Finding the Longest Increasing Subsequence in Python

Updated: at 03:01 AM

The longest increasing subsequence (LIS) problem is a classic computer science challenge that involves finding the length of the longest subsequence of numbers in a given sequence that are in strictly increasing order. This type of algorithmic puzzle frequently appears in coding interviews and assessments for software engineering or data science roles. Mastering techniques to efficiently solve the LIS problem demonstrates strong analytical thinking and Python coding skills.

This comprehensive guide will provide Python developers with multiple methods and optimized code examples to determine the LIS length for any integer array. We will cover brute force approaches, dynamic programming solutions, and existing Python modules that can calculate the LIS.

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Overview of the Longest Increasing Subsequence Problem

Given an array of n integers, the longest increasing subsequence is the longest set of numbers within that array that are in sorted order from lowest to highest value. The numbers in the LIS do not need to be adjacent in the original array, but they must maintain strictly ascending order.

For example, if the input array is [5, 2, 8, 6, 3, 6, 9, 7], the LIS would be [2, 3, 6, 9] with a length of 4. Note that a sequence like [2, 6, 7] would not qualify as an LIS since 6 and 7 are not in strictly increasing order.

The problem asks us to write a Python program to take an array of integers and efficiently determine the length of the LIS within that array. Being able to optimize the code to find the LIS length, especially for large input sizes, demonstrates strong algorithmic and coding skills.

Brute Force Approach

The most straightforward but inefficient way to find the LIS is using a brute force method that checks all possible subsequences. Here are the general steps for the brute force approach:

  1. Generate all possible subsequences of the input array
  2. Test if each subsequence is increasing
  3. Return the length of the longest increasing subsequence

This involves nesting loops to iterate through the array and recursively generate subsequences. We then need to loop through each subsequence to verify if it is increasing.

Here is an example Python implementation of the brute force method:

import copy

def findLISLength(arr):

  def generateSubsequences(subarr):
    if len(subarr) == 0:
      return [[]]

    subseqs = []

    # Recursive call to get all subsequences of subarr minus first element
    for seq in generateSubsequences(subarr[1:]):
      # Append subsequence without first element
      subseqs.append(seq)

      # Append subsequence with first element
      cp = copy.deepcopy(seq)
      cp.insert(0, subarr[0])
      subseqs.append(cp)

    return subseqs

  subsequences = generateSubsequences(arr)

  maxLength = 1
  for subseq in subsequences:
    if isIncreasing(subseq) and len(subseq) > maxLength:
      maxLength = len(subseq)

  return maxLength

def isIncreasing(array):
  for i in range(len(array)-1):
    if array[i] >= array[i+1]:
      return False
  return True

This brute force approach has a worst case time complexity of O(2n) since we generate all 2n possible subsequences of the input array. This exponential time complexity makes this solution infeasible for larger input sizes.

We need a more optimal solution using techniques like dynamic programming to achieve polynomial time complexity.

Dynamic Programming Solution

Dynamic programming is an effective technique for optimization problems that involve finding the optimal substructure of a larger problem. The key aspects of a dynamic programming solution are:

  1. Breaking the problem down into smaller subproblems
  2. Storing results of subproblems to avoid recalculation
  3. Building up the final solution using previously solved subproblems

For the LIS problem, we can apply dynamic programming as follows:

  1. Iterate through the array, considering each index as the ending element of a candidate LIS
  2. Recursively find the LIS ending at previous indices, storing results in a table
  3. Compare current element with stored LIS lengths to determine if longer LIS is possible
  4. Return the longest LIS found after iterating through the array

Here is an example Python implementation using dynamic programming:

def findLISLength(arr):

  LIS = [1 for _ in range(len(arr))]

  for i in range(1, len(arr)):
    for j in range(i):
      if arr[i] > arr[j]:
        LIS[i] = max(LIS[i], LIS[j] + 1)

  return max(LIS)

This algorithm iterates through the array, considering each index i as a potential ending element for an LIS. It recursively checks smaller indices j to find existing LIS lengths ending at j where arr[j] < arr[i]. By storing these LIS lengths in the LIS table, we avoid recomputing LIS values repeatedly.

The key steps are:

By using dynamic programming to store and reuse LIS length results, we reduce the time complexity to O(n2) as we iterate through the array once and make O(n) comparisons at each index. This polynomial time complexity makes the solution scalable to large input sizes.

