How do you solve Split Array With Minimum Difference on LeetCode?

Split Array With Minimum Difference (LeetCode #3698) 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: Strictly increasing and decreasing conditions are crucial.

What is the time complexity of Split Array With Minimum Difference?

The optimal solution for Split Array With Minimum Difference runs in O(n) time and uses O(n) space. The optimal solution runs in linear time due to the single pass for prefix and suffix arrays, making it efficient.

What is the brute force solution for Split Array With Minimum Difference?

The brute force approach for Split Array With Minimum Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible splits of the array and validate if the left is increasing and the right is decreasing. Calculate the absolute difference for valid splits. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Split Array With Minimum Difference?

Split Array With Minimum Difference uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Two Pointers.

What are the interview tips for Split Array With Minimum Difference?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice similar problems to build confidence.

What are common mistakes when solving Split Array With Minimum Difference?

Forgetting to check strict conditions. Not handling edge cases where no valid split exists.

#3698

Split Array With Minimum Difference

Medium

ArrayPrefix 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)

Use prefix and suffix arrays to track increasing and decreasing sequences, allowing efficient validation of splits.

⚙️

Algorithm

3 steps

1Step 1: Create a boolean array 'inc' to track if nums[0..i] is strictly increasing.
2Step 2: Create a boolean array 'dec' to track if nums[i..n-1] is strictly decreasing.
3Step 3: Iterate through split points, calculate sums, and find the minimum absolute difference for valid splits.

solution.py16 lines

1def minAbsDifference(nums):
2    n = len(nums)
3    inc = [True] * n
4    dec = [True] * n
5    for i in range(1, n):
6        inc[i] = inc[i-1] and nums[i] > nums[i-1]
7    for i in range(n-2, -1, -1):
8        dec[i] = dec[i+1] and nums[i] > nums[i+1]
9    min_diff = float('inf')
10    left_sum = 0
11    for i in range(n - 1):
12        left_sum += nums[i]
13        if inc[i] and dec[i + 1]:
14            right_sum = sum(nums[i + 1:])
15            min_diff = min(min_diff, abs(left_sum - right_sum))
16    return min_diff if min_diff != float('inf') else -1

ℹ

Complexity note: The optimal solution runs in linear time due to the single pass for prefix and suffix arrays, making it efficient.

1Strictly increasing and decreasing conditions are crucial.
2Using prefix and suffix arrays allows efficient validation.

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