518. Coin Change II
Explanation
To solve this problem, we can use dynamic programming. We will create a 1D array dp where dp[i] represents the number of combinations to make up amount i. We initialize dp[0] as 1, since there is one way to make up amount 0, which is by using no coins. Then, for each coin, we iterate through all amounts from the coin value to the target amount and update the number of combinations accordingly.
Algorithm:
- Initialize a 1D array
dpof sizeamount + 1with all values set to 0, exceptdp[0] = 1. - Iterate through each coin in the
coinsarray. - For each coin, iterate through all amounts from the coin value to the target amount.
- Update
dp[j] = dp[j] + dp[j - coin]to accumulate the number of combinations.
Time Complexity: O(amount * n), where n is the number of coins Space Complexity: O(amount)
class Solution {
public int change(int amount, int[] coins) {
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
for (int j = coin; j <= amount; j++) {
dp[j] += dp[j - coin];
}
}
return dp[amount];
}
}Code Editor (Testing phase)
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