Longest Increasing Subsequence

Problem Statement

• Example 1

Input: {4,2,3,6,10,1,12}
Output: 5
Explanation: The LIS is {2,3,6,10,12}.

• Example 2

Input: {-4,10,3,7,15}
Output: 4
Explanation: The LIS is {-4,3,7,15}.

Basic Solution

A basic brute-force solution could be to try all the subsequences of the given number sequence. We can process one number at a time, so we have two options at any step:

• If the current number is greater than the previous number that we included, we can increment our count and make a recursive call for the remaining array.

• We can skip the current number to make a recursive call for the remaining array.

The length of the longest increasing subsequence will be the maximum number returned by the two recurse calls from the above two options.

Code

Here is the code:

def find_LIS_length(nums):
    return find_LIS_length_recursive(nums, 0, -1)


def find_LIS_length_recursive(nums, currentIndex,  previousIndex):
    if currentIndex == len(nums):
        return 0

    # include nums[currentIndex] if it is larger than the last included number
    c1 = 0
    if previousIndex == -1 or nums[currentIndex] > nums[previousIndex]:
        c1 = 1 + \
            find_LIS_length_recursive(nums, currentIndex + 1, currentIndex)

    # excluding the number at currentIndex
    c2 = find_LIS_length_recursive(nums, currentIndex + 1, previousIndex)

    return max(c1, c2)


def main():
    print(find_LIS_length([4, 2, 3, 6, 10, 1, 12]))
    print(find_LIS_length([-4, 10, 3, 7, 15]))


main()

Top-down Dynamic Programming with Memoization

To overcome the overlapping subproblems, we can use an array to store the already solved subproblems.

The two changing values for our recursive function are the current and the previous index. Therefore, we can store the results of all subproblems in a two-dimensional array. (Another alternative could be to use a hash-table whose key would be a string (currentIndex + “|” + previousIndex)).

Code

Here is the code:

def find_LIS_length(nums):
    n = len(nums)
    dp = [[-1 for _ in range(n+1)] for _ in range(n)]
    return find_LIS_length_recursive(dp, nums, 0, -1)


def find_LIS_length_recursive(dp, nums, currentIndex, previousIndex):
    if currentIndex == len(nums):
        return 0

    if dp[currentIndex][previousIndex + 1] == -1:
        # include nums[currentIndex] if it is larger than the last included number
        c1 = 0
        if previousIndex == -1 or nums[currentIndex] > nums[previousIndex]:
            c1 = 1 + \
                find_LIS_length_recursive(
                    dp, nums, currentIndex + 1, currentIndex)

        c2 = find_LIS_length_recursive(
            dp, nums, currentIndex + 1, previousIndex)
        dp[currentIndex][previousIndex + 1] = max(c1, c2)

    return dp[currentIndex][previousIndex + 1]


def main():
    print(find_LIS_length([4, 2, 3, 6, 10, 1, 12]))
    print(find_LIS_length([-4, 10, 3, 7, 15]))


main()

What is the time and space complexity of the above solution?

Bottom-up Dynamic Programming

The above algorithm tells us two things:

• If the number at the current index is bigger than the number at the previous index, we increment the count for LIS up to the current index.

• But if there is a bigger LIS without including the number at the current index, we take that.

So we need to find all the increasing subsequences for the number at index i, from all the previous numbers (i.e. number till index i-1), to eventually find the longest increasing subsequence.

If i represents the currentIndex and j represents the previousIndex, our recursive formula would look like:

if num[i] > num[j] => dp[i] = dp[j] + 1 if there is no bigger LIS for 'i'

Let’ s draw this visually for {-4,10,3,7,15}. Start with a subsequence of length 1, as every number will be a LIS of length 1:

From the above visualization, we can clearly see that the longest increasing subsequence is of length 4 – as shown by dp[4].

Code

Here is the code for our bottom-up dynamic programming approach:

def find_LIS_length(nums):
    n = len(nums)
    dp = [0 for _ in range(n)]
    dp[0] = 1

    maxLength = 1
    for i in range(1, n):
        dp[i] = 1
        for j in range(i):
            if nums[i] > nums[j] and dp[i] <= dp[j]:
                dp[i] = dp[j] + 1
                maxLength = max(maxLength, dp[i])

    return maxLength


def main():
    print(find_LIS_length([4, 2, 3, 6, 10, 1, 12]))
    print(find_LIS_length([-4, 10, 3, 7, 15]))


main()

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