**Making Change Sequel - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers denominations and an integer amount, find the minimum number of coins needed to make amount. Return -1 if there's no way to make amount. Constraints n ≤ 10 where n is the length of denominations. amount ≤ 500,000. Example 1 Input denominations = [1, 5, 10, 25] amount = 60 Output 3 Explanation We can make 60 with 2 quarters and 1 dime. Example 2 Input denominations = [3, 7, 10] amount = 8 Output -1 Explanation We can't make 8 with any of the denominations given.

### Solution :

` ````
Solution in C++ :
vector<vector<int>> dp;
int helper(vector<int>& d, int i, int a) {
int ans = INT_MAX;
if (a == 0) {
return 0;
}
if (i >= d.size() || a < 0) {
int x = 1e9;
return x;
}
if (dp[i][a] != -1) {
return dp[i][a];
}
if (d[i] != 0) {
ans = min(ans, helper(d, i, a - d[i]) + 1);
}
ans = min(ans, helper(d, i + 1, a));
return dp[i][a] = ans;
}
int solve(vector<int>& denominations, int amount) {
int ans, x = 1e9;
dp.clear();
dp.resize(denominations.size(), vector<int>(amount + 1, -1));
ans = helper(denominations, 0, amount);
if (ans == x) {
return -1;
}
return ans;
}
```

` ````
Solution in Python :
class Solution:
def solve(self, denominations, amount):
dp = [float("inf")] * (amount + 1)
dp[0] = 0
for y in range(0, amount + 1):
for coin in denominations:
if y - coin < 0:
continue
dp[y] = min(dp[y], dp[y - coin] + 1)
if dp[-1] == float("inf"):
return -1
return dp[-1]
```

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