Solutions Added

Author: architg-op

Array/Given an array A[] and a number x, check for pair in A[] with sum as x (Using HashSet.java

@@ -0,0 +1,35 @@
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+class GFG
+ {
+	public static void main (String[] args)
+	 {
+	     Scanner sc = new Scanner(System.in);
+	     int t = sc.nextInt();
+	     while (t-- != 0) {
+	         int n = sc.nextInt();
+	         int x = sc.nextInt();
+	         int a[] = new int[n];
+	         for (int i = 0; i < n; i++) {
+	             a[i] = sc.nextInt();
+	         }
+	         int f = 0;
+	         Set<Integer> s = new HashSet<>();
+	         for (int i = 0; i < a.length; i++) {
+	             if (s.contains(x - a[i])) {
+	                 f = 1;
+	                 break;
+	             }
+	             s.add(a[i]);
+	         }
+	         if (f == 1) {
+	             System.out.println("Yes");
+	         }
+	         else {
+	             System.out.println("No");
+	         }
+	     }
+	     
+	 }
+}

Array/Given an array A[] and a number x, check for pair in A[] with sum as x O(m+n).java

@@ -0,0 +1,44 @@
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+class GFG
+ {
+	public static void main (String[] args)
+	 {
+	     Scanner sc = new Scanner(System.in);
+	     int t = sc.nextInt();
+	     while (t-- != 0) {
+	         int n = sc.nextInt();
+	         int x = sc.nextInt();
+	         int a[] = new int[n];
+	         for (int i = 0; i < n; i++) {
+	             a[i] = sc.nextInt();
+	         }
+	         Arrays.sort(a);
+	         int l = 0;
+	         int r = n - 1;
+	         int f = 0, sum = 0;
+	         while (l < r) {
+	             sum = a[l] + a[r];
+	             if (sum == x) {
+	                 f = 1;
+	                 break;
+	             }
+	             else if (sum < x) {
+	                 l++;
+	             }
+	             else {
+	                 r--;
+	             }
+	         }
+	         if (f == 1){
+	             System.out.println("Yes");
+	         }
+	         else {
+	             System.out.println("No");
+	         }
+	         
+	     }
+	     
+	 }
+}

Binary Tree/Bottom View of Binary Tree.java

@@ -0,0 +1,29 @@
+class GfG
+{
+    // Should print bottom view of tree with given root
+    public void bottomView(Node root)
+    {
+        Map<Integer, Integer> tm = new TreeMap<>();
+        int hd = 0;
+        Queue<Node> q = new LinkedList<>();
+        q.add(root);
+        root.hd = 0;
+        while (!q.isEmpty()) {
+            Node temp = q.poll();
+            hd = temp.hd;
+            tm.put(hd, temp.data);
+            if (temp.left != null) {
+                q.add(temp.left);
+                temp.left.hd = hd - 1;
+            }
+            if (temp.right != null) {
+                q.add(temp.right);
+                temp.right.hd = hd + 1;
+            }
+        }
+        for (Map.Entry<Integer, Integer> m : tm.entrySet()) {
+            System.out.print(m.getValue() + " ");
+        }
+    }
+}
+	

String/Remove characters from the first string which are present in the second string.java

@@ -0,0 +1,33 @@
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+class GFG
+ {
+	public static void main (String[] args)
+	 {
+	     Scanner sc = new Scanner(System.in);
+	     int t = sc.nextInt();
+	     while (t-- != 0) {
+	         String s1 = sc.next();
+	         String s2 = sc.next();
+	         char ch3[] = new char[s1.length()];
+	         int count[] = new int[257];
+	         int k = 0;
+	         for (int i = 0; i < 257; i++) {
+	             count[i] = 0;
+	         }
+	         for (int i = 0; i < s2.length(); i++) {
+	             count[s2.charAt(i)]++;
+	         }
+	         for (int i = 0; i < s1.length(); i++) {
+	             if (count[s1.charAt(i)] == 0) {
+	                 ch3[k++] = s1.charAt(i);
+	             }
+	         }
+	         for (int i = 0; i < k; i++) {
+	             System.out.print(ch3[i]);
+	         }
+	         System.out.println();
+	     }
+	 }
+}