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2604. Minimum Time to Eat All Grains πŸ”’

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

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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class Solution:
    def minimumTime(self, hens: List[int], grains: List[int]) -> int:
        def check(t):
            j = 0
            for x in hens:
                if j == m:
                    return True
                y = grains[j]
                if y <= x:
                    d = x - y
                    if d > t:
                        return False
                    while j < m and grains[j] <= x:
                        j += 1
                    while j < m and min(d, grains[j] - x) + grains[j] - y <= t:
                        j += 1
                else:
                    while j < m and grains[j] - x <= t:
                        j += 1
            return j == m

        hens.sort()
        grains.sort()
        m = len(grains)
        r = abs(hens[0] - grains[0]) + grains[-1] - grains[0] + 1
        return bisect_left(range(r), True, key=check)

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class Solution {
    private int[] hens;
    private int[] grains;
    private int m;

    public int minimumTime(int[] hens, int[] grains) {
        m = grains.length;
        this.hens = hens;
        this.grains = grains;
        Arrays.sort(hens);
        Arrays.sort(grains);
        int l = 0;
        int r = Math.abs(hens[0] - grains[0]) + grains[m - 1] - grains[0];
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    private boolean check(int t) {
        int j = 0;
        for (int x : hens) {
            if (j == m) {
                return true;
            }
            int y = grains[j];
            if (y <= x) {
                int d = x - y;
                if (d > t) {
                    return false;
                }
                while (j < m && grains[j] <= x) {
                    ++j;
                }
                while (j < m && Math.min(d, grains[j] - x) + grains[j] - y <= t) {
                    ++j;
                }
            } else {
                while (j < m && grains[j] - x <= t) {
                    ++j;
                }
            }
        }
        return j == m;
    }
}

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class Solution {
public:
    int minimumTime(vector<int>&</