Added codes for 18 May

GeeksForGeeks/May/18-5-24/GFG.java

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+//{ Driver Code Starts
+import java.io.*;
+import java.lang.*;
+import java.util.*;
+
+class GFG {
+    public static void main(String[] args) throws IOException {
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        int T = Integer.parseInt(br.readLine().trim());
+        while (T-- > 0) {
+            String s = br.readLine().trim();
+            int n = Integer.parseInt(s);
+            String S = br.readLine();
+            String[] s1 = S.split(" ");
+            List<Integer> a = new ArrayList<Integer>();
+            for (int i = 0; i < n; i++) {
+                a.add(Integer.parseInt(s1[i]));
+            }
+            Solution ob = new Solution();
+            int ans = ob.findPeakElement(a);
+            System.out.println(ans);
+        }
+    }
+}
+
+// } Driver Code Ends
+
+
+class Solution {
+    public int findPeakElement(List<Integer> a) {
+        // Code here
+
+        // Collections.sort(a);
+        // return a.get(a.size()-1);
+
+        int left = 0;
+        int right = a.size() - 1;
+
+        while (left < right) 
+        {
+            int mid = left + (right - left) / 2;
+
+            if (a.get(mid) > a.get(right))
+                right = mid;
+            else
+                left = mid + 1;
+        }
+
+        return a.get(left);
+    }
+}

GeeksForGeeks/May/18-5-24/README.md

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

LeetCode/May/18-5-24/README.md

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

LeetCode/May/18-5-24/Solution.java

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+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution 
+{
+    public int distributeCoins(TreeNode root) 
+    {
+        dfs(root);
+        return ans;
+    }
+
+    private int ans = 0;
+
+    // Returns the number of coins I can give (positive) / take (negative).
+    private int dfs(TreeNode root) 
+    {
+        if (root == null)
+            return 0;
+
+        int l = dfs(root.left);
+        int r = dfs(root.right);
+        ans += Math.abs(l) + Math.abs(r);
+
+        return (root.val - 1) + l + r;
+    }
+}