What is the time complexity of Maximum Number of Non-Overlapping Subarrays With Sum Equals Target?

The optimal solution for Maximum Number of Non-Overlapping Subarrays With Sum Equals Target runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the array once, and the space complexity is O(n) due to the HashMap storing prefix sums.

What is the brute force solution for Maximum Number of Non-Overlapping Subarrays With Sum Equals Target?

The brute force approach for Maximum Number of Non-Overlapping Subarrays With Sum Equals Target is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if its sum equals the target. While this is straightforward, it can be very slow for larger arrays due to its nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Non-Overlapping Subarrays With Sum Equals Target?

Maximum Number of Non-Overlapping Subarrays With Sum Equals Target uses the following concepts: Array, Hash Table, Greedy, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Non-Overlapping Subarrays With Sum Equals Target?

Always consider edge cases, such as arrays with all negative numbers or zeros. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice identifying patterns in problems, as they often lead to optimal solutions.

What are common mistakes when solving Maximum Number of Non-Overlapping Subarrays With Sum Equals Target?

Not resetting the current sum and prefix sums after finding a valid subarray. Failing to account for negative numbers in the array.

#1546

Maximum Number of Non-Overlapping Subarrays With Sum Equals Target

Medium

ArrayHash TableGreedyPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a HashMap to keep track of prefix sums, allowing us to efficiently determine if a subarray sum equals the target. By greedily forming valid subarrays as soon as we find them, we ensure they are non-overlapping.

⚙️

Algorithm

3 steps

1Step 1: Initialize a HashMap to store prefix sums and a variable for the current sum.
2Step 2: Iterate through the array, updating the current sum and checking if (current sum - target) exists in the HashMap.
3Step 3: If it exists, increment the count and reset the current sum and HashMap to start a new subarray.

solution.py13 lines

1def maxNonOverlapping(nums, target):
2    count = 0
3    current_sum = 0
4    prefix_sum = {0: 1}
5    for num in nums:
6        current_sum += num
7        if (current_sum - target) in prefix_sum:
8            count += 1
9            current_sum = 0  # Reset for non-overlapping
10            prefix_sum.clear()  # Clear the map
11            prefix_sum[0] = 1  # Reset prefix sum
12        prefix_sum[current_sum] = 1
13    return count

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once, and the space complexity is O(n) due to the HashMap storing prefix sums.

1Using prefix sums allows us to efficiently check for subarray sums.
2Greedily forming subarrays as soon as we find valid ones ensures they are non-overlapping.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.