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:
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.length1 <= n <= 1001 <= fruits[i], baskets[i] <= 1000class Solution {
public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
int rem = fruits.length;
for(int f: fruits){
for(int i=0; i<baskets.length; i++)
if(baskets[i]>=f){
rem--;
baskets[i] = 0;
break;
}
}
return rem;
}
}