How do you solve Check if it is Possible to Split Array on LeetCode?

Check if it is Possible to Split Array (LeetCode #2811) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sum of elements is crucial for determining if an array can be split into good arrays.

What is the time complexity of Check if it is Possible to Split Array?

The optimal solution for Check if it is Possible to Split Array runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the array once, accumulating sums and counting good arrays.

What is the brute force solution for Check if it is Possible to Split Array?

The brute force approach for Check if it is Possible to Split Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible ways to split the array into smaller arrays. We will recursively attempt to split the array and check if the resulting arrays meet the 'good' criteria. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if it is Possible to Split Array?

Check if it is Possible to Split Array uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming, Array.

What are the interview tips for Check if it is Possible to Split Array?

Clearly explain your thought process and the conditions for a 'good' array. Consider edge cases, such as arrays with all elements less than m. Practice breaking down the problem into smaller parts before coding.

What are common mistakes when solving Check if it is Possible to Split Array?

Not checking the sum of the entire array before attempting splits. Failing to reset the sum after counting a good array.

#2811

Check if it is Possible to Split Array

Medium

ArrayDynamic ProgrammingGreedyGreedyDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages a greedy approach. We can track the sum of the elements and check if we can form good arrays by accumulating elements until we reach or exceed m, then reset for the next array.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to keep track of the current sum and a counter for good arrays.
2Step 2: Iterate through the array and add each element to the current sum.
3Step 3: Whenever the current sum reaches or exceeds m, increment the good array counter and reset the current sum.
4Step 4: If the good array counter reaches n (the length of nums), return true.
5Step 5: After the loop, check if the counter is equal to n; if not, return false.

solution.py9 lines

1def canSplit(nums, m):
2    current_sum = 0
3    good_arrays = 0
4    for num in nums:
5        current_sum += num
6        if current_sum >= m:
7            good_arrays += 1
8            current_sum = 0
9    return good_arrays >= len(nums)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, accumulating sums and counting good arrays.

1The sum of elements is crucial for determining if an array can be split into good arrays.
2Tracking the number of good arrays formed helps in deciding the outcome.

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