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Coin Change Problem

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

Write a C program to find the minimum number of coins needed to make a given amount. You are given coins of different denominations and a total amount of money. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1
    
  
  • Input: coins = [1, 3, 4], amount = 6
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  • Output: 2
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  • Explanation:
    Step 1: To make amount 6, we can use coins [3, 3].
    Step 2: This uses 2 coins, which is the minimum possible.
    Step 3: Other combinations like [1, 1, 4] use 3 coins, which is more than optimal.
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Example 2
    
  
  • Input: coins = [2], amount = 3
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  • Output: -1
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  • Explanation:
    Step 1: We only have coins of denomination 2.
    Step 2: It's impossible to make amount 3 using only coins of value 2.
    Step 3: Therefore, return -1.
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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
    • 
  
  • Time Complexity: O(amount * coins.length)
    • 
  
  • Space Complexity: O(amount)
    •
ArraysDynamic Programming MicrosoftAccenture
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to solve this problem optimally.
    • 
  
  • Create a dp array where dp[i] represents minimum coins needed for amount i.
    • 
  
  • Initialize dp[0] = 0 and all other values to a large number (amount + 1).
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  • For each amount from 1 to target, try all possible coins.
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  • Update dp[amount] = min(dp[amount], dp[amount - coin] + 1).
    •

Steps to solve by this approach:

 Step 1: Create a dp array where dp[i] represents minimum coins needed for amount i.
 Step 2: Initialize dp[0] = 0 and all other values to amount + 1 (impossible value).
 Step 3: For each amount from 1 to target amount, iterate through all available coins.
 Step 4: For each coin, if coin value <= current amount, update dp[amount].
 Step 5: Set dp[amount] = min(dp[amount], dp[amount - coin] + 1).
 Step 6: After processing all amounts, check if dp[amount] is still impossible value.
 Step 7: Return dp[amount] if possible, otherwise return -1.

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