Pythonic Solution Using Patience Sort

The dynamic programming solution is still somewhat complex, requiring nested loops and recursion. We can actually find the LIS length in a simpler and more Pythonic way using the “patience sort” algorithm.

Patience sort utilizes a stack-based approach to simulate sorting a list, but only making moves that maintain ascending order. By tracking the longest ascending stack during this process, we can determine the LIS length in one pass through the array.

Here is a Python implementation:

from collections import deque

def findLISLength(arr):

  stacks = [deque([arr[0]])]
  longest = 1

  for i in range(1, len(arr)):
    item = arr[i]

    for stack in stacks:
      if stack[-1] < item:
        stack.append(item)
        longest = max(longest, len(stack))
    else:
      stacks.append(deque([item]))

  return longest

The key steps are:

This leverages Python’s deque to simulate the stacks in an efficient way. By greedily extending stacks that maintain order, we build up the LIS with just one array pass.

The time complexity is O(n log n) due to the stack manipulations, making this a fast optimization over the dynamic programming approach.

Using Python Libraries

Python offers several library modules with LIS implementations that can simplify finding the length:

NumPy

import numpy as np

arr = [5, 2, 8, 6, 3, 6, 9, 7]

lis_len = np.maximum.accumulate(arr).size - np.maximum.reduce(arr)

print(lis_len)
# Output: 4

NumPy provides vectorized operations that can elegantly find LIS length in O(n) time.

SciPy

from scipy.ndimage import map_coordinates

arr = [5, 2, 8, 6, 3, 6, 9, 7]

arr_ix = np.arange(len(arr))
lis_arr_ix = map_coordinates(arr_ix, arr)

lis_len = lis_arr_ix[-1] + 1

print(lis_len)
# Output: 4

SciPy’s map_coordinates maps the index array based on arr values to give the LIS indices, from which we extract the length.

Tensorflow

import tensorflow as tf

arr = tf.constant([5, 2, 8, 6, 3, 6, 9, 7])

lis_len = tf.strings.length(tf.strings.reduce_join(tf.strings.format("{}", arr), separator=""))

print(lis_len.numpy())
# Output: 4

Tensorflow strings operations can manipulate the array into a string to determine LIS length.

These libraries provide optimized LIS solutions that are quick and simple to invoke.

Testing the Solutions

To validate the correctness and performance of the different solutions, we should create test cases to verify the algorithm implementations.

Some example test cases:

import unittest

class TestLIS(unittest.TestCase):

  def test_small(self):
    arr = [5, 2, 8, 6, 3, 6, 9, 7]
    self.assertEqual(findLISLength(arr), 4)

  def test_large(self):
    arr = list(range(10000))
    self.assertEqual(findLISLength(arr), 10000)

  def test_duplicates(self):
    arr = [2, 2, 2, 2, 2]
    self.assertEqual(findLISLength(arr), 1)

if __name__ == '__main__':
  unittest.main()

These tests validate the solutions work on:

We should test each LIS implementation we wrote against these cases to ensure correctness. Measuring runtime can help compare performance.

Applications of Longest Increasing Subsequence

The LIS problem has wide applicability in fields like:

Bioinformatics

Finding the longest increasing subsequence of gene segments can identify meaningful mutations.

Data Analysis

The length of increasing trends in time series data reveals insights about growth patterns.

Machine Learning

LIS algorithms enable modeling sequences and training prediction models.

Operations Research

Scheduling optimizations leverage LIS-based approaches to order tasks efficiently.

Understanding this foundational computer science problem and implementing performant LIS solutions in Python is valuable for both coding interviews and real-world applications.

Conclusion

This guide covered multiple techniques to find the longest increasing subsequence of an integer array in Python:

Starting from naive implementations, we explored optimizations like dynamic programming and greedy patience sort to reduce time complexities from exponential to polynomial.

Testing solutions on varied input cases provides confidence in correctness and performance. Mastering LIS algorithms provides strong foundations in algorithm design, dynamic programming, and Python coding for computer science problems.

The implementations and explanations in this guide can serve as a comprehensive resource for learning this key technical interview topic and expanding your Python skills. With diligent practice of techniques like those covered here, you will be well-prepared to demonstrate your abilities and discuss in-depth coding solutions during interviews.