![\"\"](\"https:\/\/prepbytes-misc-images.s3.ap-south-1.amazonaws.com\/assets\/1646202336103-B_365-java.png\")
<\/p>\n<\/p>\n
public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n String str = in.nextLine();\n int length = str.length(); \n boolean isPalindrome = true; \n for(int i = 0; i < length; i++)\n {\n if(str.charAt(i) != str.charAt(length-1-i)) {\n System.out.println("Snot a palindrome.");\n isPalindrome = false;\n break;\n }\n } \n if(isPalindrome) {\n System.out.println("palindrome.");\n }\n}<\/code><\/pre>\n<\/li>\n- Count the occurrence of a given character in string.<\/strong>
\nAns.<\/p>\npublic static void main(String args[]){\nString str = "this is a string";\nchar ch = 's';\nint count = 0;\nfor(char c:strArr){\n if(c == 's'){\n count++;\n }\n}\n System.out.println(count); \n}<\/code><\/pre>\n<\/li>\n- Find the GCD of two numbers in logn time.<\/strong>
\nAns.<\/p>\npublic static int gcd(int a, int b) {\nwhile (((a > 0) && (b > 0))) {\n if ((a > b)) {\n a = (a % b);\n } else {\n b = (b % a);\n }\n}\nif ((a == 0)) {\n return b;\n} else {\n return a;\n}\n} \npublic static void main(String[] args) {\nint a = new Scanner(System.in);\nint b = new Scanner(System.in);\nSystem.out.println(GCD.gcd(a, b));\n}<\/code><\/pre>\n<\/li>\n- Can you code the Binary Search Algorithm in Java?<\/strong>
\nAns.<\/p>\npublic int binarySearch(int a[], int num)\n{\n if(num>a[a.length-1]||num<a[0])return -1;\n int start=0;\n int end =a.length-1;\n int mid=(start+end)\/2;\n while(a[mid]!=num){\n if(num<mid){\n end=mid;\n }else{\n start=mid;\n }\n mid = (start+end)\/2;\n }\n return mid;\n}<\/code><\/pre>\n<\/li>\n- Implement the merge Sort algorithm.<\/strong>
\nAns. This is a Divide and conquer algorithm.<\/p>\n\npublic static void main(String[] args) {\n int[] list = {4,5,2,1,3};\n mergeSort(list, 0, list.length - 1);\n}\npublic static void mergeSort(int[] a, int first, int last)\n {\n if(last - first == 0) \n {\n }\n else if (last - first == 1) \n {\n if(a[first] > a[last])\n {\n int temp = a[first];\n a[first] = a[last];\n a[last] = temp;\n }\n }\n else \n {\n int mid = (first + last) \/ 2;\n mergeSort(a, first, mid);\n mergeSort(a, mid + 1, last);\n merge(a, first, mid, last);\n }\n }\nprivate static void merge(int[] a, int first, int mid, int last)\n {\n int[] temp = new int[last - first + 1];\n int i = first; int j = mid + 1;\n for(int k = first; k <= last; k++)\n {\n if(i > mid || j > last)\n {\n if(i > mid && j <= last)\n {\n System.out.println("a[j]: " + a[j]);\n temp[k - first] = a[j];\n j++;\n }\n else if(i <= mid && j > last)\n {\n System.out.println("a[i]: " + a[i]);\n temp[k - first] = a[i];\n i++;\n }\n else\n {\n break;\n }\n }\n else\n {\n if(a[i] < a[j])\n {\n temp[k - first] = a[i];\n i++;\n }\n else\n {\n temp[k - first] = a[j];\n j++;\n }\n }\n }\n for(int count = 0; count < temp.length; count++)\n {\n a[first + count] = temp[count];\n }\n }<\/code><\/pre>\n<\/li>\n- \n
Implement a quicksort algorithm.<\/strong>
