The process we are illustrating above is walking us through finding the perfect subarray that obeys the conditions of having late bloomers between the blooming flowers.

The value obtained is then compared with the current value we have in the min.

When we reach the end of the window we check for the maximum of left and right elements.

If we are not then we consider the next window.

As long as we are obeying this rule we continue.

We check for each element if it is greater than left and right.

At any point of time if left=i then right=i+k+1.

We maintain a sliding window of size k+2.

We thus start with our sliding window for K Empty Slots.

The above two steps help us to proceed with the sliding window and hence are important for us to proceed.

We maintain a variable min to hold the answer.

Days-1] will hold the position i.e i+1.

Let us create an array named days to hold the position of the flower number that is blooming on that particular day.

Without any further delay let us jump right into the solution of K Empty Slots We have a wide range of problems on this topic that you can check out before we proceed on to learn further.

The approach I am about to describe uses the sliding window approach. Image illustrating the same An illustration showing the blooming pattern How to approach the problem? On the second day, the third and the fifth flower bloom leaving a flower that is still blooming. So, rather than exaggerating, I will be running through a sample for an input. The mystery behind K Empty Slots is thus unlocked, or is it? We will be finding the day on which two flowers are blooming with k non-blooming flowers between them.

Complexity Analysis What is the blooming condition?.

If the result is equal to INT_MAX return -1 else return res.

If the flag is true, set result as maximum value.

At each step traverse from 1 to k and check if days is less than the maximum value break the inner loop and set flag false.

Store the left and right values with k slots between in 2 variables and find it’s maximum. Traverse the days array with i starting from zero and update the array as days] = i+1.Initialize another array days to store days o which each flower will bloom.Initialize an array a and an integer k.

So, without taking much time we move to algorithm used to solve this problem. Now we know the problem statement of K Empty Slots LeetCode.

Java Program for K Empty Slots LeetCode.

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