2604. Minimum Time to Eat All Grains π
Description
There are n
hens and m
grains on a line. You are given the initial positions of the hens and the grains in two integer arrays hens
and grains
of size n
and m
respectively.
Any hen can eat a grain if they are on the same position. The time taken for this is negligible. One hen can also eat multiple grains.
In 1
second, a hen can move right or left by 1
unit. The hens can move simultaneously and independently of each other.
Return the minimum time to eat all grains if the hens act optimally.
Example 1:
Input: hens = [3,6,7], grains = [2,4,7,9] Output: 2 Explanation: One of the ways hens eat all grains in 2 seconds is described below: - The first hen eats the grain at position 2 in 1 second. - The second hen eats the grain at position 4 in 2 seconds. - The third hen eats the grains at positions 7 and 9 in 2 seconds. So, the maximum time needed is 2. It can be proven that the hens cannot eat all grains before 2 seconds.
Example 2:
Input: hens = [4,6,109,111,213,215], grains = [5,110,214] Output: 1 Explanation: One of the ways hens eat all grains in 1 second is described below: - The first hen eats the grain at position 5 in 1 second. - The fourth hen eats the grain at position 110 in 1 second. - The sixth hen eats the grain at position 214 in 1 second. - The other hens do not move. So, the maximum time needed is 1.
Constraints:
1 <= hens.length, grains.length <= 2*104
0 <= hens[i], grains[j] <= 109
Solutions
Solution 1: Sorting + Binary Search
First, sort the chickens and grains by their position from left to right. Then enumerate the time $t$ using binary search to find the smallest $t$ such that all the grains can be eaten up in $t$ seconds.
For each chicken, we use the pointer $j$ to point to the leftmost grain that has not been eaten, and the current position of the chicken is $x$ and the position of the grain is $y$. There are the following cases:
- If $y \leq x$, we note that $d = x - y$. If $d \gt t$, the current grain cannot be eaten, so directly return
false
. Otherwise, move the pointer $j$ to the right until $j=m$ or $grains[j] \gt x$. At this point, we need to check whether the chicken can eat the grain pointed to by $j$. If it can, continue to move the pointer $j$ to the right until $j=m$ or $min(d, grains[j] - x) + grains[j] - y \gt t$. - If $y \lt x$, move the pointer $j$ to the right until $j=m$ or $grains[j] - x \gt t$.
If $j=m$, it means that all the grains have been eaten, return true
, otherwise return false
.
Time complexity $O(n \times \log n + m \times \log m + (m + n) \times \log U)$, space complexity $O(\log m + \log n)$. $n$ and $m$ are the number of chickens and grains respectively, and $U$ is the maximum value of all the chicken and grain positions.
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