Coin Change

To solve this coding challenge, we need to determine the minimum number of coins needed to make up a given amount using an infinite supply of coins of given denominations. This problem can be approached efficiently using dynamic programming.

Explanation

The problem is a classic example of dynamic programming (DP), where we build a solution step by step using previously computed results to solve the overall problem. Here's how we can systematically tackle this:

1. Initialize Data Structures : We'll create an array
   dp[]
   where
   dp[i]
   represents the fewest number of coins required to make the amount
   i
   . The array is initialized with a high value (say,
   inf
   or a large number) indicating that initially, we assume we cannot make that amount. Additionally, set
   dp[0] = 0
   , since no coins are required to make an amount of 0.
2. Iterate through Each Coin : For each coin in our coin array, update the
   dp
   array for all amounts from that coin's value up to the target amount. This is done with a nested loop: The outer loop goes over each coin, and the inner loop updates the entries in the
   dp
   array.
3. Update DP Table : For each amount
   i
   , if we can reach it by using the current coin (i.e.,
   i >= coin
   ), then we update
   dp[i]
   to the minimum of its current value or
   dp[i - coin] + 1
   . This step incorporates the logic that if we needed
   dp[i - coin]
   coins to make up amount
   i - coin
   , then we would need one more coin (the current coin) to make the amount
   i
   .
4. Result Extraction : After all coins have been processed, the final answer will be in
   dp[amount]
   . If
   dp[amount]
   still holds the initial high value, it means it's not possible to make that amount with given coins, and we return -1. Otherwise,
   dp[amount]
   gives the minimum number of coins needed.
Below is the very detailed elaboration of the pseudocode for solving the challenge:

Step-by-Step Explanation

5. Initialization :
• Create an array
  dp
  of size
  amount + 1
  initialized to a large number (indicating infinity).
• Set
  dp[0]
  to 0 because we need zero coins to make the amount 0.
6. Iterate Over Coins :
• For each coin in
  coins
  , iterate through all amounts from the value of the coin to the target amount.
• For each amount
  i
  , if
  i
  can be reached using the current coin, update the
  dp[i]
  value.
7. Update DP Array :
• Check if
  dp[i - coin] + 1
  (i.e., using one more coin) is less than the current
  dp[i]
  .
• If so, set
  dp[i]
  to this new value.
8. Finalize Result :
• If
  dp[amount]
  is still set to the large number, return -1 (indicating not possible); otherwise, return
  dp[amount]
  (minimum coins needed).

Detailed Steps In Pseudocode

                                            
// Initialize the dp array with a large number
dp = array of size (amount + 1), each initialized to a large number

// Base case: 0 amount requires 0 coins
dp[0] = 0

// For each coin in our coin denominations
for each coin in coins:
    // For each amount from coin to the target amount
    for i from coin to amount:
        // Check if using the current coin results in fewer coins for amount i
        dp[i] = minimum(dp[i], dp[i - coin] + 1)

// If the dp[amount] is still large, it means the amount cannot be made up with the given coins
if dp[amount] == large number:
    return -1
else:
    return dp[amount]

                                        

This pseudocode captures the essence of the dynamic programming solution to the Coin Change problem, ensuring we efficiently compute the fewest number of coins needed for any given amount.