How hard is LeetCode 2604. Minimum Time to Eat All Grains?

LeetCode 2604. Minimum Time to Eat All Grains is rated Hard on LeetCode.

What is the time complexity of LeetCode 2604. Minimum Time to Eat All Grains?

The Python solution on this page runs in O(n) (after optional sort O(n log n)).

What is the space complexity of LeetCode 2604. Minimum Time to Eat All Grains?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 2604. Minimum Time to Eat All Grains cover?

LeetCode 2604. Minimum Time to Eat All Grains is tagged Array, Two Pointers, Binary Search and Sorting on LeetCode.

Is LeetCode 2604. Minimum Time to Eat All Grains a premium problem?

Yes. LeetCode 2604. Minimum Time to Eat All Grains is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Minimum Time to Eat All Grains — LeetCode 2604 Python Solution

HardLeetCode PremiumArrayTwo PointersBinary SearchSorting

Problem: #2604
Pattern: Two Pointers
Reading time: 5 min
Source: leetcode.com

The problem

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.

Example

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:

Python solution

Python

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)

Complexity

Measure	Complexity
Time	O(n) (after optional sort O(n log n))
Space	O(1) auxiliary

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 2604. Minimum Time to Eat All Grains is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Two Pointers.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2604. Minimum Time to Eat All Grains?: LeetCode 2604. Minimum Time to Eat All Grains is rated Hard on LeetCode.
What is the time complexity of LeetCode 2604. Minimum Time to Eat All Grains?: The Python solution on this page runs in O(n) (after optional sort O(n log n)).
What is the space complexity of LeetCode 2604. Minimum Time to Eat All Grains?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2604. Minimum Time to Eat All Grains cover?: LeetCode 2604. Minimum Time to Eat All Grains is tagged Array, Two Pointers, Binary Search and Sorting on LeetCode.
Is LeetCode 2604. Minimum Time to Eat All Grains a premium problem?: Yes. LeetCode 2604. Minimum Time to Eat All Grains is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.