Introduction

In this blog post, we will explore how to solve the LeetCode problem "1546" 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 array nums and an integer target, return the maximum number of non-empty non-overlapping subarrays such that the sum of values in each subarray is equal to target. Example 1: Input: nums = [1,1,1,1,1], target = 2 Output: 2 Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to target(2). Example 2: Input: nums = [-1,3,5,1,4,2,-9], target = 6 Output: 2 Explanation: There are 3 subarrays with sum equal to 6. ([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping. Constraints: 1 <= nums.length <= 105 -104 <= nums[i] <= 104 0 <= target <= 106

Explanation

Here's the approach to solve this problem:

• Prefix Sum and Hash Map: Use a hash map to store the prefix sums of the array and their corresponding indices. This allows us to efficiently check if a subarray with the target sum exists.

• Greedy Approach: Iterate through the array and keep track of the current sum. When the current sum minus the target is found in the hash map, it signifies a subarray with the target sum. Increment the count of subarrays and update the starting index for the next non-overlapping subarray.

• Optimization: Reset the current sum to zero and update the starting index to the element after the identified subarray to ensure non-overlapping subarrays.

• Complexity: Time Complexity: O(n), Space Complexity: O(n)

Code

    def maxNonOverlapping(nums, target):
    """
    Finds the maximum number of non-overlapping subarrays that sum to target.

    Args:
        nums (list[int]): The input array of integers.
        target (int): The target sum.

    Returns:
        int: The maximum number of non-overlapping subarrays with sum equal to target.
    """

    count = 0
    curr_sum = 0
    seen = {0: -1}  # Initialize with sum 0 at index -1
    last_index = -1 # Index of last included element

    for i in range(len(nums)):
        curr_sum += nums[i]

        if (curr_sum - target) in seen:
            start_index = seen[curr_sum - target]

            if start_index < last_index:
                continue

            count += 1
            last_index = i #Include this element

        seen[curr_sum] = i

    return count