Solution Approach

We implement a binary search to find the maximum ribbon length using the standard template. Since we want the maximum feasible length, we search for the first infeasible length and track the last feasible one.

Template Structure:

left, right = 1, max(ribbons) first_true_index = -1 # Tracks the answer (or 0 for this problem)  while left <= right:  mid = (left + right) // 2  if feasible(mid):  first_true_index = mid # Record this feasible length  left = mid + 1 # Try larger lengths  else:  right = mid - 1 # Try smaller lengths  return first_true_index if first_true_index != -1 else 0

Key components:

1. Initialize: Set left = 1 (minimum valid ribbon length) and right = max(ribbons). Use first_true_index = 0 to handle the case where no valid length exists.

2. Feasibility check: For each mid, count pieces: count = sum(ribbon // mid for ribbon in ribbons). If count >= k, the length is feasible.

3. Binary search logic:

   • If feasible: Record first_true_index = mid and search right (left = mid + 1) for potentially larger lengths
   • If not feasible: Search left (right = mid - 1) for smaller lengths

4. Return: first_true_index contains the maximum feasible length, or 0 if none exists.

The time complexity is O(n × log(max_ribbon)) where n is the number of ribbons. The space complexity is O(1).

Example Walkthrough

Let's walk through the solution with ribbons = [9, 7, 5] and k = 3.

Step 1: Initialize

• left = 1, right = 9 (max ribbon length)
• first_true_index = 0 (default: no valid length found)

Step 2: Binary search iterations

Iteration 1: left = 1, right = 9

• mid = (1 + 9) // 2 = 5
• Count pieces of length 5:
  • Ribbon 9: 9 // 5 = 1 piece
  • Ribbon 7: 7 // 5 = 1 piece
  • Ribbon 5: 5 // 5 = 1 piece
  • Total: 3 >= k = 3 ✓ feasible
• Update: first_true_index = 5, left = 5 + 1 = 6

Iteration 2: left = 6, right = 9

• mid = (6 + 9) // 2 = 7
• Count pieces of length 7:
  • Ribbon 9: 9 // 7 = 1 piece
  • Ribbon 7: 7 // 7 = 1 piece
  • Ribbon 5: 5 // 7 = 0 pieces
  • Total: 2 < k = 3 ✗ not feasible
• Update: right = 7 - 1 = 6

Iteration 3: left = 6, right = 6

• mid = (6 + 6) // 2 = 6
• Count pieces of length 6:
  • Ribbon 9: 9 // 6 = 1 piece
  • Ribbon 7: 7 // 6 = 1 piece
  • Ribbon 5: 5 // 6 = 0 pieces
  • Total: 2 < k = 3 ✗ not feasible
• Update: right = 6 - 1 = 5

Step 3: Loop terminates

• Now left = 6 > right = 5, so loop exits
• Return first_true_index = 5

The maximum length where we can obtain at least 3 pieces is 5.

Time and Space Complexity

The time complexity is O(n × log M), where n is the number of ribbons and M is the maximum length among all ribbons.

• The binary search runs for O(log M) iterations, as it searches through the range [0, max(ribbons)] where max(ribbons) represents the maximum ribbon length M.
• In each iteration of the binary search, we calculate sum(x // mid for x in ribbons), which takes O(n) time as it iterates through all n ribbons.
• Therefore, the overall time complexity is O(n × log M).

The space complexity is O(1).

• The algorithm only uses a constant amount of extra space for variables left, right, mid, and cnt.
• The sum operation with generator expression sum(x // mid for x in ribbons) doesn't create additional data structures and computes the sum on-the-fly.
• No additional data structures that scale with input size are created.

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

Common Pitfalls

1. Using the Wrong Binary Search Template

The Pitfall: Using while left < right with left = mid instead of the standard while left <= right template with first_true_index tracking.

Why It Happens: There are multiple binary search variants. The left < right with left = mid variant requires careful mid = (left + right + 1) // 2 calculation to avoid infinite loops. The standard template is more straightforward.

Solution: Use the standard template:

left, right = 1, max(ribbons) first_true_index = 0  while left <= right:  mid = (left + right) // 2  if feasible(mid):  first_true_index = mid  left = mid + 1  else:  right = mid - 1  return first_true_index

2. Division by Zero Error

The Pitfall: If left starts at 0, then mid could be 0, causing division by zero when calculating ribbon // mid.

Solution: Start left = 1 since ribbon length must be at least 1. The template naturally avoids this issue.

3. Not Handling the Case Where k > Total Ribbon Length

The Pitfall: Not considering that if k is larger than the sum of all ribbon lengths, it's impossible to cut k pieces even of length 1.

Solution: The binary search naturally handles this by returning first_true_index = 0 when no feasible length exists. You could add an early check for optimization:

if sum(ribbons) < k:  return 0