5 Best Ways to Find Maximum Number of Non-Overlapping Subarrays with Sum Equals Target in Python

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πŸ’‘ Problem Formulation: Given an array of integers and a target sum, the goal is to find the maximum number of non-overlapping subarrays that add up to the target. For example, if the input is [1, 1, 1, 1, 1] and the target is 2, the output would be 2 because there are two non-overlapping subarrays ([1, 1] and [1, 1]) that sum to the target.

Method 1: Greedy Approach

This approach scans the array from left to right and maintains a cumulative sum. When the sum equals the target, we increment our counter and reset the sum to zero to start considering a new subarray. This method is efficient for ensuring non-overlapping subarrays as it starts a new subarray search immediately after finding a valid one.

Here’s an example:

def maxNonOverlapping(nums, target):
    count = sum = 0
    prefix_sums = {0}
    for num in nums:
        sum += num
        if sum - target in prefix_sums:
            count += 1
            sum = 0
            prefix_sums = {0}
        else:
            prefix_sums.add(sum)
    return count

print(maxNonOverlapping([1, 1, 1, 1, 1], 2))

Output: 2

In this code snippet, we use a greedy approach to iterate through the array while tracking the cumulative sum and the occurrences of this sum. Whenever the current cumulative sum minus the target is found in our prefix sums set, we found a non-overlapping subarray with the sum equal to the target. We then reset the sum and prefix sums to start finding the next subarray.

Method 2: Using a Hash Map

Here we utilize a hash map to keep track of prefix sums encountered through our array traversal. Similar to Method 1, we use the appearance of a sum equal to the current sum minus the target to identify a valid subarray. The hash map records the frequencies which assist in determining whether a particular sum has been seen before.

Here’s an example:

def maxNonOverlapping(nums, target):
    count = sum = 0
    prefix_sums = {0: 1}
    for num in nums:
        sum += num
        if (sum - target) in prefix_sums:
            count += 1
            sum = 0
            prefix_sums.clear()
            prefix_sums[0] = 1
        else:
            prefix_sums[sum] = prefix_sums.get(sum, 0) + 1
    return count

print(maxNonOverlapping([1, 1, 1, 1, 1], 2))

Output: 2

The code applies a hash map to record the prefix sums and their counts. Upon reaching a sum that allows for the target subarray sum, we increment our subarray count and reset everything to search for the next subarray. This approach ensures no overlapping as we clear the current state every time a valid subarray is found.

Method 3: Two-Pointer Technique

The two-pointer technique involves using two pointers to track the start and end of the current subarray. If the sum of the current subarray equals the target, we increment our counter and move our pointers to search for the next non-overlapping subarray.

Here’s an example:

def maxNonOverlapping(nums, target):
    count = start = sum = 0
    for end in range(len(nums)):
        sum += nums[end]
        while sum > target:
            sum -= nums[start]
            start += 1
        if sum == target:
            count += 1
            sum = 0
            start = end + 1
    return count

print(maxNonOverlapping([1, 1, 1, 1, 1], 2))

Output: 2

With the two-pointer method, we maintain a sliding window that either expands by moving the ‘end’ pointer or contracts by shifting the ‘start’ pointer. Once the target sum is reached within that window, we increase the count and reset the pointers to proceed with the search for the next subarray.

Method 4: Dynamic Programming

Dynamic programming can be utilized to solve this problem by looking at each index and deciding whether to include the current number in a new subarray or extend the previous subarray if the sum matches the target. This method generally provides a clear conceptual understanding of the solution space.

Here’s an example:

def maxNonOverlapping(nums, target):
    dp = [0] * (len(nums) + 1)
    sum = 0
    for i, num in enumerate(nums):
        sum += num
        dp[i+1] = dp[i]
        if sum == target:
            dp[i+1] += 1
            sum = 0
        # Check if sum can be formed by subtracting target from any prefix sum
        for j in range(i):
            if sum - nums[j] == target:
                dp[i+1] = max(dp[i+1], dp[j+1] + 1)
                sum -= nums[j]
    return dp[len(nums)]

print(maxNonOverlapping([1, 1, 1, 1, 1], 2))

Output: 2

The dynamic programming approach here builds on the idea of optimal substructure by using a dp array to keep track of the maximum subarrays up to each index. We iterate over the numbers and update the dp array with the count of non-overlapping subarrays that fulfill our criteria at each step.

Bonus One-Liner Method 5: Pythonic Expressiveness

Often, Python’s expressive power allows us to condense complex operations into concise one-liners. This is generally not the most efficient or maintainable approach but demonstrates Python’s ability to solve problems in unique ways.

Here’s an example:

from itertools import accumulate

maxNonOverlapping = lambda nums, target: len({sum(j) for i in range(len(nums)) for j in itertools.combinations(nums[i:], 2) if sum(j) == target})

Output: Depends on input; designed to showcase Python syntax rather than perform on all inputs efficiently.

This one-liner uses a comprehension to create all possible combinations of subarrays from the list and filters them based on the target sum. This is impractical for large arrays but serves as an exercise in Python’s expressive capabilities.

Summary/Discussion

  • Method 1: Greedy Approach. This method is fast and straightforward, making it suitable for situations where time complexity is crucial. However, it relies on a specialized condition that may not generalize well to variations of the problem.
  • Method 2: Using a Hash Map. The hash map offers a clear and efficient way to keep track of multiple sums, enhancing the algorithm’s ability to handle varying array sizes. However, the creation and resetting of hash maps can be more memory intensive.
  • Method 3: Two-Pointer Technique. It is quite efficient and intuitive for array traversal problems, particularly when subarray boundaries need to be adjusted dynamically. That said, this method can become complex with additional constraints.
  • Method 4: Dynamic Programming. Dynamic programming is a powerful tool for solving complex problems and can simplify the understanding of the solution process. However, it can lead to increased time and space complexities which may not be ideal for larger inputs.
  • Method 5: Pythonic Expressiveness. This showcases the language’s succinct syntax but is usually not practical for real-world applications due to poor scalability and performance.