Coin Change Coding Interview Question

November 30, 2021 admin Algorithm, interview coding questions 0

Front-End Interview

Coin Change is popular coding interview question.

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.

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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 = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

Input: coins = [1], amount = 2
Output: 2

Constraints:

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

Approach Dynamic programming – Bottom up

Algorithm

For the iterative solution, we think in bottom-up manner. Before calculating F(i), we have to compute all minimum counts for amounts up to ii. On each iteration i of the algorithm F(i) is computed as         minj=0…n−1F(i−cj)+1
coin change 
In the example above you can see that:

F(3) = min{F(3−c1),F(3−c2),F(3−c3)}+1
     = min{F(3−1),F(3−2),F(3−3)}+1
     = min{F(2),F(1),F(0)}+1
     = min{1,1,0}+1
     = 1


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So now we know the steps to solve the coding challenge, let’s add logic together. Here is the working solution!

public class Solution {
  public int coinChange(int[] coins, int amount) {
    int max = amount + 1;
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, max);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
      for (int j = 0; j < coins.length; j++) {
        if (coins[j] <= i) {
          dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
        }
      }
    }
    return dp[amount] > amount ? -1 : dp[amount];
  }
}
function coinChange(coins, amount) {
  let finalResult = FindMinCount(coins, 0, 0, amount, 0, amount + 1);

  function FindMinCount(coins, i, currentSum, amount, count, result) {
    if (amount == 0) {
      return 0;
    }
    if (currentSum > amount) {
      return result;
    }
    if (currentSum == amount) {
      result = Math.min(count, result)
      return result
    }
    if (i <= coins.length - 1) {
      count++;
      result = FindMinCount(coins, i, currentSum + coins[i], amount, count, result);
      count--;
      result = FindMinCount(coins, i + 1, currentSum, amount, count, result);

    }
    return result;
  }
  return finalResult

}