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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].

Solution:​

  int totalFruit(vector<int>& fruits) {
        map<int, int> typeMap; // stores fruit type → count in current window
        int left = 0, right = 0;
        int n = fruits.size();
        int ans_max = INT_MIN;

        while (right < n) {
            typeMap[fruits[right]]++; // add fruit at 'right' to map

            // Shrink window if more than 2 types
            while (typeMap.size() > 2) {
                typeMap[fruits[left]]--; // reduce count of left fruit
                if (typeMap[fruits[left]] == 0) {
                    typeMap.erase(fruits[left]); // remove type if count is 0
                }
                left++; // shrink window from left
            }

            // update max length of valid window
            ans_max = max(ans_max, right - left + 1);
            right++;
        }

        return ans_max;
    }

How the code works​

  • Use a map to count each fruit type in the current window.
  • Expand right to include new fruits.
  • If the map holds more than 2 types, shrink window from left until only 2 remain.
  • Track max valid window size as right - left + 1.

⚠️ Edge Cases​

  • All same type → whole array valid.
  • Sudden 3rd type → triggers cleanup from left.
  • Ensure erase is done only when count becomes zero.

📉 Complexity​

  • Time: O(n) — each element added/removed once.
  • Space: O(1) — map stores at most 3 entries (bounded by unique fruit types, max 3).
💬

Discussion & Doubts