# Three Equal Parts Problem

## Description

LeetCode Problem 927.

You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value.

If it is possible, return any [i, j] with i + 1 < j, such that:

• arr[0], arr[1], …, arr[i] is the first part,
• arr[i + 1], arr[i + 2], …, arr[j - 1] is the second part, and
• arr[j], arr[j + 1], …, arr[arr.length - 1] is the third part.
• All three parts have equal binary values.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

Example 1:

``````1
2
Input: arr = [1,0,1,0,1]
Output: [0,3]
``````

Example 2:

``````1
2
Input: arr = [1,1,0,1,1]
Output: [-1,-1]
``````

Example 3:

``````1
2
Input: arr = [1,1,0,0,1]
Output: [0,2]
``````

Constraints:

• 3 <= arr.length <= 3 * 10^4
• arr[i] is 0 or 1

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution {
public:
vector<int> threeEqualParts(vector<int>& A) {
vector<int> dp;
for (int i = 0 ; i < A.size(); i++)
// this loop is used to store the index of all 1s
if (A[i])
dp.push_back(i);

if (dp.size() % 3)
// if the number of 1s cannot be devided perfectly by 3, the input is invalid
return {-1, -1};

if (dp.empty())
// if the number of 1 is zero, then it is natually valid, return {0, 2}
return {0,2};

//if we want to devide into 3 parts, the distribution pattern of 1s in three parts should be the same
int l1 = 0, l2 = dp.size() / 3, l3 = l2 * 2;
for (int i = 1; i < l2; i++ ) {
int diff = dp[i] - dp[i-1];
if (dp[l2+i] - dp[l2+i-1] != diff || dp[l3+i] - dp[l3+i-1] != diff)
//unmatched pattern
return {-1, -1};
}

// calculate how many 0s tail
int tail0 = A.size() - dp.back();
if (dp[l3] - dp[l3-1] < tail0 ||   dp[l2] - dp[l2-1] < tail0)
// all three parts should tail with the same number of 0s with that in the last part
return {-1,-1};

return {dp[l2-1] + tail0 - 1, dp[l3-1] + tail0};
}
};
``````