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Longest Increasing Subsequence

Certification: Intermediate Level Accuracy: 0% Submissions: 1 Points: 15

Write a C++ program to find the Longest Increasing Subsequence (LIS) in an array. Given an integer array nums, implement a function lengthOfLIS(vector<int>& nums) that returns the length of the longest strictly increasing subsequence. A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements.

Example 1
    

  • Input: nums = [10, 9, 2, 5, 3, 7, 101, 18]
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  • Output: 4
    • 

  • Explanation: 

      

    • The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4.
      • 

    • Other possible increasing subsequences are [10, 101], [9, 101], [2, 5, 7, 101], etc., but none are longer than 4 elements.
      • 

    

    •
Example 2
    

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

  • Explanation: 

      

    • The longest increasing subsequence is [0, 1, 2, 3], therefore the length is 4.
      • 

    • Note that even though [0, 1, 0, 3] is an increasing subsequence, it's not the longest one.
      • 

    

    •
Constraints
    

  • 1 ≤ nums.length ≤ 2500
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  • -10^4 ≤ nums[i] ≤ 10^4
    • 

  • Time Complexity: O(n²) for dynamic programming, O(n log n) for optimal solution
    • 

  • Space Complexity: O(n)
    •
ArraysDynamic Programming PwCSnowflake
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Hint
Solution

Solution Hints

    
  
  • Dynamic programming can be used to solve this problem with O(n²) time complexity
    • 
  
  • Create an array to store the length of LIS ending at each position
    • 
  
  • For each element, find the longest subsequence that can be formed by including it
    • 
  
  • An optimized solution using binary search can achieve O(n log n) time complexity
    • 
  
  • For the optimized solution, maintain a list that represents the smallest element seen so far that ends an increasing subsequence of length i
    •

Steps to solve by this approach:

 Step 1: Create a DP array of the same length as the input array, initialized with 1 (as each element is an increasing subsequence of length 1).
 Step 2: For each element at position i, look back at all elements at positions j < i.
 Step 3: If nums[i] > nums[j], then the element at position i can be appended to the increasing subsequence ending at position j.
 Step 4: Update dp[i] to be the maximum of its current value and dp[j] + 1.
 Step 5: After filling the entire dp array, find the maximum value in it, which represents the length of the longest increasing subsequence.
 Step 6: For the optimized O(n log n) solution, maintain a sorted subsequence and use binary search to find the correct position to place each element.
 Step 7: The length of this list at the end gives the length of the LIS.

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