LeetCode 2189: Number of Ways to Build House of Cards Solution

Master LeetCode problem 2189 (Number of Ways to Build House of Cards), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2189. Number of Ways to Build House of Cards

Medium

Math Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. Let's define a 2D array dp[i][j] to represent the number of ways to build a house of cards with i floors and j cards. The base case is dp[0][0] = 1.

To calculate dp[i][j], we need to consider two cases:

If we use the current card to form the roof of the house, then the remaining cards can be used to build a house with i-1 floors and j-1 cards. So, dp[i][j] += dp[i-1][j-1].
If we use the current card to form the walls of the house, then the remaining cards can be used to build a house with i-1 floors and j cards. So, dp[i][j] += dp[i-1][j].

The final answer will be the sum of all dp[n][k], where n is the number of floors and k is the total number of cards.

Time Complexity:

The time complexity of this approach is O(n*k), where n is the number of floors and k is the total number of cards.

Space Complexity:

The space complexity of this approach is O(n*k) for the 2D array dp.

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Solution Code

class Solution {
    public int buildHouseOfCards(int n, int k) {
        int mod = (int)1e9 + 7;
        int[][] dp = new int[n+1][k+1];
        dp[0][0] = 1;
        
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) {
                dp[i][j] = (dp[i-1][j-1] + dp[i-1][j]) % mod;
            }
        }
        
        int ways = 0;
        for (int i = 0; i <= k; i++) {
            ways = (ways + dp[n][i]) % mod;
        }
        
        return ways;
    }
}

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