2025-08-05 Daily Challenge
Today I have done leetcode's August LeetCoding Challenge with cpp
.
August LeetCoding Challenge 5
Description
Fruits Into Baskets II
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 inbaskets[1] = 5
.fruits[1] = 2
is placed inbaskets[0] = 3
.fruits[2] = 5
cannot be placed inbaskets[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 inbaskets[0] = 6
.fruits[1] = 6
cannot be placed inbaskets[1] = 4
(insufficient capacity) but can be placed in the next available basket,baskets[2] = 7
.fruits[2] = 1
is placed inbaskets[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
Solution
class Solution {
public:
int numOfUnplacedFruits(vector<int>& fruits, vector<int>& baskets) {
int answer = 0;
for(auto f : fruits) {
for(auto &b : baskets) {
if(b < f) continue;
b = 0;
f = 0;
break;
}
answer += !!f;
}
return answer;
}
};
// Accepted
// 904/904 cases passed (1 ms)
// Your runtime beats 62.18 % of cpp submissions
// Your memory usage beats 75.84 % of cpp submissions (48.8 MB)