Coin Change

Coin Change

Question:

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the fewest 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 may assume that you have an infinite number of each kind of coin.

How to Solve:

This is a dynamic programming problem. We maintain an array dp of size "amount + 1", where each element corresponds to the fewest number of coins needed to make up the amount index. Therefore, dp[amount] will be our eventual answer.

We initialize dp[0] as 0 and iterate from 1 to amount, gradually filling up the DP array. Within each iteration, we iterate through the variety of coins, and if the coin value is smaller than or equal to the amount (somewhere between 1 and amount) at hand, we perform the following update:

dp[i] = min(dp[i], 1 + dp[i - val]);

That is, we either keep the current value of dp[i], or update it with one plus the number of coins needed to make up the amount i - val, whichever is smaller. The "one plus" comes from the fact that we will be using the current coin value once, totaling the amount i. Essentially, for every amount leading up to amount, we try all the coin face values and update the DP array repeatedly.

My C++ Solution:

class Solution {
 public:
  int coinChange(vector<int> &coins, int amount) {
    vector<long> dp(amount + 1, INT_MAX);
    dp[0] = 0;  // 0 coins required to achieve 0 sum
    for (int i = 1; i <= amount; ++i) {
      for (const auto &val : coins) {
        if (i >= val) {
          dp[i] = min(dp[i], 1 + dp[i - val]);
        }
      }
    }
    return dp[amount] == INT_MAX ? -1 : dp[amount];
  }
};