0904 - Fruit Into Baskets (Medium)

Problem Link

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

Problem Statement

You are visiting a farm that has a single row of fruit trees arranged from left to right. The trees are represented by an integer array fruits where fruits[i] is the type of fruit the ith tree produces.

You want to collect as much fruit as possible. However, the owner has some strict rules that you must follow:

• You only have two baskets, and each basket can only hold a single type of fruit. There is no limit on the amount of fruit each basket can hold.
• Starting from any tree of your choice, you must pick exactly one fruit from every tree (including the start tree) while moving to the right. The picked fruits must fit in one of your baskets.
• Once you reach a tree with fruit that cannot fit in your baskets, you must stop.

Given the integer array fruits, return the maximum number of fruits you can pick.

Example 1:

Input: fruits = [1,2,1]
Output: 3
Explanation: We can pick from all 3 trees.

Example 2:

Input: fruits = [0,1,2,2]
Output: 3
Explanation: We can pick from trees [1,2,2].
If we had started at the first tree, we would only pick from trees [0,1].

Example 3:

Input: fruits = [1,2,3,2,2]
Output: 4
Explanation: We can pick from trees [2,3,2,2].
If we had started at the first tree, we would only pick from trees [1,2].

Constraints:

• 1 <= fruits.length <= 10^5
• 0 <= fruits[i] < fruits.length

Approach 1: Sliding Window

C++

Written by @wingkwong

class Solution {
public:
    int totalFruit(vector<int>& fruits) {
        int n = fruits.size(), ans = 0;
        // count the frequency of fruits
        unordered_map<int, int> m;
        // two pointers 
        // - l is the pointer to the starting index of the window
        // - r is the pointer to the ending index of the window
        for (int l = 0, r = 0; r < n; r++) {
            // add fruits[r] to a hashmap
            m[fruits[r]]++;
            // if there is more than two types
            if (m.size() > 2) {
                // then we need to substract one from the freq of leftmost element, i.e. fruits[l]
                // if it is 0, then we can erase it
                if (--m[fruits[l]] == 0) m.erase(fruits[l]);
                // shrink the window by moving the `l` to the right
                l += 1;
            }
            // the maximum number of fruits we can pick is simply the window size
            ans = max(ans, r - l + 1);
        }
        return ans;
    }
};

Python

Written by @wingkwong

class Solution:
    def totalFruit(self, fruits: List[int]) -> int:
        ans = 0
        # count the frequency of fruits
        cnt = Counter()
        # two pointers 
        # - l is the pointer to the starting index of the window
        # - r is the pointer to the ending index of the window
        l = 0
        for r in range(len(fruits)):
            # add fruits[r] to a counter
            cnt[fruits[r]] += 1
            # if there is more than two types
            if len(cnt) > 2:
                # then we need to substract one from the freq of leftmost element, i.e. fruits[l]
                cnt[fruits[l]] -= 1
                # if it is 0, then we can erase it
                if cnt[fruits[l]] == 0: cnt.pop(fruits[l])
                # shrink the window by moving the `l` to the right
                l += 1
            # the maximum number of fruits we can pick is simply the window size
            ans = max(ans, r - l + 1)
        return ans