algoadvance

## Clarifying Questions

1. Input Format:
   • Is the input always a list of integers?
   • Can the list contain negative numbers?
   • What is the minimum and maximum length of the list?
2. Output Format:
   • Should the result be a single integer representing the count of the longest increasing subsequences?
3. Edge Cases:
   • What should be returned if the list is empty?
   • How do we handle cases where all elements are the same?

Assuming the input is a list of integers with standard constraints (like the length can be up to 2000), we can proceed.

Strategy

1. Dynamic Programming Approach:
   • Use two arrays, lengths and counts:
     • lengths[i] will store the length of the longest increasing subsequence that ends with the element at index i.
     • counts[i] will store the number of LIS that end with the element at index i.
   • Initialize both arrays with 1s because each element by itself is an increasing subsequence of length 1.
2. Update Arrays:
   • Iterate through the array. For each element at index i, check the elements before it (from index 0 to i-1).
   • If nums[j] < nums[i], it means that nums[i] can be appended to the increasing subsequence ending at j.
     • Update the lengths and counts arrays accordingly.
3. Count the LIS:
   • The length of the longest increasing subsequences is the maximum value in the lengths array.
   • To find the number of such subsequences, sum the values in the counts array where the corresponding value in lengths array is equal to the maximum length.

Code

def findNumberOfLIS(nums):
    if not nums:
        return 0
    
    n = len(nums)
    lengths = [1] * n
    counts = [1] * n
    
    for i in range(n):
        for j in range(i):
            if nums[j] < nums[i]:
                if lengths[j] + 1 > lengths[i]:
                    lengths[i] = lengths[j] + 1
                    counts[i] = counts[j]
                elif lengths[j] + 1 == lengths[i]:
                    counts[i] += counts[j]
    
    max_length = max(lengths)
    return sum(c for l, c in zip(lengths, counts) if l == max_length)

# Example usage:
nums = [1, 3, 5, 4, 7]
print(findNumberOfLIS(nums))  # Output: 2

Time Complexity

• Time Complexity: (O(n^2)) because we have a nested loop: the outer loop iterates over each element, and the inner loop iterates over all previous elements.
• Space Complexity: (O(n)) for the lengths and counts arrays.

This solution efficiently calculates and counts the longest increasing subsequences in a given list.

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