Fruit Into Baskets

Medium

In a row of trees, thei-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 1:

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

Example 2:

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].

Example 3:

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].

Example 4:

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.

Note:

  1. 1 <= tree.length <= 40000
  2. 0 <= tree[i] < tree.length

Analysis

复杂的问题背景描述之下,其实是考验转化问题的能力:

The question is actually:

What is the length of longest subarray that contains up to two distinct integers?

参考:https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/

& Longest Substring with At Most K Distinct Characters

用快慢指针解决即可,辅助HashMap存sliding window中每个distinct integer的出现次数,每次向前移动慢指针,就给慢指针对应元素的计数减1,如果减为0,则remove慢指针对应元素。

Solution

Sliding Window - Two Pointers Fast-Slow - O(n) time, O(n) space - (66 ms, faster than 44.17%)

class Solution {
    public int totalFruit(int[] tree) {
        HashMap<Integer, Integer> count = new HashMap<Integer, Integer>();
        int firstIdx = 0;
        int totalMax = 0;

        for (int i = 0; i < tree.length; i++) {
            count.put(tree[i], count.getOrDefault(tree[i], 0) + 1);
            while (count.size() > 2) {
                count.put(tree[firstIdx], count.get(tree[firstIdx]) - 1);
                if (count.get(tree[firstIdx]) == 0) {
                    count.remove(tree[firstIdx]);
                }
                firstIdx++;
            }
            totalMax = Math.max(totalMax, i - firstIdx + 1);
        }
        return totalMax;
    }
}

Reference

https://leetcode.com/problems/fruit-into-baskets/solution/

https://leetcode.com/problems/fruit-into-baskets/discuss/170745/Problem%3A-Longest-Subarray-With-2-Elements

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