904. Fruit Into Baskets

Description

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^th 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 <= 10⁵
0 <= fruits[i] < fruits.length

Solutions

Solution 1: Hash Table + Sliding Window

Thinking

We need the longest subarray with at most two fruit types. \(n\le 10^5\), so enumerating endpoints and recounting types is too slow. After the right end absorbs a fruit, the left end must move if the window has more than two types.

A counter tracks frequencies; the number of keys is the number of types. Shrink from the left while that exceeds \(2\), and keep the maximum window length.

We use a hash table \(cnt\) to maintain the types and corresponding quantities of fruits in the current window, and use two pointers \(j\) and \(i\) to maintain the left and right boundaries of the window.

We traverse the \(\textit{fruits}\) array, add the current fruit \(x\) to the window, i.e., \(cnt[x]++\), then judge whether the types of fruits in the current window exceed \(2\). If it exceeds \(2\), we need to move the left boundary \(j\) of the window to the right until the types of fruits in the window do not exceed \(2\). Then we update the answer, i.e., \(ans = \max(ans, i - j + 1)\).

After the traversal ends, we can get the final answer.

1 2 3 2 2 1 4
^   ^
j   i


1 2 3 2 2 1 4
  ^ ^
  j i


1 2 3 2 2 1 4
  ^     ^
  j     i

The time complexity is \(O(n)\), and the space complexity is \(O(1)\). Here, \(n\) is the length of the \(\textit{fruits}\) array.

Python3JavaC++GoTypeScriptRustC#

class Solution:
    def totalFruit(self, fruits: List[int]) -> int:
        cnt = Counter()
        ans = j = 0
        for i, x in enumerate(fruits):
            cnt[x] += 1
            while len(cnt) > 2:
                y = fruits[j]
                cnt[y] -= 1
                if cnt[y] == 0:
                    cnt.pop(y)
                j += 1
            ans = max(ans, i - j + 1)
        return ans

class Solution {
    public int totalFruit(int[] fruits) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int ans = 0;
        for (int i = 0, j = 0; i < fruits.length; ++i) {
            int x = fruits[i];
            cnt.merge(x, 1, Integer::sum);
            while (cnt.size() > 2) {
                int y = fruits[j++];
                if (cnt.merge(y, -1, Integer::sum) == 0) {
                    cnt.remove(y);
                }
            }
            ans = Math.max(ans, i - j + 1);
        }
        return ans;
    }
}

class Solution {
public:
    int totalFruit(vector<int>& fruits) {
        unordered_map<int, int> cnt;
        int ans = 0;
        for (int i = 0, j = 0; i < fruits.size(); ++i) {
            int x = fruits[i];
            ++cnt[x];
            while (cnt.size() > 2) {
                int y = fruits[j++];
                if (--cnt[y] == 0) {
                    cnt.erase(y);
                }
            }
            ans = max(ans, i - j + 1);
        }
        return ans;
    }
};

func totalFruit(fruits []int) int {
    cnt := map[int]int{}
    ans, j := 0, 0
    for i, x := range fruits {
        cnt[x]++
        for ; len(cnt) > 2; j++ {
            y := fruits[j]
            cnt[y]--
            if cnt[y] == 0 {
                delete(cnt, y)
            }
        }
        ans = max(ans, i-j+1)
    }
    return ans
}

function totalFruit(fruits: number[]): number {
    const n = fruits.length;
    const cnt: Map<number, number> = new Map();
    let ans = 0;
    for (let i = 0, j = 0; i < n; ++i) {
        cnt.set(fruits[i], (cnt.get(fruits[i]) || 0) + 1);
        for (; cnt.size > 2; ++j) {
            cnt.set(fruits[j], cnt.get(fruits[j])! - 1);
            if (!cnt.get(fruits[j])) {
                cnt.delete(fruits[j]);
            }
        }
        ans = Math.max(ans, i - j + 1);
    }
    return ans;
}

use std::collections::HashMap;

impl Solution {
    pub fn total_fruit(fruits: Vec<i32>) -> i32 {
        let mut cnt = HashMap::new();
        let mut ans = 0;
        let mut j = 0;

        for (i, &x) in fruits.iter().enumerate() {
            *cnt.entry(x).or_insert(0) += 1;

            while cnt.len() > 2 {
                let y = fruits[j];
                j += 1;
                *cnt.get_mut(&y).unwrap() -= 1;
                if cnt[&y] == 0 {
                    cnt.remove(&y);
                }
            }

            ans = ans.max(i - j + 1);
        }

        ans as i32
    }
}

public class Solution {
    public int TotalFruit(int[] fruits) {
        var cnt = new Dictionary<int, int>();
        int ans = 0;
        for (int i = 0, j = 0; i < fruits.Length; ++i) {
            int x = fruits[i];
            if (cnt.ContainsKey(x)) {
                cnt[x]++;
            } else {
                cnt[x] = 1;
            }
            while (cnt.Count > 2) {
                int y = fruits[j++];
                if (cnt.ContainsKey(y)) {
                    cnt[y]--;
                    if (cnt[y] == 0) {
                        cnt.Remove(y);
                    }
                }
            }
            ans = Math.Max(ans, i - j + 1);
        }
        return ans;
    }
}

