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The core idea
Define dp[i] = minimum coins needed to make amount i.
For each amount i and each coin c, if c \le i, we can use coin c and then need dp[i - c] more coins.
dp[i] = 1 + min(dp[i - c] for each coin c where c \le i).
Base: dp[0] = 0 (zero coins for amount zero).
If dp[amount] stays at infinity, return .