\nAns. <\/p>\npublic int partition(ArrayList list, int start, int end) {\ndouble pivot = list.get(end);\nint i = (start - 1); \nfor (int j = start; j < end; j++) {\nif (list.get(j) <= pivot) { \n i++;\n double temp = list.get(i);\n list.set(i, list.get(j));\n list.set(j, temp);\n }\n} double temp = list.get(i + 1); \n list.set((i + 1), list.get(end));\n list.set(end, temp);\n return i + 1;\n}\npublic ArrayList sort(ArrayList list, int start, int end) { \nif (start < end) {\nint pi = partition(list, start, end); \nsort(list, start, pi - 1); \nsort(list, pi + 1, end);\n } return list;\n} \npublic ArrayList Solve(double[] nums, int a, int b, int c) {\nArrayList newList = new ArrayList();\ndouble xyz;\nfor (int i = 0; i < nums.length; i++) {\n xyz = ((a * nums[i]) * nums[i]) + b * nums[i] + c;\n newList.add(xyz);\n } return newList; \n} \npublic static void main (String[] args) {\nQuickSort obj = new QuickSort();\ndouble nums[] = {1,2,4, 9,5,3,0};\nint a = 1, b = 0, c = 0;\nArrayList list = obj.Solve(nums, a, b, c);\nlist = obj.sort(list, 0, (nums.length - 1));\nfor (int i = 0; i \");\n current = current.next();\n count++;\n if(count == 10)\n break;\n } \n return sb.toString();\n}\npublic static class Node {\n private Node next;\n private String data;\n public Node(String data) {\n this.data = data;\n }\n public String data() { \n return data; \n } \n public void setData(String data) {\n this.data = data;\n }\n public Node next() { \n return next; \n } \n public void setNext(Node next) {\n this.next = next; \n }\n @Override\n public String toString() {\n return this.data;\n }\n}\n}\npublic class LoopInLinkedListTest { \npublic static void main(String args[]) { \n LinkedList linkedList = new LinkedList();\n if(false){\n linkedList.appendIntoTail(new LinkedList.Node(\"101\")); \n linkedList.appendIntoTail(new LinkedList.Node(\"201\")); \n linkedList.appendIntoTail(new LinkedList.Node(\"301\")); \n linkedList.appendIntoTail(new LinkedList.Node(\"401\")); \n System.out.println(\"Linked List : \" + linkedList); \n }else{\n linkedList.appendIntoTail(new LinkedList.Node(\"101\"));\n LinkedList.Node cycle = new LinkedList.Node(\"201\");\n linkedList.appendIntoTail(cycle);\n linkedList.appendIntoTail(new LinkedList.Node(\"301\"));\n linkedList.appendIntoTail(new LinkedList.Node(\"401\"));\n linkedList.appendIntoTail(cycle); \n System.out.println(\"Linked List : \" + linkedList);\n } \n if(linkedList.isCyclic()){ \n System.out.println(\"contains loop\"); \n }\n else{ \n System.out.println(\"no loop\"); \n } \n\n}\n}<\/code><\/pre>\n<\/li>\n- Check if a linked list is a palindrome or not.<\/strong>
\nAns.<\/p>\nclass Node {\n int data;\n Node next;\n public Node(int data) {\n this.data = data;\n this.next = null;\n }\n}\npublic int size;\npublic Node head = null;\npublic Node tail = null;\npublic void addNode(int data) {\n Node node = new Node(data);\n if(head == null) {\n head = node;\n tail = node;\n } else {\n tail.next = node;\n tail = node;\n }\n size++;\n}\npublic Node reverseList(Node temp) {\n Node current = temp;\n Node prevNode = null;\n Node nextNode = null;\n while(current != null) {\n nextNode = current.next;\n current.next = prevNode;\n prevNode = current;\n current = nextNode;\n }\n return prevNode; \n}\npublic void isPalindrome() {\n Node current = head;\n boolean flag = true;\n int mid = (size % 2 == 0) ? (size\/2) : ((size+1)\/2);\n for(int i = 1; i < mid; i++) {\n current = current.next;\n }\n Node revHead = reverseList(current.next);\n while(head != null && revHead != null) {\n if(head.data != revHead.data) {\n flag = false;\n break;\n }\n head = head.next;\n revHead = revHead.next;\n }\n if(flag) {\n System.out.println("\\nPalondrome !");\n } else {\n System.out.println("\\nnot Palindrome !");\n }\n}\npublic void display() {\n Node current = head;\n while(current != null) {\n System.out.print(current.data + " ");\n current = current.next;\n }\n System.out.println(" ");\n}\npublic static void main(String[] args) {\n LinkedListCheckPalindrome linkedList = new LinkedListCheckPalindrome();\n linkedList.addNode(1);\n linkedList.addNode(2);\n linkedList.addNode(3);\n linkedList.addNode(2);\n linkedList.addNode(1);\n linkedList.display();\n linkedList.isPalindrome();\n}<\/code><\/pre>\n<\/li>\n- Find pairs in the array such that its sum is closest to 0.<\/strong>
\nAns.<\/p>\npublic static void PairWithMinSum(int arr[]) {\n int minimumSum = arr[0] + arr[1];\n int pair1stIndex = 0;\n int pair2ndIndex = 1;\n for (int i = 0; i < arr.length; i++) {\n for (int j = i + 1; j < arr.length; j++) {\n int tempSum = arr[i] + arr[j];\n if (Math.abs(tempSum) < Math.abs(minimumSum)) {\n pair1stIndex = i;\n pair2ndIndex = j;\n minimumSum = tempSum;\n }\n }\n }\n System.out.println( arr[pair1stIndex] + " " + arr[pair2ndIndex]);\n}\npublic static void main(String[] args) {\n int[] arr = { 2, 3, -1, 5, 6, -3};\n PairWithMinSum(arr);\n}<\/code><\/pre>\n<\/li>\n- Given an array of 0\u2019s and 1\u2019s in random order, you need to separate 0\u2019s and 1\u2019s in an array.<\/strong>
\nAns. <\/p>\npublic static int[] separate0s1sSolution1(int arr[]){\nint count=0;\nfor (int i = 0; i < arr.length; i++) {\nif(arr[i]==0)\n{\ncount++;\n}\n}\nfor (int i = 0; i < count; i++) {\narr[i]=0;\n}\nfor (int i = count; i < arr.length; i++) {\narr[i]=1;\n}\nreturn arr;\n}<\/code><\/pre>\n<\/li>\n- You are given an array, find the subarray with a given sum.<\/strong>
\nAns. <\/p>\npublic void findSubarraySum(int arr[], int target) { \n int left = 0;\n int right = left + 1;\n int sum = 0;\n if (arr.length == 0) {\n System.out.println("-1");\n return;\n } \n if (arr.length == 1) { \n if (arr[0] == target) {\n System.out.println((arr[0] + 1) + " " + (arr[0] + 1));\n } else {\n System.out.println("-1");\n }\n }\n sum += arr[left] + arr[right];\n while (right < arr.length) { \n if (sum == target) {\n System.out.println((left + 1) + " " + (right + 1));\n return;\n } else if (sum < target) { \n right++;\n if (right = 0) {\n int n = in.nextInt();\n int target = in.nextInt();\n int[] arr = new int[n];\n for (int i = 0; i < arr.length; i++) {\n arr[i] = in.nextInt();\n } \n ss.findSubarraySum(arr, target);\n }\n}<\/code><\/pre>\n<\/li>\n- You are given an array, find the maximum subarray sum.<\/strong>
\nAns. We will use Kadane\u2019s algorithm here <\/p>\npublic static void main(String[] args) {\n int[] Arr = {5,2,-1,-2,3,-5,1};\n findMaxSubArray(Arr);\n}\npublic static void findMaxSubArray(int[] inputArray) {\n int maxStartIndex = 0;\n int maxEndIndex = 0;\n int maxSum = Integer.MIN_VALUE;\n int cumulativeSum = 0;\n int maxStartIndexUntilNow = 0;\n for (int currentIndex = 0; currentIndex maxSum) {\n maxSum = cumulativeSum;\n maxStartIndex = maxStartIndexUntilNow;\n maxEndIndex = currentIndex;\n } \n if (cumulativeSum < 0) {\n maxStartIndexUntilNow = currentIndex + 1;\n cumulativeSum = 0;\n }\n }\n System.out.println(maxSum);\n}<\/code><\/pre>\n<\/li>\n- Write a function to add two numbers represented by two linked list.<\/strong>
