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| 1 | +class Solution { |
| 2 | + public: |
| 3 | + // Helper function to add two integers |
| 4 | + int add(int a, int b){ |
| 5 | + return (a + b); // Returns the sum of a and b |
| 6 | + } |
| 7 | + |
| 8 | + // Helper function to multiply two integers |
| 9 | + int multi(int a, int b){ |
| 10 | + return (a * b); // Returns the product of a and b |
| 11 | + } |
| 12 | + |
| 13 | + // Function to calculate the number of ways to paint the fence with n posts and k colors |
| 14 | + int solve(int n, int k){ |
| 15 | + // Base case: If there is only one post, there are k ways to paint it with k colors |
| 16 | + if(n == 1) return k; |
| 17 | + |
| 18 | + // Base case: If there are two posts, there are k * k ways to paint them |
| 19 | + // (each post can be painted with any of the k colors) |
| 20 | + if(n == 2) return add(k, multi(k, k-1)); |
| 21 | + |
| 22 | + // Recurrence relation: |
| 23 | + // The number of ways to paint n posts is calculated as: |
| 24 | + // dp[n] = (dp[n-1] * (k-1)) + (dp[n-2] * (k-1)) |
| 25 | + // This is based on the fact that we can paint the nth post in (k-1) ways to avoid |
| 26 | + // repeating the color of the (n-1)th post. |
| 27 | + int ans = add(multi(solve(n-2, k), k-1), multi(solve(n-1, k), k-1)); |
| 28 | + |
| 29 | + // Return the calculated number of ways to paint n posts |
| 30 | + return ans; |
| 31 | + } |
| 32 | + |
| 33 | + // Main function to initiate the calculation and return the result |
| 34 | + int countWays(int n, int k) { |
| 35 | + return solve(n, k); // Call the solve function to calculate the result |
| 36 | + } |
| 37 | +}; |
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