Solution 2: Monotonic Variable-Length Sliding Window

Thinking

Method 1 shrinks the window and updates the answer on every step, yet only the maximum length is required, so shorter historical windows can be ignored. Grow the window monotonically: when a third type appears, advance the left end once. The final \(n-j\) is the largest feasible window.

In Solution 1, we find that the window size sometimes increases and sometimes decreases, which requires us to update the answer each time.

But what this problem actually asks for is the maximum number of fruits, that is, the "largest" window. We don't need to shrink the window, we just need to let the window monotonically increase. So the code omits the operation of updating the answer each time, and only needs to return the size of the window as the answer after the traversal ends.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the length of the \(\textit{fruits}\) array.

Python3JavaC++GoTypeScriptRustC#

class Solution:
    def totalFruit(self, fruits: List[int]) -> int:
        cnt = Counter()
        j = 0
        for x in fruits:
            cnt[x] += 1
            if len(cnt) > 2:
                y = fruits[j]
                cnt[y] -= 1
                if cnt[y] == 0:
                    cnt.pop(y)
                j += 1
        return len(fruits) - j

class Solution {
    public int totalFruit(int[] fruits) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int j = 0, n = fruits.length;
        for (int x : fruits) {
            cnt.merge(x, 1, Integer::sum);
            if (cnt.size() > 2) {
                int y = fruits[j++];
                if (cnt.merge(y, -1, Integer::sum) == 0) {
                    cnt.remove(y);
                }
            }
        }
        return n - j;
    }
}

class Solution {
public:
    int totalFruit(vector<int>& fruits) {
        unordered_map<int, int> cnt;
        int j = 0, n = fruits.size();
        for (int& x : fruits) {
            ++cnt[x];
            if (cnt.size() > 2) {
                int y = fruits[j++];
                if (--cnt[y] == 0) {
                    cnt.erase(y);
                }
            }
        }
        return n - j;
    }
};

func totalFruit(fruits []int) int {
    cnt := map[int]int{}
    j := 0
    for _, x := range fruits {
        cnt[x]++
        if len(cnt) > 2 {
            y := fruits[j]
            cnt[y]--
            if cnt[y] == 0 {
                delete(cnt, y)
            }
            j++
        }
    }
    return len(fruits) - j
}

function totalFruit(fruits: number[]): number {
    const n = fruits.length;
    const cnt: Map<number, number> = new Map();
    let j = 0;
    for (const x of fruits) {
        cnt.set(x, (cnt.get(x) || 0) + 1);
        if (cnt.size > 2) {
            cnt.set(fruits[j], cnt.get(fruits[j])! - 1);
            if (!cnt.get(fruits[j])) {
                cnt.delete(fruits[j]);
            }
            ++j;
        }
    }
    return n - j;
}

use std::collections::HashMap;

impl Solution {
    pub fn total_fruit(fruits: Vec<i32>) -> i32 {
        let mut cnt = HashMap::new();
        let mut j = 0;
        let n = fruits.len();

        for &x in &fruits {
            *cnt.entry(x).or_insert(0) += 1;
            if cnt.len() > 2 {
                let y = fruits[j];
                j += 1;
                *cnt.get_mut(&y).unwrap() -= 1;
                if cnt[&y] == 0 {
                    cnt.remove(&y);
                }
            }
        }

        (n - j) as i32
    }
}

public class Solution {
    public int TotalFruit(int[] fruits) {
        var cnt = new Dictionary<int, int>();
        int j = 0, n = fruits.Length;
        foreach (int x in fruits) {
            if (cnt.ContainsKey(x)) {
                cnt[x]++;
            } else {
                cnt[x] = 1;
            }

            if (cnt.Count > 2) {
                int y = fruits[j++];
                if (cnt.ContainsKey(y)) {
                    cnt[y]--;
                    if (cnt[y] == 0) {
                        cnt.Remove(y);
                    }
                }
            }
        }
        return n - j;
    }
}

904. Fruit Into Baskets

Description

Solutions

Solution 1: Hash Table + Sliding Window

Solution 2: Monotonic Variable-Length Sliding Window

Comments