3790. Fruits Into Baskets Ii¶

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3790. Fruits Into Baskets II

Easy

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:

Example 2:

Constraints:

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

Solution¶

class Solution {
    public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
        int n = fruits.length;
        int vis[] = new int[n];
        int count = 0;
        for (int i = 0; i < n; i++) {
            boolean flag = false;
            for (int j = 0; j < n; j++) {
                if (vis[j] == 0 && baskets[j] >= fruits[i]) {
                    vis[j] = 1;
                    flag = true;
                    break;
                }
            }
            if (flag == false) count++;
        }
        return count;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here