3477. Fruits Into Baskets II

3477. Fruits Into Baskets II

Difficulty: Easy

Topics: Array, Binary Search, Segment Tree, Simulation, Ordered Set, Weekly Contest 440

You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the ith type of fruit, and baskets[j] represents the capacity of the jth basket.

From left to right, place the fruits according to these rules:

• Each fruit type must be placed in the leftmost available basket with a capacity greater than or equal to the quantity of that fruit type.
• Each basket can hold only one type of fruit.
• If a fruit type cannot be placed in any basket, it remains unplaced.

Return the number of fruit types that remain unplaced after all possible allocations are made.

Example 1:

• Input: fruits = [4,2,5], baskets = [3,5,4]
• Output: 1
• Explanation:
  • fruits[0] = 4 is placed in baskets[1] = 5.
  • fruits[1] = 2 is placed in baskets[0] = 3.
  • fruits[2] = 5 cannot be placed in baskets[2] = 4.

Since one fruit type remains unplaced, we return 1.

Example 2:

• Input: fruits = [3,6,1], baskets = [6,4,7]
• Output: 0
• Explanation:
  • fruits[0] = 3 is placed in baskets[0] = 6.
  • fruits[1] = 6 cannot be placed in baskets[1] = 4 (insufficient capacity) but can be placed in the next available basket, baskets[2] = 7.
  • fruits[2] = 1 is placed in baskets[1] = 4.

Since all fruits are successfully placed, we return 0.

Constraints:

• n == fruits.length == baskets.length
• 1 <= n <= 100
• 1 <= fruits[i], baskets[i] <= 1000

Hint:

1. Simulate the operations for each fruit as described

Solution:

We need to determine the number of fruit types that remain unplaced after attempting to place each fruit type into the leftmost available basket that can accommodate it. The key challenge is efficiently matching each fruit to the earliest basket that meets the capacity requirement while ensuring each basket is used only once.

Approach

1. Problem Analysis: The problem involves two arrays, fruits and baskets, each of the same length. Each fruit must be placed in the leftmost basket that has sufficient capacity (i.e., the basket's capacity is at least the fruit's quantity) and is not already used. If no such basket is available, the fruit remains unplaced.
2. Intuition: For each fruit, we scan the baskets from left to right to find the first available basket that can hold the fruit. Once found, we mark the basket as used and proceed to the next fruit. If no basket is found, the fruit is counted as unplaced.
3. Algorithm Selection: We use a simple simulation approach. We maintain a boolean array to track which baskets have been used. For each fruit, we iterate through the baskets in order, checking for the first available basket that meets the capacity requirement. The algorithm efficiently handles the constraints given the problem size (n ≤ 100).
4. Complexity Analysis: The algorithm involves a nested loop where each fruit is processed in O(n) time, leading to an overall time complexity of O(n²). The space complexity is O(n) for the boolean array tracking used baskets.

Let's implement this solution in PHP: 3477. Fruits Into Baskets II

<?php
/**
 * @param Integer[] $fruits
 * @param Integer[] $baskets
 * @return Integer
 */
function numOfUnplacedFruits($fruits, $baskets) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// === Example usage ===

// Example 1:
$fruits1 = array(4, 2, 5);
$baskets1 = array(3, 5, 4);
$result1 = numOfUnplacedFruits($fruits1, $baskets1);
echo "Example 1 Output: " . $result1 . "\n"; // expected 1

// Example 2:
$fruits2 = array(3, 6, 1);
$baskets2 = array(6, 4, 7);
$result2 = numOfUnplacedFruits($fruits2, $baskets2);
echo "Example 2 Output: " . $result2 . "\n"; // expected 0
?>

Explanation:

1. Initialization: We start by determining the number of fruits (or baskets) n. We initialize a boolean array taken to keep track of which baskets have been used, setting all entries to false initially. The variable unplaced counts the number of fruits that cannot be placed.
2. Processing Fruits: For each fruit in the array:
   • We scan the baskets from left to right.
   • For each basket, if it is not taken and its capacity is sufficient for the current fruit, we mark it as taken and move to the next fruit.
   • If no suitable basket is found during the scan, we increment the unplaced count.
3. Result: After processing all fruits, the value of unplaced gives the number of fruit types that could not be placed into any basket, which is returned as the result.

This approach efficiently matches each fruit to the earliest available basket, ensuring optimal placement while adhering to the problem constraints. The solution is straightforward and leverages direct simulation to achieve the desired outcome.

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