i], dp[j] + 1);

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Afloor

Bround

Cmax

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min which gives the shortest length

Using Math.floor or Math.round which are unrelated

We want the longest subsequence, so use Math.max to update dp[i].

4fill in blank

hard

Fill both blanks to find the maximum length in dp array after processing.

DSA Typescript

const maxLength = dp.[1]((a, b) => a [2] b ? a : b);

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Areduce

Bmax

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using max as a method (not valid on arrays)

Using < which finds minimum

Use reduce with > to find the maximum value in dp.

5fill in blank

hard

Fill all three blanks to complete the LIS function that returns the length of the longest increasing subsequence.

DSA Typescript

function lengthOfLIS(nums: number[]): number {
  const dp = new Array(nums.length).fill([1]);
  for (let i = 1; i < nums.length; i++) {
    for (let j = 0; j < i; j++) {
      if (nums[j] [2] nums[i]) {
        dp[i] = Math.[3](dp[i], dp[j] + 1);
      }
    }
  }
  return dp.reduce((a, b) => a > b ? a : b);
}

Drag options to blanks, or click blank then click option'

A1

B<

Cmax

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1 for dp initialization

Using > instead of < in condition

Using Math.min instead of Math.max

Initialize dp with 1, check if nums[j] < nums[i], and update dp[i] with Math.max.