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:
-
3 <= arr.length <= 5 * 104
-
-104 <= arr[i] <= 104
SOLUTION:
class Solution:
def canThreePartsEqualSum(self, arr: List[int]) -> bool:
n = len(arr)
total = sum(arr)
curr = 0
f = -1
s = -1
for i in range(n):
curr += arr[i]
if f < 0 and curr * 3 == total:
f = i
if curr * 3 == 2 * total:
s = i
if f >= 0 and s >= 0 and f < s and s < n - 1:
return True
return False
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