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Coin Change Problem

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 10

Write a C# program to find the minimum number of coins required to make up a given amount. You are given an array of positive integers representing coin denominations and a total amount of money. Return the fewest number of coins needed to make up that amount. If the amount cannot be made up by any combination of the coins, return -1.

Example 1
    
  
  • Input: coins = [1, 2, 5], amount = 11
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  • Output: 3
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  • Explanation: 11 = 5 + 5 + 1, so you need at least 3 coins.
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Example 2
    
  
  • Input: coins = [2], amount = 3
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  • Output: -1
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  • Explanation: There's no way to make the amount 3 using only coins of denomination 2.
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Constraints
    
  
  • 1 ≤ coins.length ≤ 12
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  • 1 ≤ coins[i] ≤ 2^31 - 1
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  • 0 ≤ amount ≤ 10^4
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  • Time Complexity: O(amount * coins.length)
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  • Space Complexity: O(amount)
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to solve this problem
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  • Create a dp array to store the minimum number of coins needed for each amount
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  • Initialize the dp array with amount + 1 (a value larger than any possible answer)
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  • Set dp[0] = 0 since no coins are needed to make amount 0
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  • For each coin and each possible amount, update dp[amount] if using the current coin results in fewer total coins
    •

Steps to solve by this approach:

 Step 1: Create a dp array of size amount + 1 to store the minimum number of coins needed for each amount.
 Step 2: Initialize dp[0] = 0 (base case - no coins needed for amount 0) and dp[1...amount] = amount + 1 (a value larger than any possible answer).
 Step 3: For each coin denomination, iterate through all possible amounts from the coin value to the target amount.
 Step 4: For each amount j, update dp[j] = min(dp[j], dp[j - coins[i]] + 1), meaning we either don't use the current coin or use it and add 1 to the solution for amount j - coins[i].
 Step 5: After filling the dp array, if dp[amount] is still amount + 1, it means the amount can't be made, so return -1.
 Step 6: Otherwise, return dp[amount] which contains the minimum number of coins needed.
 Step 7: The final answer represents the optimal solution for the coin change problem.

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