You are given coins of different denominations and a total amount of money amount. Write a function to compute 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`.

Example 1:

```Input: coins = `[1, 2, 5]`, amount = `11`
Output: `3`
Explanation: 11 = 5 + 5 + 1```

Example 2:

```Input: coins = `[2]`, amount = `3`
Output: -1
```

Note:
You may assume that you have an infinite number of each kind of coin.

## Solution : Dynamic Programming

Top Down Approach : Recursion + Memoization

```class Solution {
public int coinChange(int[] coins, int amount) {

//Base case
if(amount == 0)
return 0;

int[] dp = new int[amount+1];

int res = minCoins(amount,coins,coins.length,dp);

return res == Integer.MAX_VALUE ? -1 : res;
}

public int minCoins(int n,int[] coins,int T,int[] dp)
{
//Base case
if(n == 0)
return 0;

//Lookup
if(dp[n] != 0)
return dp[n];

//Recursive Case
int ans = Integer.MAX_VALUE;

for(int i=0;i<T;i++)
{
if(n-coins[i] >= 0)
{
int subProblem = minCoins(n-coins[i],coins,T,dp);

if(subProblem != Integer.MAX_VALUE)
ans = Math.min(ans,subProblem + 1);
}
}
dp[n] = ans;
return dp[n];
}
}
```

Time Complexity: O(T * N)

Space Complexity: O(N)

## Bottom Up Approach

```class Solution {
public int coinChange(int[] coins, int amount) {
if(amount == 0)
return 0;
int res = findMinCoinsBU(amount,coins,coins.length);

return res == Integer.MAX_VALUE ? -1 : res;

}

public int findMinCoinsBU(int N,int[] coins,int T)
{
int[] dp = new int[N+1];

for(int n=1;n<=N;n++)
{
dp[n] = Integer.MAX_VALUE;

for(int i=0;i<T;i++)
{
if(n - coins[i] >= 0)
{
int subProblem = dp[n-coins[i]];
if(subProblem != Integer.MAX_VALUE)
dp[n] = Math.min(subProblem+1, dp[n]);

}
}
}
return dp[N];
}

}
```

Time Complexity: O(T * N)

Space Complexity: O(N)