How do you solve Partition Array Into Three Parts With Equal Sum on LeetCode?

Partition Array Into Three Parts With Equal Sum (LeetCode #1013) 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 total sum must be divisible by 3 for a valid partition to exist.

What is the time complexity of Partition Array Into Three Parts With Equal Sum?

The optimal solution for Partition Array Into Three Parts With Equal Sum runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the array once, maintaining a running sum and a count.

What is the brute force solution for Partition Array Into Three Parts With Equal Sum?

The brute force approach for Partition Array Into Three Parts With Equal Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to partition the array into three parts and checks if their sums are equal. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition Array Into Three Parts With Equal Sum?

Partition Array Into Three Parts With Equal Sum uses the following concepts: Array, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Partition Array Into Three Parts With Equal Sum?

Always start by analyzing the problem constraints and requirements. Think about edge cases, such as arrays with negative numbers or zeros. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Partition Array Into Three Parts With Equal Sum?

Not checking if the total sum is divisible by 3 before attempting to partition. Failing to account for the requirement that each partition must be non-empty.

#1013

Partition Array Into Three Parts With Equal Sum

Easy

ArrayGreedyHash MapArray

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 approach leverages the fact that if we can find two partitions that equal one-third of the total sum, the remaining elements will automatically sum to the same value. This reduces the problem to a linear scan of the array.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total sum of the array.
2Step 2: If the total sum is not divisible by 3, return false.
3Step 3: Set a target sum as total sum divided by 3.
4Step 4: Traverse the array while maintaining a running sum and a count of how many times we reach the target sum.
5Step 5: If we reach the target sum twice before the last element, return true.

solution.py20 lines

1# Full working Python code
2
3def can_partition(arr):
4    total_sum = sum(arr)
5    if total_sum % 3 != 0:
6        return False
7    target = total_sum // 3
8    current_sum = 0
9    count = 0
10    for i in range(len(arr)):
11        current_sum += arr[i]
12        if current_sum == target:
13            count += 1
14            current_sum = 0
15        if count == 2 and i < len(arr) - 1:
16            return True
17    return False
18
19# Example usage
20print(can_partition([0,2,1,-6,6,-7,9,1,2,0,1]))  # Output: True

ℹ

Complexity note: This complexity is linear because we only traverse the array once, maintaining a running sum and a count.

1The total sum must be divisible by 3 for a valid partition to exist.
2Finding the first two partitions that equal one-third of the total sum is sufficient to confirm the third.

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