07. Build Array Where You Can Find The Maximum Exactly K Comparisons

The problem can be found at the following link: Question Link

My Approach

1. Define a private function called "solve" with the following parameters:

   • n: The remaining number of elements to be placed in the array.

   • m: The maximum value that can be used to fill an element in the array.

   • k: The remaining allowed changes to the previously placed value.

   • prev: The previously placed value in the array.

   • dp: A three-dimensional vector used for memoization.

2. Check if n is 0 and k is 0, indicating that we have successfully constructed the array. Return 1 in this case.

3. Check if k is less than 0 or n is less than 0, indicating an invalid state. Return 0 in these cases.

4. Check if the result for the current state (n, k, prev) is already calculated and stored in the memoization table dp. If yes, return the stored result.

5. Initialize a variable ans to 0 to keep track of the number of valid arrays.

6. Iterate through values from 1 to m (inclusive) representing the next element to be placed in the array.

7. If prev is less than the current value i, recursively call the "solve" function with reduced n, k, and prev updated to i. Add the result to ans.

8. If prev is greater than or equal to the current value i, recursively call the "solve" function with reduced n and k, keeping prev unchanged. Add the result to ans.

9. Calculate the final result as ans % mod, where mod is defined as 1e9+7.

10. Store the calculated result in the memoization table dp for the current state (n, k, prev).

11. In the public function "numOfArrays," check if k is greater than n. If yes, return 0 because it's not possible to construct the array with more allowed changes than the remaining elements.

12. Initialize a three-dimensional vector dp of size (n+1) x (k+1) x (m+1) with all values set to -1 for memoization.

Time and Auxiliary Space Complexity

• Time Complexity: O(n)

• Auxiliary Space Complexity: O(n)

Code (C++)

class Solution {
private:
    int solve(int n, int m, int k, int prev, vector<vector<vector<int>>> &dp){
        if(n == 0 && k == 0){
            return 1;
        }
        if(k < 0 || n < 0)
            return 0;
        if(dp[n][k][prev] != -1)
            return dp[n][k][prev];

        int ans = 0;
        for(int i = 1; i <= m; i++){
            if(prev < i)
                ans = (ans + solve(n - 1, m, k - 1, i, dp)) % mod;
            else
                ans = (ans + solve(n - 1, m, k, prev, dp)) % mod;
        }
        return dp[n][k][prev] = ans % mod;
    }
public:
    long long mod = 1e9+7;
    int numOfArrays(int n, int m, int k) {
        if(k > n)
            return 0;
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>> (k + 1, vector<int> (m + 1, -1)));
        return solve(n, m, k, 0, dp);
    }
};

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