BASIC AND EASIEST APPROACH
Intuition\n Describe your first thoughts on how to solve this problem. \nJust do what mentioned in the question. Traverse all the the possiblites and calculate 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust do what mentioned in the question. Traverse all the the possiblites and calculate the probability.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Base Cases:**\n\n- if (i < 0 || i >= n || j < 0 || j >= n): This condition checks if the current position (i, j) is outside the chessboard. If the knight is outside the board, it returns probability 0 since the knight falls off the board.\n- if (k == 0) return 1;: This condition checks if the number of moves (k) has reached 0. If so, it means the knight has completed all moves and remains on the board, so it returns probability 1.\n\n**Move Calculation:**\n\n- The function uses two arrays arr1 and arr2 to represent the relative positions of the knight\'s moves (using row and column offsets). Each array contains eight elements, representing the eight possible moves a knight can make on a chessboard.\n- It iterates over all eight possible moves of the knight and recursively calls itself (probab) with the updated position (adji, adjy) and k decremented by 1.\n- It multiplies the recursive result by 0.125 (1/8) since each move has an equal probability of happening.\n- It accumulates the results from all eight moves in the ans variable.\n\n\n**The function returns the computed probability ans for the current position (i, j) after making k moves.**\n****\n\n\n# Code\n```\nclass Solution {\npublic:\n    double dp[26][26][101];\n    double probab(int n,int k,int i,int j){\n        if(i<0 || i>=n || j<0 || j>=n){\n            return 0;\n        }\n        if(k==0)return 1;\n        if(dp[i][j][k]>-0.9)return dp[i][j][k];\n        int arr1[]={-2,-1,1,2,2,1,-1,-2};\n        int arr2[]={-1,-2,-2,-1,1,2,2,1};\n        double ans=0;\n        for(int t=0;t<8;t++){\n            int adji=i+arr1[t];\n            int adjy=j+arr2[t];\n            ans+=probab(n,k-1,adji,adjy)*0.125;\n        }\n        return dp[i][j][k]=ans;\n    }\n    double knightProbability(int n, int k, int row, int column) {\n        memset(dp,-1,sizeof(dp));\n        return probab(n,k,row,column);\n    }\n};\n```
['Recursion', 'Memoization', 'C++']