Solution Approach

The solution implements binary search to find the minimum eating speed. Here's how the algorithm works:

1. Set up the binary search boundaries:

• Left boundary l = 1 (minimum possible eating speed)
• Right boundary r = max(piles) (maximum useful eating speed)

2. Define the check function: The helper function check(k) determines if Koko can finish all bananas at speed k within h hours:

def check(k: int) -> bool:
    return sum((x + k - 1) // k for x in piles) <= h

• For each pile x, calculate hours needed: (x + k - 1) // k (ceiling division)
• Sum all hours needed across all piles
• Return True if total hours ≤ h, otherwise False

3. Apply binary search: The code uses Python's bisect_left function in an elegant way:

return 1 + bisect_left(range(1, max(piles) + 1), True, key=check)

This works because:

• bisect_left searches for the leftmost position where True can be inserted in a sorted sequence
• The key=check parameter transforms each speed value into a boolean (can Koko finish in time?)
• For speeds too slow, check returns False; for speeds fast enough, it returns True
• Since we want speeds that work (True values), and False < True in Python, the sequence looks like: [False, False, ..., False, True, True, ..., True]
• bisect_left finds the first True position, which corresponds to the minimum valid speed
• Adding 1 adjusts for the 0-based indexing of bisect_left on the range starting from 1

Time Complexity: O(n * log(max(piles))) where n is the number of piles

• Binary search runs in O(log(max(piles))) iterations
• Each iteration calls check, which takes O(n) time to sum over all piles

Space Complexity: O(1) - only using constant extra space for variables

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: piles = [3, 6, 7, 11], h = 8

Step 1: Initialize binary search boundaries

• Left boundary: l = 1 (minimum eating speed)
• Right boundary: r = max([3, 6, 7, 11]) = 11 (maximum useful speed)
• Search range: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

Step 2: Apply binary search with the check function

The check(k) function calculates if speed k allows finishing within 8 hours:

• For each pile, we calculate hours needed using (pile + k - 1) // k

Let's evaluate check for different speeds:

• k = 1: Hours needed = (3+0)//1 + (6+0)//1 + (7+0)//1 + (11+0)//1 = 3 + 6 + 7 + 11 = 27 → 27 > 8 → False
• k = 2: Hours needed = (3+1)//2 + (6+1)//2 + (7+1)//2 + (11+1)//2 = 2 + 4 + 4 + 6 = 16 → 16 > 8 → False
• k = 3: Hours needed = (3+2)//3 + (6+2)//3 + (7+2)//3 + (11+2)//3 = 1 + 3 + 3 + 5 = 12 → 12 > 8 → False
• k = 4: Hours needed = (3+3)//4 + (6+3)//4 + (7+3)//4 + (11+3)//4 = 1 + 2 + 2 + 3 = 8 → 8 ≤ 8 → True
• k = 5: Hours needed = (3+4)//5 + (6+4)//5 + (7+4)//5 + (11+4)//5 = 1 + 2 + 2 + 3 = 8 → 8 ≤ 8 → True

The sequence of check results looks like:

Speed:  [1,     2,     3,     4,    5,    6,    7,    8,    9,    10,   11]
Result: [False, False, False, True, True, True, True, True, True, True, True]

Step 3: Find the minimum valid speed

Using bisect_left:

• We search for the leftmost position where True can be inserted
• The first True appears at index 3 (corresponding to speed 4)
• Since we're searching in range(1, 12), we add 1 to convert from 0-based index to actual speed
• Result: 1 + 3 = 4

Verification for k = 4:

• Pile 1 (3 bananas): needs ⌈3/4⌉ = 1 hour
• Pile 2 (6 bananas): needs ⌈6/4⌉ = 2 hours
• Pile 3 (7 bananas): needs ⌈7/4⌉ = 2 hours
• Pile 4 (11 bananas): needs ⌈11/4⌉ = 3 hours
• Total: 1 + 2 + 2 + 3 = 8 hours ✓

The minimum eating speed is 4 bananas per hour.

Time and Space Complexity

The time complexity is O(n × log M), where n is the length of the array piles and M is the maximum value in piles. This is because:

• The binary search using bisect_left searches through the range [1, max(piles)], which has M elements, taking O(log M) iterations
• For each iteration of the binary search, the check function is called, which computes the sum of (x + k - 1) // k for all elements in piles, taking O(n) time
• Therefore, the total time complexity is O(n × log M)

The space complexity is O(1). The algorithm only uses a constant amount of extra space for:

• The variable k in the check function
• The internal variables used by bisect_left
• The generator expression in the sum function doesn't create a list but computes values on-the-fly

Note that while range(1, max(piles) + 1) is passed to bisect_left, the implementation of bisect_left with a key function doesn't materialize the entire range into memory but treats it as virtual bounds for the binary search.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

The most critical mistake is using a different binary search variant that doesn't match the standard template. The template-compliant approach uses first_true_index to track the answer, while left <= right, and right = mid - 1 when feasible.

Wrong approach (alternative variant):

left, right = 1, max(piles)
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid  # WRONG: should be right = mid - 1
    else:
        left = mid + 1
return left  # WRONG: should track first_true_index

Correct template-compliant approach:

left, right = 1, max(piles)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index

2. Incorrect Binary Search Boundaries

A common mistake is setting the right boundary as sum(piles) or sum(piles) // h instead of max(piles). While these might seem logical, they're inefficient and can cause issues:

Why it's wrong:

• If k > max(piles), Koko still takes 1 hour per pile (she can't eat faster than the pile size)
• Setting r = sum(piles) creates an unnecessarily large search space
• Using sum(piles) // h might be too small if piles are unevenly distributed

Correct approach:

left, right = 1, max(piles)

3. Incorrect Ceiling Division

Many developers incorrectly calculate the hours needed for each pile:

Common mistakes:

# Wrong: Simple division truncates, underestimating hours needed
hours = pile // k  # This would give 2 for pile=7, k=3 (should be 3)

# Wrong: Using math.ceil without importing
hours = ceil(pile / k)  # NameError if math not imported

Correct approach:

# Efficient ceiling division without imports
hours = (pile + k - 1) // k

4. Integer Overflow in Other Languages

While not an issue in Python, developers coming from other languages might worry about overflow:

Safe approach:

mid = left + (right - left) // 2