ORIGIN

LeetCode 2335. Minimum Amount of Time to Fill Cups

2023-02-05

ACM Math Heap 2 mins378 words

Problem Description

2335. Minimum Amount of Time to Fill Cups

Analysis

Denote the column value as a >= b >= c

From the above formula, we can get two situations:

a >= b + c

In this situation, we can easily get that, first, use $b$ steps, then use $c$, steps, then use $a - b - c$ steps.

So the total steps are $a$ steps.
a < b + c

define $b := c + x$, $a := b + y = c + x + y$

So the stones are [c+x+y, c+x, c]

In order to achieve min steps, we should distribute the a properly to b and c.
1. first, we use x steps to remove from the first two stones.
  
  Now the stones are [c+y, c, c]
  
  since a >= b >= c, so c+x+y < c+x + c => y < c
2. if c+y is even
  
  define c+y = 2k because c+y is even.
  
  because y < c, so k < c
  
  so the stones are [2k, c, c]. So the steps are:
  - k steps for the first to stone, [k, c-k, c]
  - k steps for the first and third stone, [0, c-k, c-k]
  - c-k steps for the last two stones: [0, 0, 0]
  So in this case, the total steps are:
  $$
  x + k + k + c - k \
  = (b - c) + c + k \
  = b + k
  = b + \frac{c + y}{2} = b + \frac{c+a-b}{2} \
  = \frac{a + b + c}{2}
  $$
3. if c+y is even
  
  define c+y = 2k + 1, k < c
  
  so the stones are [2k + 1, c, c] and the steps are:
  - k steps for first two: [k + 1, c - k, c - k]
  - k steps for first and third stone : [1, c-k, c-k]
  - c - k steps for last two: [1, 0, 0]
  - 1 step for first stone
  So the total steps are:
  $$
  x + k + k + c - k + 1 = \frac{a+b+c}{2} + 1 = ceil(\frac{a+b+c}{2})
  $$

Code

class Solution {
public:
    int fillCups(vector<int>& amount) {
        int sum = 0;
        int maxN = 0;
        for(int i : amount) {
            sum += i;
            maxN = max(maxN, i);
        }
        return max((sum + 1) / 2, maxN);
    }
};