Leetcode #2604: Minimum Time to Eat All Grains
In this guide, we solve Leetcode #2604 Minimum Time to Eat All Grains in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Array, Two Pointers, Binary Search, Sorting
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
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:
- 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.
Python Solution
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
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.