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.


n ≤ 10 where n is the length of denominations.
amount ≤ 500,000.

Example 1


denominations = [1, 5, 10, 25]
amount = 60




We can make 60 with 2 quarters and 1 dime.

Example 2


denominations = [3, 7, 10]
amount = 8




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.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:
                dp[y] = min(dp[y], dp[y - coin] + 1)
        if dp[-1] == float("inf"):
            return -1
        return dp[-1]

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