3790 Fruits Into Baskets Ii
3790 Fruits Into Baskets Ii
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:
1
2
3
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:
1
2
3
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:
1
2
3
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:
1
2
3
n == fruits.length == baskets.length
1 <= n <= 100
1 <= fruits[i], baskets[i] <= 1000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
impl Solution {
pub fn num_of_unplaced_fruits(fruits: Vec<i32>, baskets: Vec<i32>) -> i32 {
// START BY PLACING THE largest fruit
// if largest fruit is placed in a so can all other fruits be placed
// no it says leftmost basket
// in that case simply assign to each basket
// one way is 2 for loop
// what if we keep a track of largest basket to the left
let mut result = 0;
let mut baskets_mut = baskets.clone();
for k in fruits {
let mut found = false;
for l in baskets_mut.iter_mut() {
if k <= *l {
*l = 0;
found = true;
break;
}
}
if ! found {
result += 1;
}
}
return result;
}
}
This post is licensed under CC BY 4.0 by the author.