927. Three Equal Parts

Problem Link

Description

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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.
• All three parts have equal binary values.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

 

Example 1:

Input: arr = [1,0,1,0,1]
Output: [0,3]

Example 2:

Input: arr = [1,1,0,1,1]
Output: [-1,-1]

Example 3:

Input: arr = [1,1,0,0,1]
Output: [0,2]

 

Constraints:

• 3 <= arr.length <= 3 * 104
• arr[i] is 0 or 1

Solution

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Python3

class Solution:
	def threeEqualParts(self, A: List[int]) -> List[int]:
 
		onesCountToIndex = {}
		count = 0
		for i, a in enumerate(A):
			if a != 0:
				count += 1
				onesCountToIndex[count] = i
		
		if count % 3 != 0:
			return [-1, -1]
		
		if count == 0:
			return [0, len(A)-1]
		
		numOnesPerPart = count // 3
		
		left = onesCountToIndex[1]
		mid = onesCountToIndex[1 + numOnesPerPart]
		right = onesCountToIndex[1 + numOnesPerPart*2]
		
		while left < mid < right < len(A) and A[left] == A[mid] == A[right]:
			left += 1
			mid += 1
			right += 1
		if right == len(A):
			return [left-1, mid]
		return [-1, -1]