Hard
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, andarr[j], arr[j + 1], ..., arr[arr.length - 1]
is the third part.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
public class Solution {
public int[] threeEqualParts(int[] arr) {
int ones = 0;
for (int num : arr) {
ones += num;
}
if (ones == 0) {
return new int[] {0, 2};
} else if (ones % 3 != 0) {
return new int[] {-1, -1};
}
ones /= 3;
int index1 = -1;
int index2 = -1;
int index3 = -1;
int totalOnes = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == 0) {
continue;
}
totalOnes += arr[i];
if (totalOnes == 1) {
index1 = i;
} else if (totalOnes == ones + 1) {
index2 = i;
} else if (totalOnes == 2 * ones + 1) {
index3 = i;
}
}
while (index3 < arr.length) {
if (arr[index1] == arr[index3] && arr[index2] == arr[index3]) {
++index1;
++index2;
++index3;
} else {
return new int[] {-1, -1};
}
}
return new int[] {index1 - 1, index2};
}
}