Added codes for 9 April

GeeksForGeeks/April/09-4-24/GFG.java

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+//{ Driver Code Starts
+// Initial Template for Java
+import java.io.*;
+import java.util.*;
+
+class GfG {
+    public static void main(String args[]) {
+        Scanner sc = new Scanner(System.in);
+        int t = sc.nextInt();
+        while (t-- > 0) {
+            int m = sc.nextInt();
+            int n = sc.nextInt();
+            int points[][] = new int[m][n];
+            for (int i = 0; i < m; i++)
+                for (int j = 0; j < n; j++) points[i][j] = sc.nextInt();
+            Solution ob = new Solution();
+            System.out.println(ob.minPoints(points, m, n));
+        }
+    }
+}
+// } Driver Code Ends
+
+
+// User function Template for Java
+
+class Solution 
+{
+    public int minPoints(int points[][], int m, int n) 
+    {
+        // Your code goes here
+        int[][] dp = new int[m][n];
+
+        dp[m-1][n-1] = Math.max(1, 1 - points[m-1][n-1]);
+
+        for (int i = m - 1; i >= 0; i--) 
+        {
+            for (int j = n - 1; j >= 0; j--) 
+            {
+                if (i == m - 1 && j == n - 1)
+                    continue;
+
+                int minPointsRequired = Integer.MAX_VALUE;
+
+                if (i + 1 < m)
+                    minPointsRequired = Math.min(minPointsRequired, dp[i + 1][j]);
+
+                if (j + 1 < n)
+                    minPointsRequired = Math.min(minPointsRequired, dp[i][j + 1]);
+
+                dp[i][j] = Math.max(1, minPointsRequired - points[i][j]);
+            }
+        } 
+        return dp[0][0];
+    }
+}

GeeksForGeeks/April/09-4-24/README.md

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

LeetCode/April/09-4-24/README.md

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+Time complexity - O(n)
+Space complexity - O(1)

LeetCode/April/09-4-24/Solution.java

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+class Solution 
+{
+    public int timeRequiredToBuy(int[] tickets, int k) 
+    {
+        int ans = 0;
+
+        for (int i = 0; i < tickets.length; ++i)
+        {
+            if (i <= k)
+                ans += Math.min(tickets[i], tickets[k]);
+            else
+                ans += Math.min(tickets[i], tickets[k] - 1);
+        }
+
+        return ans;
+    }
+}