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Coin Change – Solution & Complexity

Interview solutions
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Coin Change

1. Understand the Goal

  • We need the minimum number of coins that sums exactly to amount.
  • Each coin denomination can be used unlimited times.
  • If no combination can make the amount, return -1.
  • The answer for amount 0 is 0 because no coins are needed.

2. Consider the Brute Force Search

  • A naive recursive solution tries every coin at every remaining amount.
  • For each choice, it solves the smaller problem remaining - coin.
  • This repeats the same remaining amounts many times, so it becomes exponential without caching.
def coin_change(coins, amount):
    if amount == 0:
        return 0
    if amount < 0:
        return -1

    best = float('inf')
    for coin in coins:
        result = coin_change(coins, amount - coin)
        if result != -1:
            best = min(best, result + 1)

    return -1 if best == float('inf') else best

3. Use Dynamic Programming

  • Let dp[x] mean the fewest coins needed to make amount x.
  • We know dp[0] = 0.
  • For every amount, try each coin and reuse the already-computed answer for amount - coin.
  • This turns repeated recursive work into a table of subproblems.

4. Build the Table Bottom-Up

  • Initialize every amount as impossible using a large sentinel value.
  • For each target amount from 1 to amount, check every coin that fits.
  • If coin c fits, a candidate answer is dp[current - c] + 1.
dp = [float('inf')] * (amount + 1)
dp[0] = 0

for current in range(1, amount + 1):
    for coin in coins:
        if coin <= current:
            dp[current] = min(dp[current], dp[current - coin] + 1)

5. Final Solution and Complexity

  • After filling the table, dp[amount] is the best answer.
  • If it is still infinity, the amount cannot be formed.
  • Time complexity is O(amount × number of coins).
  • Space complexity is O(amount).
def coin_change(coins: list[int], amount: int) -> int:
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0

    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(dp[current], dp[current - coin] + 1)

    return -1 if dp[amount] == float('inf') else dp[amount]

FAQ

See also:

Practice Coin Change online (run code in the browser)Best Time to Buy and Sell Stock solution (easy)House Robber solution (medium)Jump Game solution (medium)Jump Game II solution (hard)All coding interview problems
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