grid47 Exploring patterns and algorithms
Oct 8, 2024
5 min read
Solution to LeetCode 300: Longest Increasing Subsequence Problem
Given an integer array ’nums’, find the length of the longest strictly increasing subsequence. A subsequence is a sequence derived by deleting some elements without changing the order of the remaining elements.
Problem
Approach
Steps
Complexity
Input: The input is an array of integers, where the length of the array can be between 1 and 2500, and the values range from -10^4 to 10^4.
Example: nums = [10, 9, 2, 5, 3, 7, 101, 18]
Constraints:
• 1 <= nums.length <= 2500
• -10^4 <= nums[i] <= 10^4
Output: The output is the length of the longest strictly increasing subsequence.
Example: For nums = [10, 9, 2, 5, 3, 7, 101, 18], the output is 4.
Constraints:
Goal: To find the length of the longest strictly increasing subsequence in the array.
Steps:
• Initialize a dp array to store the length of the longest subsequence ending at each index.
• Iterate through the array, comparing each element with previous elements to update the dp array.
• Return the maximum value from the dp array as the length of the longest subsequence.
Goal: The solution should handle the input size and time complexity constraints efficiently.
Steps:
• The array can have up to 2500 elements.
• The elements can range between -10^4 and 10^4.
Assumptions:
• The solution should account for the possibility of duplicate values in the array.
• Input: nums = [10, 9, 2, 5, 3, 7, 101, 18]
• Explanation: The longest increasing subsequence is [2, 3, 7, 101], and its length is 4.
• Input: nums = [0, 1, 0, 3, 2, 3]
• Explanation: The longest increasing subsequence is [0, 1, 2, 3], and its length is 4.
• Input: nums = [7, 7, 7, 7, 7, 7, 7]
• Explanation: Since all the elements are the same, the longest increasing subsequence has a length of 1.
Approach: To solve this problem, we need to find the length of the longest increasing subsequence using dynamic programming (DP). We will use a DP array to store the length of the longest subsequence at each index.
Observations:
• A dynamic programming approach is ideal for this problem as we can build the solution incrementally.
• To optimize, we can reduce the time complexity by using a binary search approach, which helps in finding the longest subsequence in O(n log n) time.
Steps:
• Initialize a DP array where each element starts with a value of 1.
• Iterate through the array and for each element, compare it with previous elements to update the DP array.
• For each element, use binary search to find the appropriate position to update the DP array efficiently.
• Return the maximum value from the DP array.
Empty Inputs:
• If the array is empty, the result should be 0.
Large Inputs:
• The algorithm should be optimized to handle arrays up to 2500 elements efficiently.
Special Values:
• If the array contains only one element, the longest increasing subsequence has a length of 1.
Constraints:
• Make sure to implement the solution within the O(n log n) time complexity to handle large inputs.
Defines the function `lengthOfLIS`, which computes the length of the longest increasing subsequence in the given vector `nums`.
2 : Initialize Variables
int n = nums.size(), mx =1;
Initializes the size of the input array `n` and the variable `mx` to track the length of the longest increasing subsequence.
3 : Initialize DP Array
vector<int> dp(n, 1);
Creates a dynamic programming array `dp` where each element is initialized to 1, representing the length of the longest increasing subsequence ending at that index.
4 : Outer Loop - Iterating Through Array
for(int i =0; i < n; i++) {
Begins the outer loop, which iterates through each element of the array `nums`.
5 : Inner Loop - Comparing Elements
for(int j =0; j < i; j++)
Starts the inner loop to compare each element `nums[j]` with `nums[i]` to check if an increasing subsequence can be extended.
6 : Check for Increasing Subsequence
if(nums[j] < nums[i]) {
Checks if the element at index `j` is smaller than the element at index `i`. If true, this means `nums[i]` can extend the subsequence ending at `nums[j]`.
7 : Update DP Array
dp[i] = max(dp[i], dp[j] +1);
Updates the `dp[i]` value, taking the maximum of the current value and the value of `dp[j] + 1`, indicating that `nums[i]` extends the subsequence ending at `nums[j]`.
8 : Track Maximum LIS Length
mx = max(mx, dp[i]);
Updates the `mx` variable to store the length of the longest subsequence found so far.
9 : End Outer Loop
}
Ends the outer loop after iterating through all elements in the array `nums`.
10 : Return Result
return mx;
Returns the value of `mx`, which holds the length of the longest increasing subsequence found.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n^2)
Description: In the best case, the time complexity is O(n log n), achieved using binary search. The worst case is O(n^2) with a dynamic programming approach.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n), where n is the size of the input array, due to the space used for the DP array.
