How do you solve Ways to Split Array Into Three Subarrays on LeetCode?

Ways to Split Array Into Three Subarrays (LeetCode #1712) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using prefix sums allows for efficient subarray sum calculations.

What is the brute force solution for Ways to Split Array Into Three Subarrays?

The brute force approach for Ways to Split Array Into Three Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to split the array into three contiguous subarrays. We will iterate through all possible split points and check the conditions for a good split. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Ways to Split Array Into Three Subarrays?

Ways to Split Array Into Three Subarrays uses the following concepts: Array, Two Pointers, Binary Search, Prefix Sum. The recommended patterns to study are: Prefix Sum, Two Pointers.

What are the interview tips for Ways to Split Array Into Three Subarrays?

Explain your thought process clearly when discussing your approach. Consider edge cases and constraints while designing your solution. Practice writing clean and efficient code under time constraints.

What are common mistakes when solving Ways to Split Array Into Three Subarrays?

Not using prefix sums to optimize sum calculations. Overlooking the need for non-empty subarrays.

#1712

Ways to Split Array Into Three Subarrays

Medium

ArrayTwo PointersBinary SearchPrefix SumPrefix SumTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a prefix sum array and two pointers to efficiently count valid splits. By iterating through the right boundary of the mid subarray, we can find the valid left boundaries using a single pass.

⚙️

Algorithm

4 steps

1Step 1: Calculate the prefix sum array for the input array.
2Step 2: Initialize two pointers: one for the left boundary and one for the mid boundary.
3Step 3: For each position of the mid boundary, adjust the left boundary to maintain the conditions for a good split.
4Step 4: Count the number of valid splits for each mid boundary position.

solution.py18 lines

1# Full working Python code
2
3def waysToSplit(nums):
4    MOD = 10**9 + 7
5    n = len(nums)
6    prefix_sum = [0] * (n + 1)
7    for i in range(n):
8        prefix_sum[i + 1] = prefix_sum[i] + nums[i]
9    count = 0
10    left = 0
11    for mid in range(1, n - 1):
12        while left < mid and prefix_sum[left + 1] <= prefix_sum[mid + 1] - prefix_sum[1]:
13            left += 1
14        right = mid
15        while right < n - 1 and prefix_sum[mid + 1] <= prefix_sum[right + 1]:
16            right += 1
17        count += right - mid
18    return count % MOD

ℹ

Complexity note: The time complexity is O(n) because we only pass through the array a constant number of times. The space complexity is O(n) due to the prefix sum array storing cumulative sums.

1Using prefix sums allows for efficient subarray sum calculations.
2Two pointers can help maintain the conditions for valid splits without redundant checks.

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