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] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])
Example 1:
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
Example 2:
Input: arr = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:
Input: arr = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Constraints:
一個陣列,當裡面的內容相加可以分成完全相等的三個部分,並且內容都不是空值就會傳 true
否則回傳 false
可以理解成三等分的每一份恰好等於陣列總和的三分之一,所以不能整除三的就必定是 false
從陣列開頭累加,當等於三分之一的總和,則切為一份
繼續計算下一份的累加值,當累加值等於三分之二的總和時,剩下的值總和必然是另外的三分之一,所以可以回傳 true
若累加沒有達到三分之二則回傳 false
class Solution:
def canThreePartsEqualSum(self, arr: List[int]) -> bool:
s = sum(arr)
if s % 3 != 0:
return False
target = s // 3
n, i, cur = len(arr), 0, 0
while i < n:
cur += arr[i]
if cur == target:
break
i += 1
if cur != target:
return False
j = i + 1
while j + 1 < n:
cur += arr[j]
if cur == target * 2:
return True
j += 1
return False