\nAns. <\/p>\npublic ListNode addTwoNumbers(ListNode l1, ListNode l2) {\n if (l1 == null || l2 == null) {\n return l1 == null ? l2 : l1;\n }\n Stack s1 = new Stack();\n Stack s2 = new Stack();\n Stack result = new Stack();\n while (l1 != null) {\n s1.push(l1);\n l1 = l1.next;\n }\n while (l2 != null) {\n s2.push(l2);\n l2 = l2.next;\n }\n int carry = 0;\n while (!s1.isEmpty() || !s2.isEmpty()) {\n int sum = carry;\n if (!s1.isEmpty()) {\n sum += s1.pop().val;\n }\n if (!s2.isEmpty()) {\n sum += s2.pop().val;\n }\n carry = sum \/ 10;\n sum = sum % 10;\n result.push(new ListNode(sum));\n }\n if (carry != 0) {\n result.push(new ListNode(carry));\n }\n ListNode node = new ListNode(-1);\n ListNode temp = node;\n while (!result.isEmpty()) {\n node.next = result.pop();\n node = node.next;\n }\n return temp.next;\n}<\/code><\/pre>\n<\/li>\n- Write a program to find the first non-repeating character of a string.<\/strong>
\nAns. <\/p>\npublic static charsolve(String word) {\n Set duplicate = new HashSet();\n List nonDuplicate = new ArrayList();\n for (int i = 0; i < word.length(); i++) {\n char letter = word.charAt(i);\n if (duplicate.contains(letter)) {\n continue;\n }\n if (nonDuplicate.contains(letter)) {\n nonDuplicate.remove((Character) letter);\n duplicate.add(letter);\n } else {\n nonDuplicate.add(letter);\n }\n }\n return nonDuplicate.get(0);\n}<\/code><\/pre>\n<\/li>\n- Write a program to find the duplicate characters in a string.<\/strong>
\nAns. <\/p>\npublic static void main(String[] args) {\n String str = new String("duplicatestring");\n int count = 0;\n char[] chars = str.toCharArray();\n for (int i=0; i<str.length();i++) {\n for(int j=i+1; j<str.length();j++) {\n if (chars[i] == chars[j]) {\n System.out.println(chars[j]);\n count++;\n break;\n }\n }\n }\n }<\/code><\/pre>\n<\/li>\n- Print all the substrings of a string.<\/strong>
\nAns. <\/p>\npublic static void PrintSubstring(String str, int length) {\n for (int i = 0; i < length; i++) {\n for (int j = i + 1; j -1; i--) {\n sb.append(arr[i]).append(\" \");\n }\n return sb.toString().trim();\n}\npublic static void main(String args[]) {\n Scanner in = new Scanner(System.in);\n String s = in.nextLine();\n in.close();\n System.out.println(reverseLine(s));\n}<\/code><\/pre>\n<\/li>\n- Implement an algorithm of counting sort.<\/strong>
\nAns. <\/p>\npublic static void countingSort(int[] input, int k) {\n int counter[] = new int[k + 1];\n for(int i : input) {\n counter[i]++;\n }\n int ndx = 0;\n for(int i = 0; i < counter.length; i++) {\n while( 0 < counter[i]) {\n input[ndx++] = i;\n counter[i]--;\n }\n }\n}\npublic static void main(String[] args) {\n int[] input = { 60, 40, 30, 20, 10, 40, 30, 60, 60, 20, 40, 30, 40 };\n int k = 60;\n countingSort(input, k);\n System.out.println(Arrays.toString(input));\n}<\/code><\/pre>\n<\/li>\n- Give a recursive function for Inorder traversal of a Binary Tree.<\/strong>
\nAns.<\/p>\nvoid inOrderTraversal2(TreeNode node) {\n if (node == null) {\n return;\n }\n inOrderTraversal2(node.left);\n System.out.println(node.val);\n inOrderTraversal2(node.right);\n}<\/code><\/pre>\n<\/li>\n- Check if a given Binary tree is height balanced or not. You will be given a root to that tree.<\/strong>
