Solution Explanation: Cutting Ribbons

This problem asks to find the maximum length x such that we can cut at least k ribbons of length x from a given array of ribbon lengths. The solution leverages binary search for efficiency.

Approach:

The core idea is that the problem exhibits a monotonicity property: if we can cut k ribbons of length x, we can also cut k ribbons of any length less than x. This allows us to efficiently search for the maximum length using binary search.

1. Initialization: We initialize the search space. left is 1 (minimum ribbon length), and right is the maximum ribbon length in the input array.

2. Binary Search: The binary search iteratively narrows down the search space. In each iteration:

   • mid is calculated as the middle point of the current search space.
   • We count the total number of ribbons of length mid that can be obtained from the input ribbons. This is done by integer division (x // mid in Python, x / mid in other languages) for each ribbon length x and summing up the results.
   • If cnt (the total count) is greater than or equal to k, it means we can obtain at least k ribbons of length mid. Thus, we update left to mid, aiming for a potentially larger length.
   • Otherwise, we update right to mid - 1, since we can't obtain enough ribbons of length mid.

3. Result: The loop continues until left and right converge. The final value of left is the maximum length x satisfying the condition.

Time Complexity Analysis:

• The binary search iterates O(log M) times, where M is the maximum ribbon length.
• Inside the loop, we iterate through the ribbons array once to calculate cnt, which takes O(n) time, where n is the number of ribbons.
• Therefore, the overall time complexity is O(n log M).

Space Complexity Analysis:

The algorithm uses only a constant amount of extra space to store variables like left, right, mid, and cnt. Thus, the space complexity is O(1).

Code Examples (with explanations):

The code examples provided in the original response are well-structured and efficient implementations of this binary search approach. Each example's comments should clarify its steps. Here is a breakdown to highlight key points:

• left = 0, right = max(ribbons): Sets up the initial search space. Note that the left boundary could be 1, as ribbons must have a positive length. Using 0 doesn't impact correctness but slightly changes the mid calculation.

• mid = (left + right + 1) // 2: Uses integer division to ensure mid is always rounded up, favoring larger lengths within the search.

• cnt = sum(x // mid for x in ribbons): Efficiently sums up the number of ribbons obtainable of length mid.

• if cnt >= k: left = mid else: right = mid - 1: The core logic of the binary search updating the boundaries.

These codes demonstrate the efficient solution using the binary search optimization. The choice of programming language mainly influences the syntax and minor implementation details but the core algorithm remains the same.