What is the time complexity of Maximum Number That Sum of the Prices Is Less Than or Equal to K?

The optimal solution for Maximum Number That Sum of the Prices Is Less Than or Equal to K runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the binary search over n and the linear accumulation of prices, which is much more efficient than the brute force approach.

What is the brute force solution for Maximum Number That Sum of the Prices Is Less Than or Equal to K?

The brute force approach for Maximum Number That Sum of the Prices Is Less Than or Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number from 1 to n, calculating its price based on the set bits at specified positions, and accumulating these prices until we exceed k. This is straightforward but inefficient for larger values of k. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Number That Sum of the Prices Is Less Than or Equal to K?

Maximum Number That Sum of the Prices Is Less Than or Equal to K uses the following concepts: Math, Binary Search, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Binary Search, Dynamic Programming.

What are the interview tips for Maximum Number That Sum of the Prices Is Less Than or Equal to K?

Always start with a brute force solution to understand the problem before optimizing. Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice calculating accumulated prices efficiently, as this is a common requirement in similar problems.

What are common mistakes when solving Maximum Number That Sum of the Prices Is Less Than or Equal to K?

Not correctly implementing the bit counting logic, leading to incorrect price calculations. Failing to handle edge cases where k is very small or very large.

#3007

Maximum Number That Sum of the Prices Is Less Than or Equal to K

Medium

MathBinary SearchDynamic ProgrammingBit ManipulationBinary SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

Using binary search allows us to efficiently find the greatest cheap number without checking every number up to k. We can leverage the properties of the accumulated price and the price calculation to narrow down our search space.

⚙️

Algorithm

5 steps

1Step 1: Define a function to calculate the accumulated price for a given number num.
2Step 2: Use binary search to find the maximum number where the accumulated price is less than or equal to k.
3Step 3: In each iteration of the binary search, calculate the accumulated price for the mid-point.
4Step 4: If the accumulated price is less than or equal to k, move the left pointer up; otherwise, move the right pointer down.
5Step 5: Return the left pointer as the result, which represents the greatest cheap number.

solution.py26 lines

1# Full working Python code
2
3def count_set_bits(num, x):
4    count = 0
5    pos = x
6    while pos <= num:
7        if num & pos:
8            count += 1
9        pos <<= 1
10    return count
11
12def accumulated_price(num, x):
13    total = 0
14    for i in range(1, num + 1):
15        total += count_set_bits(i, x)
16    return total
17
18def max_cheap_number(k, x):
19    left, right = 1, k
20    while left < right:
21        mid = (left + right + 1) // 2
22        if accumulated_price(mid, x) <= k:
23            left = mid
24        else:
25            right = mid - 1
26    return left

ℹ

Complexity note: The time complexity is O(n log n) due to the binary search over n and the linear accumulation of prices, which is much more efficient than the brute force approach.

1Understanding how to calculate the price based on set bits is crucial for both approaches.
2Binary search significantly reduces the number of checks needed to find the maximum cheap number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.