Question
Given an array A
of integers, return true
if and only 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 (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])
Example 1:
Input: [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: [0,2,1,-6,6,7,9,-1,2,0,1] Output: false
Example 3:
Input: [3,3,6,5,-2,2,5,1,-9,4] Output: true Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Note:
3 <= A.length <= 50000
-10000 <= A[i] <= 10000
Difficulty:Easy
Category:
Solution
Time complexity: O(n), Space complexity: O(1)
class Solution {
public:
bool canThreePartsEqualSum(vector<int>& A) {
int sum = 0;
for (int& a : A) sum += a;
if (sum % 3 != 0) return false;
int target = sum / 3;
int preSum = 0, ind1 = -1, ind2 = -1;
for (int i = 0; i < A.size(); ++i) {
preSum += A[i];
if (preSum == target && ind1 == -1) ind1 = 1;
else if (preSum == 2 * target && ind1 == 1) {
ind2 = 1;
break;
}
}
if (ind1 == 1 && ind2 == 1) return true;
return false;
}
};