# 256. Paint House

![](/files/-Mey5iUaoxOp_P46ZWgW)

![](/files/-Mey5m3LxlgH3QXXbNkr)

## n colors solutions, traditional DP

time: O(3m), costs' matrix size(3 colors), if n colors, it will be O(mn)

space: O(3m)

```java
class Solution {
    public int minCost(int[][] costs) {
        // one of three colors, R G B
        // that no two adjacent houses have the "same" color.
        // costs[house_index][color_type]
        
        // dp[i][0] = costs[i][0] + dp[i][1 or 2]
        // dp[i][1] = costs[i][1] + dp[i][0 or 2]
        // dp[i][2] = costs[i][2] + dp[i][1 or 3] 
        // ... j color       
        // dp[i][j] = costs[i][j] + dp[i][color_type excludes j]
        
        if (costs == null || costs.length == 0) return 0;
        
        int m = costs.length;
        int n = costs[0].length;
        int dp[][] = new int[m][n];
                
        for(int k = 0; k < n ; k++) {
            dp[0][k] = costs[0][k];
        }

        for (int i = 1; i < m; i++) {
            for(int j = 0; j < n ; j++) { // 0 1 2,   n colors
                dp[i][j] = costs[i][j] + findMinColor(i-1, j, dp); // use j color
            }
        }
        
        return findMinColor(m-1, -1, dp); // use -1 means, no needs to exclude colors
    }
    private int findMinColor(int i, int j, int dp[][]) { // cant use j color
        int min = Integer.MAX_VALUE;
        for (int k = 0; k < dp[0].length; k++) { // loop n colors again
            if (k != j) {
                min = Math.min(min, dp[i][k]);
            }
        } 
        return min;
    }
}
```

## use variables

time: O(m)

space: O(1)

```java
class Solution {
    public int minCost(int[][] costs) {
        // one of three colors, R G B
        // that no two adjacent houses have the "same" color.
        // costs[house_index][color_type]
        // dp[i-1][0] + costs[i][1~2]
        
        if (costs == null || costs.length == 0) return 0;
        
        int m = costs.length;
        
        int r = 0;
        int g = 0;
        int b = 0;
    
        for (int i = 0; i < m; i++) {
            int rr = r, gg = g, bb = b;
            r = costs[i][0] + Math.min(gg, bb);
            g = costs[i][1] + Math.min(rr, bb);
            b = costs[i][2] + Math.min(rr, gg);
        }        
        return Math.min(r, Math.min(g, b));
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://timmybeeflin.gitbook.io/cracking-leetcode/dynamic-programming/256.-paint-house.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.