Solution Explanation: Coin Change

The problem asks for the minimum number of coins needed to make up a given amount, given an array of coin denominations. This is a classic dynamic programming problem.

Approach: Dynamic Programming

We can solve this using dynamic programming by building a table (or array) where dp[i] represents the minimum number of coins needed to make up the amount i.

1. Initialization:

   • dp[0] = 0 (0 coins needed to make amount 0).
   • Initialize the rest of the dp array to a large value (infinity) to represent that we haven't yet found a solution for those amounts.

2. Iteration: Iterate through each coin denomination (coin) in the coins array and for each amount i from coin up to the target amount:

   • If dp[i - coin] is not infinity (meaning we have a solution for i - coin), then we can potentially improve the solution for i by using one coin in addition to the solution for i - coin:
     • dp[i] = min(dp[i], dp[i - coin] + 1)

3. Result: After iterating through all coins and amounts, dp[amount] will contain the minimum number of coins needed to make up the target amount. If dp[amount] remains infinity, it means no solution exists.

Time and Space Complexity

• Time Complexity: O(amount * number of coins). We iterate through the dp array once for each coin.
• Space Complexity: O(amount). We use a dp array of size amount + 1.

Code Implementation (Python)

This solution uses a 1D dp array for space optimization:

def coinChange(coins, amount):
    dp = [float('inf')] * (amount + 1)  # Initialize dp array with infinity
    dp[0] = 0  # Base case: 0 coins needed for amount 0
 
    for coin in coins:
        for i in range(coin, amount + 1):
            if dp[i - coin] != float('inf'):
                dp[i] = min(dp[i], dp[i - coin] + 1)
 
    return dp[amount] if dp[amount] != float('inf') else -1  #Return -1 if no solution found
 

Code Implementation (Other Languages)

The logic remains the same across different programming languages. Here are examples in Java and C++:

Java:

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1); // Initialize with a large value
        dp[0] = 0;
 
        for (int coin : coins) {
            for (int i = coin; i <= amount; i++) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

C++:

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount + 1, amount + 1); // Initialize with a large value
        dp[0] = 0;
 
        for (int coin : coins) {
            for (int i = coin; i <= amount; i++) {
                dp[i] = min(dp[i], dp[i - coin] + 1);
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};

These implementations follow the same dynamic programming approach, efficiently finding the minimum number of coins or returning -1 if no solution exists. The choice of language only affects the syntax and specific library functions used (like Arrays.fill in Java or vector and min in C++). The core algorithm is identical.