Fruits into Baskets (medium)

We'll cover the following

- - Problem Statement
  - Try it yourself
- - Solution
  - Code
- - - Time Complexity
    - Space Complexity
  - Similar Problems

Problem Statement #

Given an array of characters where each character represents a fruit tree, you are given two baskets and your goal is to put maximum number of fruits in each basket. The only restriction is that each basket can have only one type of fruit.

You can start with any tree, but once you have started you can’t skip a tree. You will pick one fruit from each tree until you cannot, i.e., you will stop when you have to pick from a third fruit type.

Write a function to return the maximum number of fruits in both the baskets.

Example 1:

Input: Fruit=['A', 'B', 'C', 'A', 'C']
Output: 3
Explanation: We can put 2 'C' in one basket and one 'A' in the other from the subarray ['C', 'A', 'C']

Example 2:

Input: Fruit=['A', 'B', 'C', 'B', 'B', 'C']
Output: 5
Explanation: We can put 3 'B' in one basket and two 'C' in the other basket. 
This can be done if we start with the second letter: ['B', 'C', 'B', 'B', 'C']

Try it yourself #

Try solving this question here:

Java

Python3

C++

Solution #

This problem follows the Sliding Window pattern and is quite similar to Longest Substring with K Distinct Characters. In this problem, we need to find the length of the longest subarray with no more than two distinct characters (or fruit types!). This transforms the current problem into Longest Substring with K Distinct Characters where K=2.

Code #

Here is what our algorithm will look like, only the highlighted lines are different from Longest Substring with K Distinct Characters:

Java

Python3

C++

import java.util.*;
 
class MaxFruitCountOf2Types {
  public static int findLength(char[] arr) {
    int windowStart = 0, maxLength = 0;
    Map<Character, Integer> fruitFrequencyMap = new HashMap<>();
    // try to extend the range [windowStart, windowEnd]
    for (int windowEnd = 0; windowEnd < arr.length; windowEnd++) {
      fruitFrequencyMap.put(arr[windowEnd], fruitFrequencyMap.getOrDefault(arr[windowEnd], 0) + 1);
      // shrink the sliding window, until we are left with '2' fruits in the frequency map
      while (fruitFrequencyMap.size() > 2) {
        fruitFrequencyMap.put(arr[windowStart], fruitFrequencyMap.get(arr[windowStart]) - 1);
        if (fruitFrequencyMap.get(arr[windowStart]) == 0) {
          fruitFrequencyMap.remove(arr[windowStart]);
        }
        windowStart++; // shrink the window
      }
      maxLength = Math.max(maxLength, windowEnd - windowStart + 1);
    }
 
    return maxLength;
  }
 
  public static void main(String[] args) {
    System.out.println("Maximum number of fruits: " + 
                          MaxFruitCountOf2Types.findLength(new char[] { 'A', 'B', 'C', 'A', 'C' }));
    System.out.println("Maximum number of fruits: " + 
                          MaxFruitCountOf2Types.findLength(new char[] { 'A', 'B', 'C', 'B', 'B', 'C' }));
  }
}
 

Output

3.333s

Maximum number of fruits: 3 Maximum number of fruits: 5

Time Complexity #

The time complexity of the above algorithm will be $O(N)$ where ‘N’ is the number of characters in the input array. The outer for loop runs for all characters and the inner while loop processes each character only once, therefore the time complexity of the algorithm will be $O(N+N)$ which is asymptotically equivalent to $O(N)$ .

Space Complexity #

The algorithm runs in constant space $O(1)$ as there can be a maximum of three types of fruits stored in the frequency map.