# 927. Three Equal Parts

## Description

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:

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

Example 2:

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

Example 3:

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

Constraints:

• 3 <= arr.length <= 3 * 104
• arr[i] is 0 or 1

## Solutions

• class Solution {
private int[] arr;

public int[] threeEqualParts(int[] arr) {
this.arr = arr;
int cnt = 0;
int n = arr.length;
for (int v : arr) {
cnt += v;
}
if (cnt % 3 != 0) {
return new int[] {-1, -1};
}
if (cnt == 0) {
return new int[] {0, n - 1};
}
cnt /= 3;

int i = find(1), j = find(cnt + 1), k = find(cnt * 2 + 1);
for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {
}
return k == n ? new int[] {i - 1, j} : new int[] {-1, -1};
}

private int find(int x) {
int s = 0;
for (int i = 0; i < arr.length; ++i) {
s += arr[i];
if (s == x) {
return i;
}
}
return 0;
}
}

• class Solution {
public:
vector<int> threeEqualParts(vector<int>& arr) {
int n = arr.size();
int cnt = accumulate(arr.begin(), arr.end(), 0);
if (cnt % 3) return {-1, -1};
if (!cnt) return {0, n - 1};
cnt /= 3;

auto find = [&](int x) {
int s = 0;
for (int i = 0; i < n; ++i) {
s += arr[i];
if (s == x) return i;
}
return 0;
};
int i = find(1), j = find(cnt + 1), k = find(cnt * 2 + 1);
for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {}
return k == n ? vector<int>{i - 1, j} : vector<int>{-1, -1};
}
};

• class Solution:
def threeEqualParts(self, arr: List[int]) -> List[int]:
def find(x):
s = 0
for i, v in enumerate(arr):
s += v
if s == x:
return i

n = len(arr)
cnt, mod = divmod(sum(arr), 3)
if mod:
return [-1, -1]
if cnt == 0:
return [0, n - 1]

i, j, k = find(1), find(cnt + 1), find(cnt * 2 + 1)
while k < n and arr[i] == arr[j] == arr[k]:
i, j, k = i + 1, j + 1, k + 1
return [i - 1, j] if k == n else [-1, -1]

• func threeEqualParts(arr []int) []int {
find := func(x int) int {
s := 0
for i, v := range arr {
s += v
if s == x {
return i
}
}
return 0
}
n := len(arr)
cnt := 0
for _, v := range arr {
cnt += v
}
if cnt%3 != 0 {
return []int{-1, -1}
}
if cnt == 0 {
return []int{0, n - 1}
}
cnt /= 3
i, j, k := find(1), find(cnt+1), find(cnt*2+1)
for ; k < n && arr[i] == arr[j] && arr[j] == arr[k]; i, j, k = i+1, j+1, k+1 {
}
if k == n {
return []int{i - 1, j}
}
return []int{-1, -1}
}

• /**
* @param {number[]} arr
* @return {number[]}
*/
var threeEqualParts = function (arr) {
function find(x) {
let s = 0;
for (let i = 0; i < n; ++i) {
s += arr[i];
if (s == x) {
return i;
}
}
return 0;
}
const n = arr.length;
let cnt = 0;
for (const v of arr) {
cnt += v;
}
if (cnt % 3) {
return [-1, -1];
}
if (cnt == 0) {
return [0, n - 1];
}
cnt = Math.floor(cnt / 3);
let [i, j, k] = [find(1), find(cnt + 1), find(cnt * 2 + 1)];
for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {}
return k == n ? [i - 1, j] : [-1, -1];
};