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(vector<int>& nums) {
int maxStreak = 0, currStreak = 0, lastStreak=0;
for(int i=0; i<nums.size(); i++){ // Modded Sliding Window
if(!nums[i]){
maxStreak = max(maxStreak, currStreak+lastStreak);
lastStreak = currStreak;
currStreak = 0;
}
else currStreak++;
}
int ans = max(maxStreak, currStreak+lastStreak); //Last streak might not be flanked by 0 at the end.
return ans==nums.size()?ans-1:ans; //Atleast 1 element to be removed
}
};