Problem

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 i<sup>th</sup> 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<sup>5</sup>
• 0 <= fruits[i] < fruits.length

Solution

Brute force Approach

Start at each index and extend the subarray while it contains at most two fruit types. Stop when a third type appears, then try the next starting index. Keep the maximum length.

Time: O(n²) in the worst case.

Extra space: O(1) — the set holds at most three types before the loop stops.

class Solution {
    public int totalFruit(int[] fruits) {
        
        int maxLength = 0;

        int n = fruits.length;


        for(int i=0; i<n; i++){
            Set<Integer> set = new HashSet<>();
            int length = 0;
            for(int j=i; j<n; j++){
                set.add(fruits[j]);
                if(set.size()>2){
                    break;
                }
                length += 1;
                maxLength = Math.max(maxLength, length);
            }
        }

        return maxLength;
    }
}

Better Approach

Use a sliding window with at most two fruit types. A frequency map tracks how many fruits of each type are in the window.

1. Add the fruit at r.
2. If a third type appears, move l until only two types remain. Remove a type from the map when its count reaches zero.
3. Update the maximum length using r - l + 1.

Time: O(n) expected — each pointer moves forward at most n times.

Extra space: O(1) — the map holds at most three types before shrinking.

class Solution {
    public int totalFruit(int[] fruits) {
        
        int maxLength = 0;

        int n = fruits.length;

        Map<Integer, Integer> map = new HashMap<>();

        int l = 0;
        int r = 0;

        while(r<n){
            
            map.put(fruits[r], map.getOrDefault(fruits[r], 0) + 1);

            while(map.size() > 2){

                if(map.containsKey(fruits[l])){
                    int freq = map.get(fruits[l]);
                    if(freq == 1){
                        map.remove(fruits[l]);
                    }
                    else{
                        map.put(fruits[l], map.getOrDefault(fruits[l], 0) - 1);
                    }
                    l+=1;
                }
            }

 
             maxLength = Math.max(maxLength, r-l+1);

            r+=1;
        }

        return maxLength;
    }
}

Optimal Approach

If the window has more than two types, move l only once instead of shrinking fully. The window may still be invalid, so update the maximum only when map.size() <= 2.

Each step adds one fruit and removes at most one, so the window length never decreases. This keeps searching for a longer valid window.

Time: O(n) expected — the same time complexity as the better approach.

Extra space: O(n) in the worst case — an invalid window can retain more than two types in the map.

class Solution {
    public int totalFruit(int[] fruits) {
        
        int maxLength = 0;

        int n = fruits.length;

        Map<Integer, Integer> map = new HashMap<>();

        int l = 0;
        int r = 0;

        while(r<n){
            
            map.put(fruits[r], map.getOrDefault(fruits[r], 0) + 1);

            if(map.size() > 2){

                if(map.containsKey(fruits[l])){
                    int freq = map.get(fruits[l]);
                    if(freq == 1){
                        map.remove(fruits[l]);
                    }
                    else{
                        map.put(fruits[l], map.getOrDefault(fruits[l], 0) - 1);
                    }
                    l+=1;
                }
            }

            if(map.size() <= 2)
                maxLength = Math.max(maxLength, r-l+1);

            r+=1;
        }

        return maxLength;
    }
}