\nAns. <\/p>\nclass TreeNode {\n int value;\n TreeNode left, right;\n}\npublic boolean isBalanced(TreeNode root) {\n if (root == null)\n return true;\n int left = getHeight(root.left);\n int right = getHeight(root.right);\n if (Math.abs(left - right) <= 1)\n return isBalanced(root.left) && isBalanced(root.right);\n else\n return false;\n}\npublic int getHeight(TreeNode root) {\n if (root == null)\n return 0;\n else {\n int left = getHeight(root.left);\n int right = getHeight(root.right);\n return Math.max(left, right) + 1;\n }\n}<\/code><\/pre>\n<\/li>\n- Write a program to invert a Binary Tree.<\/strong>
\nAns. <\/p>\nclass TreeNode {\n int val;\n TreeNode left;\n TreeNode right;\n TreeNode(int x) {\n val = x;\n }\n}\npublic TreeNode invertTree(TreeNode root) {\n if(root == null)\n return null;\n TreeNode result = new TreeNode(root.val); \n result.right = this.invertTree(root.left);\n result.left = this.invertTree(root.right); \n return result;\n}<\/code><\/pre>\n<\/li>\n- Write a program to sort a stack recursively.<\/strong>
\nAns. <\/p>\npublic static void main(String args[]) {\n Stack stack = new Stack();\n stack.push(1);\n stack.push(2);\n stack.push(4);\n stack.push(3);\n stack.push(-2); \n for (Integer i: stack) {\n System.out.print(i + \", \");\n }\n System.out.println(); \n sort(stack);\n for (Integer i: stack) {\n System.out.print(i + \", \");\n }\n System.out.println();\n} \npublic static void sort(Stack stack) {\n removeElement(stack);\n} \nprivate static void removeElement(Stack stack) {\n if(!stack.isEmpty()) {\n int tmp = stack.pop();\n removeElement(stack);\n insertElement(tmp, stack);\n }\n}\nprivate static void insertElement(int tmp, Stack stack) {\n if(stack.isEmpty() || tmp >= stack.peek()) {\n stack.push(tmp);\n }\n else {\n int poped = stack.pop();\n insertElement(tmp, stack);\n stack.push(poped);\n }\n}<\/code><\/pre>\n<\/li>\n- Implement a queue using two stacks.<\/strong>
\nAns. <\/p>\nStack entryStack;\nStack exitStack;\npublic MyQueue() {\n this.entryStack = new Stack();\n this.exitStack = new Stack();\n}\npublic void push(int x) {\n this.entryStack.push(x);\n} \npublic int pop() {\n while(!this.entryStack.isEmpty()){\n this.exitStack.push(this.entryStack.pop());\n }\n int retValue = this.exitStack.pop();\n while(!this.exitStack.isEmpty()){\n this.entryStack.push(this.exitStack.pop());\n }\n return retValue;\n}\npublic int peek() {\n while(!this.entryStack.isEmpty()){\n this.exitStack.push(this.entryStack.pop());\n }\n int retValue = this.exitStack.peek();\n while(!this.exitStack.isEmpty()){\n this.entryStack.push(this.exitStack.pop());\n }\n return retValue;\n} \npublic boolean empty() {\n return this.entryStack.isEmpty();\n}<\/code><\/pre>\n<\/li>\n- Implement an LRU cache in Java.<\/strong>
\nAns. <\/p>\nclass Node {\n int key;\n int value;\n Node prev;\n Node next;\n public Node(int key, int value){\n this.key = key;\n this.value = value;\n }\n} \nclass DoublyLinkedList {\n Node head;\n Node tail; \n public boolean isEmpty(){\n return this.head == null;\n } \n public void addToHead(Node node){\n if(this.isEmpty()){\n this.head = this.tail = node;\n node.prev = null;\n node.next = null;\n }\n else{\n node.prev = null;\n node.next = this.head;\n this.head.prev = node;\n this.head = node;\n }\n } \n public Node deleteFromTail(){\n Node node = this.tail;\n if(this.head == this.tail){\n this.head = this.tail = null;\n }\n else{\n this.tail = this.tail.prev;\n this.tail.next = null; \n }\n return node;\n } \n public void moveToHead(Node node){\n if(this.head == node)\n return;\n else if(this.tail == node){\n this.deleteFromTail();\n this.addToHead(node);\n }\n else{\n node.prev.next = node.next;\n if(node.next != null)\n node.next.prev = node.prev;\n node.next = null;\n node.prev = null;\n this.addToHead(node);\n }\n }\n} \nHashMap nodes;\nDoublyLinkedList list;\nint count, capacity;\npublic LRUCache(int capacity) {\n this.capacity = capacity;\n this.count = 0;\n this.nodes = new HashMap();\n this.list = new DoublyLinkedList();\n} \npublic int get(int key) {\n if(nodes.containsKey(key)){\n Node node = nodes.get(key);\n this.list.moveToHead(node);\n return node.value;\n }\n else\n return -1;\n} \npublic void put(int key, int value) {\n if(nodes.containsKey(key)){\n Node node = nodes.get(key);\n node.value = value;\n this.list.moveToHead(node);\n }\n else{\n Node node = new Node(key, value);\n nodes.put(key, node);\n this.count++;\n this.list.addToHead(node);\n if(this.count > this.capacity){\n Node toDelete = this.list.deleteFromTail();\n this.nodes.remove(toDelete.key);\n this.count--;\n }\n }\n}\n\/**\n* Your LRUCache object will be instantiated and called as such:\n* LRUCache obj = new LRUCache(capacity);\n* int param_1 = obj.get(key);\n* obj.put(key,value);\n*\/<\/code><\/pre>\n<\/li>\n- Traverse the Binary tree in zig-zag way and print the nodes.<\/strong>
\nAns. <\/p>\npublic class TreeNode {\n int val;\n TreeNode left;\n TreeNode right;\n TreeNode() {}\n TreeNode(int val) { this.val = val; }\n TreeNode(int val, TreeNode left, TreeNode right) {\n this.val = val;\n this.left = left;\n this.right = right;\n }\n }\npublic List<List> zigzagLevelOrder(TreeNode root) {\n List<List> result = new LinkedList();\n if(root == null) return result;\n Queue queue = new LinkedList();\n queue.add(root);\n boolean direction = false;\n while(queue.size() > 0){\n direction = !direction;\n Queue level = new LinkedList();\n List rlevel = new LinkedList();\n while(queue.size() > 0){\n TreeNode node = queue.poll();\n if(direction){\n rlevel.add(node.val);\n }else{\n rlevel.add(0,node.val);\n }\n if(node.left != null) level.add(node.left);\n if(node.right != null) level.add(node.right);\n }\n result.add(rlevel);\n queue = level;\n }\n return result;\n }<\/code><\/pre>\n<\/li>\n- Write a program to traverse a graph in BFS.<\/strong>
\nAns. <\/p>\npublic class Graph {\nprivate final int v;\nprivate final LinkedList[] adjacents;\npublic Graph(int v) {\n this.v = v;\n adjacents = new LinkedList[v];\n for (int i = 0; i < v; ++i) {\n adjacents[i] = new LinkedList();\n }\n}\nvoid addEdge(int v, int e) {\n adjacents[v].add(e);\n}\nvoid BFS(int start) {\n boolean visited[] = new boolean[v];\n LinkedList queue = new LinkedList();\n visited[start] = true;\n queue.add(start);\n while (!queue.isEmpty()) {\n start = queue.poll();\n System.out.print(start + \" \");\n Iterator i = adjacents[start].listIterator();\n while (i.hasNext()) {\n int n = i.next();\n if (!visited[n]) {\n visited[n] = true;\n queue.add(n);\n }\n }\n }\n}\npublic static void main(String[] args) {\n Graph graph = new Graph(4);\n graph.addEdge(0, 3);\n graph.addEdge(0, 1);\n graph.addEdge(1, 0);\n graph.addEdge(2, 3);\n graph.addEdge(3, 0);\n graph.addEdge(3, 2);\n graph.addEdge(3, 3); \n graph.BFS(0);\n}\n}<\/code><\/pre>\n<\/li>\n- \n
