carloscn
commented
1 year ago Analysis
use std::collections::BinaryHeap;
pub fn fill_cups(amount: Vec<i32>) -> i32
{
let mut ret:i32 = 0;
if amount.len() < 1 {
return ret;
}
let mut pq: BinaryHeap<_> = amount.into_iter().collect();
while let Some(x) = pq.pop() {
if x == 0 {
break;
} else if let Some(y) = pq.pop() {
ret += 1;
pq.push(x - 1);
pq.push(y - 1);
} else {
ret += x;
}
}
return ret;
}
Description
You have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water.
You are given a 0-indexed integer array amount of length 3 where amount[0], amount[1], and amount[2] denote the number of cold, warm, and hot water cups you need to fill respectively. Return the minimum number of seconds needed to fill up all the cups.
Example 1:
Input: amount = [1,4,2] Output: 4 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup and a warm cup. Second 2: Fill up a warm cup and a hot cup. Second 3: Fill up a warm cup and a hot cup. Second 4: Fill up a warm cup. It can be proven that 4 is the minimum number of seconds needed.
Example 2:
Input: amount = [5,4,4] Output: 7 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup, and a hot cup. Second 2: Fill up a cold cup, and a warm cup. Second 3: Fill up a cold cup, and a warm cup. Second 4: Fill up a warm cup, and a hot cup. Second 5: Fill up a cold cup, and a hot cup. Second 6: Fill up a cold cup, and a warm cup. Second 7: Fill up a hot cup. Example 3:
Input: amount = [5,0,0] Output: 5 Explanation: Every second, we fill up a cold cup.
Constraints:
amount.length == 3 0 <= amount[i] <= 100