# Coin Change (Leetcode 322) - Coding Interview Question Mani
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You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.

Return 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`.

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

Example 1:

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

Example 2:

```Input: coins = , amount = 3
Output: -1
```

Example 3:

```Input: coins = , amount = 0
Output: 0
```

Example 4:

```Input: coins = , amount = 1
Output: 1
```

Example 5:

```Input: coins = , amount = 2
Output: 2
```

Constraints:

• `1 <= coins.length <= 12`
• `1 <= coins[i] <= 231 - 1`
• `0 <= amount <= 104`

Solution:

``````1
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import java.util.Arrays;

/**
* This is a variation of Unbounded Knapsack DP.
* Important points to note:
* Initialization for dp[i] = 0 for all i.
* If an element is not included it means its not included in any of the counts.
* Note the subtle difference in code between 0/1 Knapsack vs Unbounded Knapsack.
* Time Complexity: O(N ^ 2)
* Space Complexit: O(N ^ 2)
*/
class Solution {
public int coinChange(int[] w, int c) {
int[][] dp = new int[w.length + 1][c + 1];
for (int[] d : dp) {
Arrays.fill(d, Integer.MAX_VALUE / 2);
}
dp = 0;
for (int i = 1; i <= w.length; i++) {
dp[i] = 0;
}
for (int i = 1; i <= w.length; i++) {
for (int j = 1; j <= c; j++) {
if (w[i - 1] <= j) {
dp[i][j] = Math.min(1 + dp[i][j - w[i - 1]], dp[i - 1][j]);
} else {
dp[i][j] = dp[i - 1][j];
}
}
}

return dp[w.length][c] == Integer.MAX_VALUE / 2 ? -1 : dp[w.length][c];
}
}
``````
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