Description
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
Solutions
Solution 1
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18 | class Solution:
def canThreePartsEqualSum(self, arr: List[int]) -> bool:
s = sum(arr)
if s % 3 != 0:
return False
i, j = 0, len(arr) - 1
a = b = 0
while i < len(arr):
a += arr[i]
if a == s // 3:
break
i += 1
while ~j:
b += arr[j]
if b == s // 3:
break
j -= 1
return i < j - 1
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28 | class Solution {
public boolean canThreePartsEqualSum(int[] arr) {
int s = 0;
for (int v : arr) {
s += v;
}
if (s % 3 != 0) {
return false;
}
int i = 0, j = arr.length - 1;
int a = 0, b = 0;
while (i < arr.length) {
a += arr[i];
if (a == s / 3) {
break;
}
++i;
}
while (j >= 0) {
b += arr[j];
if (b == s / 3) {
break;
}
--j;
}
return i < j - 1;
}
}
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25 | class Solution {
public:
bool canThreePartsEqualSum(vector<int>& arr) {
int s = 0;
for (int v : arr) s += v;
if (s % 3) return false;
int i = 0, j = arr.size() - 1;
int a = 0, b = 0;
while (i < arr.size()) {
a += arr[i];
if (a == s / 3) {
break;
}
++i;
}
while (~j) {
b += arr[j];
if (b == s / 3) {
break;
}
--j;
}
return i < j - 1;
}
};
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26 | func canThreePartsEqualSum(arr []int) bool {
s := 0
for _, v := range arr {
s += v
}
if s%3 != 0 {
return false
}
i, j := 0, len(arr)-1
a, b := 0, 0
for i < len(arr) {
a += arr[i]
if a == s/3 {
break
}
i++
}
for j >= 0 {
b += arr[j]
if b == s/3 {
break
}
j--
}
return i < j-1
}
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