Introduction

In this blog post, we will explore how to solve the LeetCode problem "3116" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array coins representing coins of different denominations and an integer k. You have an infinite number of coins of each denomination. However, you are not allowed to combine coins of different denominations. Return the kth smallest amount that can be made using these coins. Example 1: Input: coins = [3,6,9], k = 3 Output: 9 Explanation: The given coins can make the following amounts: Coin 3 produces multiples of 3: 3, 6, 9, 12, 15, etc. Coin 6 produces multiples of 6: 6, 12, 18, 24, etc. Coin 9 produces multiples of 9: 9, 18, 27, 36, etc. All of the coins combined produce: 3, 6, 9, 12, 15, etc. Example 2: Input: coins = [5,2], k = 7 Output: 12 Explanation: The given coins can make the following amounts: Coin 5 produces multiples of 5: 5, 10, 15, 20, etc. Coin 2 produces multiples of 2: 2, 4, 6, 8, 10, 12, etc. All of the coins combined produce: 2, 4, 5, 6, 8, 10, 12, 14, 15, etc. Constraints: 1 <= coins.length <= 15 1 <= coins[i] <= 25 1 <= k <= 2 * 109 coins contains pairwise distinct integers.

Explanation

Here's the breakdown of the problem and the Python solution:

• Key Idea: The problem asks for the kth smallest amount achievable using multiples of given coin denominations individually. The core idea is to use a min-heap to keep track of the smallest amounts generated by each coin. We iteratively pop the smallest element from the heap, increment its corresponding coin's multiple, and push the new multiple back into the heap. We repeat this k times.

• Optimization: We can avoid duplicates in the min-heap by keeping track of which multiples of each coin have already been added to the heap.

• Complexity:

  • Runtime: O(k log n), where n is the number of coins.
  • Storage: O(n)

Code

    import heapq

def kth_smallest_amount(coins, k):
    """
    Finds the kth smallest amount that can be made using the given coins individually.

    Args:
        coins: A list of coin denominations.
        k: The kth smallest amount to find.

    Returns:
        The kth smallest amount.
    """

    n = len(coins)
    heap = []
    visited = set()  # Keep track of visited amounts to avoid duplicates

    for i in range(n):
        heapq.heappush(heap, (coins[i], i, 1))  # (amount, coin_index, multiplier)
        visited.add((i, 1))

    count = 0
    result = 0

    while count < k:
        amount, coin_index, multiplier = heapq.heappop(heap)
        result = amount
        count += 1

        next_multiplier = multiplier + 1
        next_amount = coins[coin_index] * next_multiplier

        if (coin_index, next_multiplier) not in visited:
            heapq.heappush(heap, (next_amount, coin_index, next_multiplier))
            visited.add((coin_index, next_multiplier))

    return result