Leetcode #927: Three Equal Parts
In this guide, we solve Leetcode #927 Three Equal Parts 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
You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value. If it is possible, return any [i, j] with i + 1 < j, such that: arr[0], arr[1], ..., arr[i] is the first part, arr[i + 1], arr[i + 2], ..., arr[j - 1] is the second part, and arr[j], arr[j + 1], ..., arr[arr.length - 1] is the third part.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: arr = [1,0,1,0,1]
Output: [0,3]
Python Solution
class Solution:
def threeEqualParts(self, arr: List[int]) -> List[int]:
def find(x):
s = 0
for i, v in enumerate(arr):
s += v
if s == x:
return i
n = len(arr)
cnt, mod = divmod(sum(arr), 3)
if mod:
return [-1, -1]
if cnt == 0:
return [0, n - 1]
i, j, k = find(1), find(cnt + 1), find(cnt * 2 + 1)
while k < n and arr[i] == arr[j] == arr[k]:
i, j, k = i + 1, j + 1, k + 1
return [i - 1, j] if k == n else [-1, -1]
Complexity
The time complexity is , where is the length of arr. 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.