2604. Minimum Time to Eat All Grains

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.

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;
    }
}

class Solution {
public:
    int minimumTime(vector<int>& hens, vector<int>& grains) {
        int m = grains.size();
        sort(hens.begin(), hens.end());
        sort(grains.begin(), grains.end());
        int l = 0;
        int r = abs(hens[0] - grains[0]) + grains[m - 1] - grains[0];
        auto check = [&](int t) -> bool {
            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 && min(d, grains[j] - x) + grains[j] - y <= t) {
                        ++j;
                    }
                } else {
                    while (j < m && grains[j] - x <= t) {
                        ++j;
                    }
                }
            }
            return j == m;
        };
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

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)

func minimumTime(hens []int, grains []int) int {
	sort.Ints(hens)
	sort.Ints(grains)
	m := len(grains)
	l, r := 0, abs(hens[0]-grains[0])+grains[m-1]-grains[0]
	check := func(t int) bool {
		j := 0
		for _, x := range hens {
			if j == m {
				return true
			}
			y := grains[j]
			if y <= x {
				d := x - y
				if d > t {
					return false
				}
				for j < m && grains[j] <= x {
					j++
				}
				for j < m && min(d, grains[j]-x)+grains[j]-y <= t {
					j++
				}
			} else {
				for j < m && grains[j]-x <= t {
					j++
				}
			}
		}
		return j == m
	}
	for l < r {
		mid := (l + r) >> 1
		if check(mid) {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

function minimumTime(hens: number[], grains: number[]): number {
    hens.sort((a, b) => a - b);
    grains.sort((a, b) => a - b);
    const m = grains.length;
    let l = 0;
    let r = Math.abs(hens[0] - grains[0]) + grains[m - 1] - grains[0] + 1;

    const check = (t: number): boolean => {
        let j = 0;
        for (const x of hens) {
            if (j === m) {
                return true;
            }
            const y = grains[j];
            if (y <= x) {
                const 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;
    };

    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

2604 - Minimum Time to Eat All Grains

2604. Minimum Time to Eat All Grains

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2604. Minimum Time to Eat All Grains

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