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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.

• 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;
}
}

############

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;
}
}

• // 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;
}
};

• 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

############

class Solution(object):
def numPairsDivisibleBy60(self, time):
"""
:type time: List[int]
:rtype: int
"""
count = collections.Counter(time)
key = list(count.keys())
N = len(key)
res = 0
for i, t in enumerate(key):
for j in range(i, N):
if (t + key[j]) % 60 == 0:
if i == j:
res += count[t] * (count[t] - 1) / 2
else:
res += count[t] * count[key[j]]
return res

• 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
}