Introduction

In this blog post, we will explore how to solve the LeetCode problem "673" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums, return the number of longest increasing subsequences. Notice that the sequence has to be strictly increasing. Example 1: Input: nums = [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7]. Example 2: Input: nums = [2,2,2,2,2] Output: 5 Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5. Constraints: 1 <= nums.length <= 2000 -106 <= nums[i] <= 106 The answer is guaranteed to fit inside a 32-bit integer.

Explanation

Here's a breakdown of the approach and the Python code:

• Dynamic Programming: Use two arrays, length and count, to store the length of the longest increasing subsequence ending at each index and the number of such subsequences, respectively.
• Iterative Calculation: Iterate through the nums array. For each element, iterate through the elements before it. If a smaller element is found, update the length and count arrays based on the lengths and counts of the subsequences ending at the smaller element.

• Result Aggregation: Find the maximum length of all increasing subsequences and sum the counts of subsequences that achieve this maximum length.

• Runtime Complexity: O(n^2), where n is the length of the input array. Storage Complexity: O(n).

Code

    def findNumberOfLIS(nums):
    n = len(nums)
    if n == 0:
        return 0

    length = [1] * n
    count = [1] * n

    for i in range(n):
        for j in range(i):
            if nums[i] > nums[j]:
                if length[j] + 1 > length[i]:
                    length[i] = length[j] + 1
                    count[i] = count[j]
                elif length[j] + 1 == length[i]:
                    count[i] += count[j]

    max_len = max(length)
    ans = 0
    for i in range(n):
        if length[i] == max_len:
            ans += count[i]

    return ans