➡️ Interviewer: Hello Sideon, thank you for coming in for this interview. I'll be your interviewer today. Can you tell me a little bit about yourself and your programming experience?
👨🦰 Sideon: Hi, thank you for having me. I'm Sideon and I'm a software developer with experience in Java. I've been programming for a few years now and I'm excited to be here today.
➡️ Interviewer: Great, let's get started then. I'd like to start by asking you a LeetCode question. Have you ever used LeetCode before?
👨🦰 Sideon: Yes, I have used LeetCode before. I find it helpful for practising my programming skills and preparing for interviews.
➡️ Interviewer: Good to know. The question I have for you today is "322. Coin Change". Have you ever heard of this problem before?
👨🦰 Sideon: Yes, I have heard of this problem. It's about finding the minimum number of coins needed to make a certain amount of money.
➡️ Interviewer: Exactly. The problem statement is as follows: You are given coins of different denominations and a total amount of money. Write a function to compute the minimum number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1. You can assume that you have an infinite number of each kind of coin.
➡️ Interviewer: So, let's start with the brute force approach. Can you tell me how you would approach this problem?
👨🦰 Sideon: Sure. The brute force approach would be to generate all possible combinations of coins and see which one adds up to the given amount of money. We would start with the smallest denomination and keep adding coins until we reach the target amount. We would keep track of the minimum number of coins required for each combination and return the minimum at the end.
➡️ Interviewer: That's correct. However, this approach has a time complexity of O(n^m) where n is the number of coins and m is the target amount. This is not very efficient. Can you think of a more optimal solution?
👨🦰 Sideon: Yes, I can think of a dynamic programming solution. We can use an array to store the minimum number of coins required for each amount up to the target amount. We would start with the base case of zero coins required for zero amount and then calculate the minimum number of coins required for each amount up to the target amount.
➡️ Interviewer: That's a good approach. Can you walk me through the algorithm in more detail?
👨🦰 Sideon: Sure. We would start by creating an array of size target+1 and initialize it to a maximum value of target+1. We set the value of the first element to zero since zero coins are required for zero amount. We then iterate over the coins array and for each coin, we check if it is less than or equal to the current amount. If it is, we subtract the coin value from the current amount and add 1 to the minimum number of coins required for that amount. We then take the minimum of the current value and the value at index (amount-coin) + 1. Finally, we return the value at the last index of the array.
➡️ Interviewer: That sounds good. Have you tested your solution with edge cases and corner cases?
👨🦰 Sideon: Yes, I have tested it with a few edge cases such as an empty coins array, a zero target amount, and a target amount that cannot be made up by any combination of coins. I also tested it with corner cases such as having only one denomination of coin and having all denominations of coin be the same.
➡️ Interviewer: Great. Does the above code follow all the coding principles and standards, and is the implementation correct with all test and edge cases
class Solution {
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
for (int j = 0; j < coins.length; j++) {
if (coins[j] <= i) {
dp[i] = Math.min(dp[i], 1 + dp[i - coins[j]]);
}
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
}
👨🦰 Sideon: Yes, I believe the code follows good coding principles and standards. I used meaningful variable names, commented my code, and wrote test cases to ensure that the code works as expected.
➡️ Interviewer: Good to hear that. Can you walk me through the time and space complexity of your solution?
👨🦰 Sideon: Sure. The time complexity of my solution is O(n * m) where n is the number of coins and m is the target amount. We iterate over the coins array for each amount up to the target amount, so the time complexity is proportional to the number of coins and the target amount.
The space complexity of my solution is also O(m) since we are only storing the minimum number of coins required for each amount up to the target amount.
➡️ Interviewer: Excellent. I'm very impressed with your solution and the way you explained it. Is there anything else you would like to add or ask me?
👨🦰 Sideon: Thank you, I appreciate that. I do have a question actually. In an interview setting, would it be more beneficial to focus on optimizing the code for time or space complexity?
➡️ Interviewer: That's a great question. It really depends on the specific problem and the constraints given. In general, it's always a good idea to optimize for both time and space complexity, but sometimes you may have to prioritize one over the other. For example, if you're working on a problem with very large input sizes, you may need to prioritize space complexity to ensure that you don't run out of memory. On the other hand, if the input size is relatively small, you may want to prioritize time complexity to ensure that the code runs quickly. It's important to ask clarifying questions and gather as much information as possible before deciding on a strategy.
👨🦰 Sideon: That makes sense. Thank you for your feedback and for answering my question.
➡️ Interviewer: Of course, it was my pleasure. Thank you for coming in for this interview and sharing your solution with me. I wish you the best of luck in your programming endeavors.
👨🦰 Sideon: Thank you, I appreciate it. It was great to go through this exercise with you and get feedback on my solution.
➡️ Interviewer: Before you go, I have one more question for you. How would you modify your solution if the given coin denominations were not unique?
👨🦰 Sideon: That's a good question. If the coin denominations were not unique, I would modify my solution by sorting the coins array and then iterating through it. During each iteration, I would check if the current coin denomination is equal to the previous one. If it is, I would skip it to avoid counting duplicate coin denominations.
➡️ Interviewer: That sounds like a good approach. Would you like to modify your code to implement this solution now?
👨🦰 Sideon: Sure, I can do that. Here is the updated code:
public int coinChange(int[] coins, int amount) {
Arrays.sort(coins);
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int coinIndex = 0; coinIndex < coins.length; coinIndex++) {
if (coinIndex > 0 && coins[coinIndex] == coins[coinIndex - 1]) {
continue;
}
int coin = coins[coinIndex];
for (int i = coin; i <= amount; i++) {
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
➡️ Interviewer: Excellent, your code looks good. You've implemented a good solution that accounts for non-unique coin denominations. I really appreciate your attention to detail and your willingness to modify your code on the spot.
👨🦰 Sideon: Thank you, I'm glad I could provide a solution. I think it's important to be flexible and adaptable in problem-solving, especially during an interview setting.
➡️ Interviewer: Absolutely, being able to adapt to new information and modify your approach is a valuable skill in programming. Once again, thank you for coming in and sharing your solution with me.
👨🦰 Sideon: Thank you, it was a pleasure to meet with you. Have a great day.
➡️ Interviewer: You too, take care.
Lets Pace Up, Keep an eye out for similar Leetcode Interview Simulations on various problems, Though we kicked off with an easy one today, always check the comments.