What is the brute force solution for Kth Smallest Amount With Single Denomination Combination?

The brute force approach for Kth Smallest Amount With Single Denomination Combination is Brute Force, with O(n²) time complexity and O(n) space complexity. We can generate all possible amounts using the given coin denominations and then sort them to find the k-th smallest. This approach is straightforward but inefficient for larger values of k. The optimal Optimal Solution approach improves this to O(n log(max(coins))).

What are the interview tips for Kth Smallest Amount With Single Denomination Combination?

Always clarify the problem constraints and edge cases before diving into coding. Think about the efficiency of your approach and how you can optimize it. Practice explaining your thought process as you code, as this is often part of the interview.

What are common mistakes when solving Kth Smallest Amount With Single Denomination Combination?

Failing to account for the distinct amounts when counting. Not recognizing the need for binary search in problems involving ordered outputs.

#3116

Kth Smallest Amount With Single Denomination Combination

Hard

ArrayMathBinary SearchBit ManipulationCombinatoricsNumber TheoryBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

Instead of generating all amounts, we can use binary search to find the k-th smallest amount by counting how many amounts can be formed up to a certain value using the coins.

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Algorithm

3 steps

1Step 1: Define a function to count how many amounts can be formed up to x using the coins.
2Step 2: Use binary search to find the smallest x such that the count of amounts is at least k.
3Step 3: Return x as the k-th smallest amount.

solution.py16 lines

1def countAmounts(coins, x):
2    count = 0
3    for coin in coins:
4        count += x // coin
5    return count
6
7
8def kthSmallest(coins, k):
9    left, right = 1, 2 * 10**9
10    while left < right:
11        mid = (left + right) // 2
12        if countAmounts(coins, mid) < k:
13            left = mid + 1
14        else:
15            right = mid
16    return left

ℹ

Complexity note: The time complexity is O(n log(max(coins))) due to the binary search over the range of possible amounts, where n is the number of coins. The space complexity is O(1) as we are only using a few variables.

1Understanding how to count distinct amounts is crucial for optimizing the solution.
2Binary search can significantly reduce the time complexity by narrowing down the search space.

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