LeetCode-in-All

322. Coin Change

Medium

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.

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

Constraints:

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

Solution

#include <stdio.h>
#include <limits.h>

int coinChange(int* coins, int coinsSize, int amount) {
    int* dp = (int*)malloc((amount + 1) * sizeof(int));

    // Initialize dp array, where dp[i] represents the minimum number of coins to make amount i.
    // Set dp[0] to 1 as the base case (1 represents that 0 coins are needed to make amount 0).
    dp[0] = 1;
    for (int i = 1; i <= amount; i++) {
        dp[i] = 0;
    }

    for (int c = 0; c < coinsSize; c++) {
        int coin = coins[c];
        for (int i = coin; i <= amount; i++) {
            int prev = dp[i - coin];
            if (prev > 0) {  // If it's possible to make amount (i - coin)
                if (dp[i] == 0) {
                    dp[i] = prev + 1;
                } else {
                    dp[i] = (dp[i] < prev + 1) ? dp[i] : prev + 1;
                }
            }
        }
    }

    int result = (dp[amount] > 0) ? dp[amount] - 1 : -1; // Subtract 1 to account for the base case offset

    free(dp); // Free allocated memory
    return result;
}

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