Formatted question description: https://leetcode.ca/all/1013.html
1013. Partition Array Into Three Parts With Equal Sum (Easy)
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: A = [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: A = [0,2,1,-6,6,7,9,-1,2,0,1] Output: false
Example 3:
Input: A = [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 <= A.length <= 50000-10^4 <= A[i] <= 10^4
Related Topics:
Array
Solution 1.
// OJ: https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool canThreePartsEqualSum(vector<int>& A) {
int total = accumulate(begin(A), end(A), 0), cnt = 0, sum = 0;
if (total % 3) return false;
total /= 3;
for (int i = 0; i < A.size() - 1; ++i) {
sum += A[i];
if (sum == total) {
sum = 0;
++cnt;
}
if (cnt == 2) return true;
}
return false;
}
};
Java
class Solution {
public boolean canThreePartsEqualSum(int[] A) {
int sum = 0;
for (int num : A)
sum += num;
if (sum % 3 != 0)
return false;
int partSum = sum / 3;
int count = 0;
int curSum = 0;
int length = A.length;
for (int i = 0; i < length; i++) {
curSum += A[i];
if (curSum == partSum * (count + 1))
count++;
}
return count >= 3;
}
}