Free AI LeetCode Communication Coach — Coin Change

The problem

Given coin denominations and an amount, return the fewest coins to make that amount, or -1 if impossible.

Example: coins = [1,2,5], amount = 11 → 3 (5+5+1)

Define dp[a] clearly.

dp = [0] + [float("inf")] * amount
for a in range(1, amount + 1):
    for c in coins:
        if c <= a:
            dp[a] = min(dp[a], dp[a - c] + 1)
return dp[amount] if dp[amount] != float("inf") else -1

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “dp[a] is the min coins for amount a.”
• “Greedy can fail — DP explores all amounts.”

Practise out loud

1. 1. Restate

   I need the fewest coins that sum to the amount, or minus one if I cannot.

2. 2. Approach

   I use DP. dp[a] is the minimum coins for amount a. For each amount I try every coin and take one plus dp of the remainder.

3. 3. Example

   With coins 1,2,5 and amount 11, dp builds up to three coins — for example 5+5+1.

4. 4. Time and space

   Time is O(amount times number of coins). Space is O(amount).

5. 5. Edge cases

   Amount zero returns zero. If no combination works I return minus one.

Common mistakes when explaining Coin Change

• Greedy “always take largest coin” without saying when it fails.
• Off-by-one on initializing dp with infinity.