Write a function to insert a node in Binary Search Tree.<\/strong>
\nAns. <\/p>\nstatic Node Insert(Node root, int value){\nNode newNode = new Node();\nnewNode.data = value;\nnewNode.left = null;\nnewNode.right = null; \nif (root == null) {\n return newNode; }\n\naddNode(root, newNode);\nreturn root;\n}\nstatic void addNode(Node root, Node newNode) {\nif (newNode.data < root.data) {\n if (root.left == null) {\n root.left = newNode;\n } else {\n addNode(root.left, newNode);\n }\n} else {\n if (root.right == null) {\n root.right = newNode;\n } else {\n addNode(root.right, newNode);\n }\n} \n}<\/code><\/pre>\n<\/li>\n- \n
Suppose you are given an array and you are asked to find the triplets whose sum will be a given target sum. If your array contains such a triplet return the sum else -1.<\/strong>
\nAns. <\/p>\npublic ArrayList<ArrayList> threeSum(int[] num) {\n ArrayList<ArrayList> result = new ArrayList<ArrayList>();\n if (num.length < 3) {\n return result;\n }\n\n Arrays.sort(num);\n\n for (int i = 0 ; i < num.length - 2 ; i++) {\n if (i != 0 && num[i] == num[i-1]) {\n continue;\n }\n int left = i + 1;\n int right = num.length - 1;\n\n while (left < right) {\n int sum = num[i] + num[left] + num[right];\n if (sum 0) {\n right--;\n } else {\n ArrayList temp = new ArrayList();\n temp.add(num[i]);\n temp.add(num[left]);\n temp.add(num[right]);\n result.add(temp);\n do {\n left++;\n } while (left left && num[right] == num[right+1]);\n }\n }\n }\n return result;\n}<\/code><\/pre>\n<\/li>\n- Find the Kth largest element in an array.<\/strong>
\nAns. <\/p>\npublic int kthLargestElement(int k, ArrayList nums) {\n if (nums == null || nums.size() == 0) {\n return 0;\n }\n return helper(nums, 0, nums.size() - 1, nums.size() - k);\n} \npublic void swap( ArrayListnums, int x, int y){\n int temp = nums.get(x);\n nums.set(x, nums.get(y));\n nums.set(y, temp);\n} \npublic int helper( ArrayList nums, int start, int end, int mid) {\n int pivot = end;\n int num = nums.get(pivot);\n int low = start;\n int high = end;\n while (low < high) {\n while(low < high && nums.get(low) < num) {\n low++;\n }\n while(low = num) {\n high--;\n }\n swap(nums, low, high);\n }\n swap(nums, low, pivot);\n if (low == mid) {\n return nums.get(low);\n } else if (low < mid) {\n return helper(nums, low + 1, end, mid);\n } else {\n return helper(nums, start, low - 1, mid);\n }\n}<\/code><\/pre>\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"Check if a given number is palindrome or not. Ans. A palindrome is a number which is when reversed give the same number. We find the reverse and check whether it is equal to the given number or not. public static void main(String[] args) { Scanner in = new Scanner(System.in); String str = in.nextLine(); int […]<\/p>\n","protected":false},"author":52,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[143],"tags":[],"class_list":["post-2663","post","type-post","status-publish","format-standard","hentry","category-java"],"yoast_head":"\n
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