Longest Subarray of 1's After Deleting One Element
Given a binary array nums, you should delete one element from it.
Return the size of the longest non-empty subarray containing only 1's in the resulting array.
Return 0 if there is no such subarray.
Example 1:
Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.
Example 2:
Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].
Example 3:
Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.
Example 4:
Input: nums = [1,1,0,0,1,1,1,0,1]
Output: 4
Example 5:
Input: nums = [0,0,0]
Output: 0
Constraints:
1 <= nums.length <= 10^5
nums[i] is either 0 or 1.
Solution:
class Solution {
public int longestSubarray(int[] nums) {
int n = nums.length;
// dp[i][j] = max length of all one subarray ending at i. j = 0 -> no element deleted so far. j = 1 -> one element is deleted so far;
int[][] dp = new int[n + 1][2];
dp[0][1] = -1;
int max = 0;
for (int i = 1; i <= n; i ++) {
if (nums[i - 1] == 0) {
dp[i][0] = 0;
dp[i][1] = dp[i - 1][0];
} else {
dp[i][0] = dp[i - 1][0] + 1;
dp[i][1] = Math.max(dp[i - 1][1] + 1, dp[i - 1][0]);
}
max = Math.max(max, dp[i][1]);
}
// for (int[] arr : dp) System.out.println(Arrays.toString(arr));
return max;
}
}