Coin Change Problem - minimum number of coins required to make change

The Coin Change Problem is best solved using dynamic programming (DP) to find the minimum number of coins. It involves building up a solution from smaller subproblems, where dp[i] represents the minimum coins needed to make amount i.

Mechanics and Why

The core idea is to initialize dp[0] = 0 (zero coins for zero amount) and all other dp[i] values to infinity. Then, iterate through each possible amount from 1 up to the target amount. For each amount, iterate through every available coin. If amount - coin is non-negative and dp[amount - coin] is not infinity, it means we can potentially form amount using 1 + dp[amount - coin] coins. We update dp[amount] with the minimum of its current value and this new possibility: dp[amount] = min(dp[amount], 1 + dp[amount - coin]).

This DP approach is necessary because a greedy strategy (always picking the largest coin possible) does not work for all coin systems. For example, with coins [1, 3, 4] and a target amount = 6, a greedy approach would pick 4, then 1, then 1 (total 3 coins). The optimal DP solution would pick 3 then 3 (total 2 coins). This problem highlights the need for systematic exploration of subproblems, a common theme in optimizing data processing tasks.

Complexity

Time Complexity: O(amount number_of_coins), as we have nested loops iterating through amounts and coins.
* Space Complexity: O(amount) for the dp array.

Example

def coin_change(coins, amount):
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    for i in range(1, amount + 1):
        for coin in coins:
            if i - coin >= 0 and dp[i - coin] != float('inf'):
                dp[i] = min(dp[i], 1 + dp[i - coin])
    return dp[amount] if dp[amount] != float('inf') else -1

In the interview, also mention…

Discuss edge cases: amount = 0 (returns 0 coins) and scenarios where no solution exists (the final dp[amount] remains infinity, typically returning -1). Emphasize that while recursive DP with memoization is an option, an iterative bottom-up approach is often preferred in production for its clarity and to avoid potential recursion depth limits. This problem is a classic example of dynamic programming, a fundamental technique for optimizing computations by storing and reusing results of subproblems, a principle applicable to various data engineering optimizations.