Added codes for 11 Feb

GeeksForGeeks/February/11-2-24/GFG.java

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+//{ Driver Code Starts
+//Initial Template for Java
+
+import java.io.*;
+import java.util.*;
+import java.lang.*;
+
+class GFG{
+    public static void main(String args[])throws IOException
+    {
+        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
+        int t = Integer.parseInt(in.readLine());
+        while(t-- > 0){
+            int n = Integer.parseInt(in.readLine());
+
+            Solution ob = new Solution();
+            ArrayList<Integer> ans = ob.recamanSequence(n);
+            for(int i = 0;i < n;i++)
+                System.out.print(ans.get(i)+" ");
+            System.out.println();
+        }
+    }
+}
+// } Driver Code Ends
+
+
+//User function Template for Java
+
+class Solution{
+    static ArrayList<Integer> recamanSequence(int n)
+    {
+        // code here
+        ArrayList<Integer>list=new ArrayList<>();
+        list.add(0);
+
+        HashSet<Integer>set=new HashSet<>();
+        set.add(0);
+
+        for(int i=1;i<=n;i++)
+        {
+            int val = list.get(i-1)-i;
+
+            if(val<0||set.contains(val))
+            {
+                val=list.get(i-1)+i;
+            }
+
+            list.add(val);
+            set.add(val);
+        }
+
+        return list;
+    }
+}

GeeksForGeeks/February/11-2-24/README.md

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+Time complexity - O(n)
+Space complexity - O(n)

LeetCode/February/11-2-24/README.md

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+Time complexity - O(m * n * n)
+Space complexity - O(m * n * n)

LeetCode/February/11-2-24/Solution.java

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+class Solution 
+{
+    public int cherryPickup(int[][] grid) 
+    {
+        final int m = grid.length;
+        final int n = grid[0].length;
+
+        // dp[x][y1][y2] := the maximum cherries we can collect
+        // where the robot #1 is on (x, y1) and the robot #2 is on (x, y2)
+
+        int[][][] dp = new int[m + 1][n][n];
+
+        for (int x = m - 1; x >= 0; --x)
+            for (int y1 = 0; y1 < n; ++y1)
+                for (int y2 = 0; y2 < n; ++y2) 
+                {
+                    int currRow = grid[x][y1] + (y1 == y2 ? 0 : 1) * grid[x][y2];
+                    for (int d1 = Math.max(0, y1 - 1); d1 < Math.min(n, y1 + 2); ++d1)
+                        for (int d2 = Math.max(0, y2 - 1); d2 < Math.min(n, y2 + 2); ++d2)
+                            dp[x][y1][y2] = Math.max(dp[x][y1][y2], currRow + dp[x + 1][d1][d2]);
+                }
+
+        return dp[0][0][n - 1];
+    }
+}