Leetcode #1013: Partition Array Into Three Parts With Equal Sum
In this guide, we solve Leetcode #1013 Partition Array Into Three Parts With Equal Sum in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an array of integers arr, return true if we can partition the array into three non-empty parts with equal sums. Formally, we can partition the array if we can find indexes i + 1 < j with (arr[0] + arr[1] + ...
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Array
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: arr = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Python Solution
class Solution:
def canThreePartsEqualSum(self, arr: List[int]) -> bool:
s, mod = divmod(sum(arr), 3)
if mod:
return False
cnt = t = 0
for x in arr:
t += x
if t == s:
cnt += 1
t = 0
return cnt >= 3
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.