LeetCode 322: Coin Change
LeetCode 322 Solution Explanation
Explanation
To solve this problem, we can use dynamic programming. We will create an array dp where dp[i] represents the fewest number of coins needed to make up the amount i. We initialize dp[0] to 0 and all other elements to a value greater than the amount (to represent infinity). Then, for each coin denomination, we iterate through the amounts from the coin value to the target amount, updating the dp array with the minimum number of coins needed for each amount. The final answer will be dp[amount] if it is less than the initial value we set or -1 if it remains unchanged.
Time complexity: O(amount * number of coins)
Space complexity: O(amount)
LeetCode 322 Solutions in Java, C++, Python
class Solution {
public int coinChange(int[] coins, int amount) {
int max = amount + 1;
int[] dp = new int[amount + 1];
Arrays.fill(dp, max);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
for (int coin : coins) {
if (coin <= i) {
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
}Interactive Code Editor for LeetCode 322
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