[Google Question][LeetCode] Fruit Into Baskets

Problem: In a row of trees, the i-th tree produces fruit with type tree[i]. You start at any tree of your choice, then repeatedly perform the following steps:

1. Add one piece of fruit from this tree to your baskets.  If you cannot, stop.
2. Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

Example:

Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].

Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2]. If we started at the first tree, we would only collect [0, 1].

Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2]. If we started at the first tree, we would only collect [1, 2].

Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2]. If we started at the first tree or the eighth tree, we would only collect 4 fruits.

Approach: Basically we can use sliding window here. We increment the right boundary till we have only 2 types. Once we see we get more than 2 types in the window, we keep incrementing the left and also we keep reduce the current count till we get only 2 types in the current window.

In the end we just need to return the max of all the current counts. That's all!

Implementation  in C#:

        public int TotalFruit(int[] tree)

        {

            Dictionary<int, int> freqMap = new Dictionary<int, int>();

            int maxCount = 0;

            int left = 0, right = 0;

            while (right < tree.Length)

            {

                if (freqMap.ContainsKey(tree[right]))

                {

                    ++freqMap[tree[right]];

                }

                else

                {

                    if (freqMap.Count == 2)

                    {

                        int i = right - 2;

                        // No need to remove continuous elements which are same as element at previous index

                        while (i >= left && tree[i] == tree[i + 1])

                        {

                            --i;

                        }

                        int tempLeft = i + 1;

                        while (i >= left)

                        {

                            --freqMap[tree[i]];

                            if (freqMap[tree[i]] == 0)

                            {

                                freqMap.Remove(tree[i]);

                            }

                            --i;

                        }

                        left = tempLeft;

                    }

                    freqMap[tree[right]] = 1;

                }

                maxCount = Math.Max(maxCount, right - left + 1);

                ++right;

            }

            return maxCount;

        }

Complexity: O(n)