0322-coin-change.cpp

/*
 Given array of coins & an amount, return fewest coins to make that amount
 Ex. coins = [1,2,5], amount = 11 -> 3, $11 = $5 + $5 + $1

 Compute all min counts for amounts up to i, "simulate" use of a coin

 Time: O(m x n) -> m = # of coins, n = amount
 Space: O(n)
*/

classSolution {
public:
intcoinChange(vector<int>& coins, int amount) {
 vector<int> dp(amount + 1, amount + 1);
 dp[0] = 0;

for (int i = 1; i < amount + 1; i++) {
for (int j = 0; j < coins.size(); j++) {
if (i - coins[j] >= 0) {
 dp[i] = min(dp[i], 1 + dp[i - coins[j]]);
 }
 }
 }

if (dp[amount] == amount + 1) {
return -1;
 }
return dp[amount];
 }
};