How do you solve Number of Longest Increasing Subsequence on LeetCode?

Number of Longest Increasing Subsequence (LeetCode #673) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding how to track multiple attributes (length and count) is crucial in dynamic programming.

What is the brute force solution for Number of Longest Increasing Subsequence?

The brute force approach for Number of Longest Increasing Subsequence is Brute Force, with O(2^n * n) time complexity and O(n) space complexity. The brute-force approach involves generating all possible subsequences of the array and checking which ones are strictly increasing. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Number of Longest Increasing Subsequence?

Explain your thought process clearly; interviewers value understanding over just getting the right answer. Consider edge cases, such as arrays with all identical elements or already sorted arrays. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Number of Longest Increasing Subsequence?

Failing to initialize the counts array correctly. Not considering all previous elements when updating lengths and counts.

#673

Number of Longest Increasing Subsequence

Medium

ArrayDynamic ProgrammingBinary Indexed TreeSegment TreeDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n * n)	O(n²)
Space	O(n)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming to keep track of the lengths of the longest increasing subsequences ending at each index, along with their counts. This significantly reduces the time complexity.

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Algorithm

3 steps

1Step 1: Initialize two arrays, 'lengths' to store the length of the longest increasing subsequence ending at each index, and 'counts' to store the number of such subsequences.
2Step 2: Iterate through the array, updating 'lengths' and 'counts' based on previous elements.
3Step 3: Find the maximum length from the 'lengths' array and sum the counts of subsequences that have this maximum length.

solution.py18 lines

1# Full working Python code
2
3def findNumberOfLIS(nums):
4    if not nums:
5        return 0
6    n = len(nums)
7    lengths = [1] * n
8    counts = [1] * n
9    for i in range(n):
10        for j in range(i):
11            if nums[i] > nums[j]:
12                if lengths[j] + 1 > lengths[i]:
13                    lengths[i] = lengths[j] + 1
14                    counts[i] = counts[j]
15                elif lengths[j] + 1 == lengths[i]:
16                    counts[i] += counts[j]
17    max_length = max(lengths)
18    return sum(counts[i] for i in range(n) if lengths[i] == max_length)

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops iterating through the array, while the space complexity is O(n) for the two arrays used to store lengths and counts.

1Understanding how to track multiple attributes (length and count) is crucial in dynamic programming.
2Recognizing patterns in subsequences can help simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.