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.
Constraints:
1 <= nums.length <= 105nums[i] is either 0 or 1.class Solution {
public int longestSubarray(int[] nums) {
int penUltStreak = 0, ultStreak = 0, ans = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 1)
ultStreak++;
else {
ans = Math.max(ans, ultStreak + penUltStreak);
penUltStreak = ultStreak;
ultStreak = 0;
}
}
ans = Math.max(ans, ultStreak + penUltStreak);
return Math.min(ans, nums.length - 1); // edge case: if all 1s, atleast one elem must be deleted
}
}