Reverse a linked list - LeetCode

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Java Solution : Iterative Method (Three pointer approach)

prev , current , next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null)
            return null;
        ListNode current = head;
        ListNode prev = null;
        ListNode next = null;
        
        while(current !=null){
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        head = prev;
        return head;
        
    }
}

Jump Game - LeetCode

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.

Java Solution :

class Solution {
    public boolean canJump(int[] nums) {
        if(nums.length <=1)
            return true;
        int jumpValue = nums[0];
        for(int i =0 ; i < nums.length; i++){
            if(jumpValue <= i && nums[i]==0) //Condition to check if can't move futher
                return false;
            if(i + nums[i] > jumpValue) //Value to jump from current postion
                jumpValue = nums[i]+i;
            if(jumpValue >= nums.length-1) //if we reach last index or beyond that we acheived.
                return true;
        }
        return false;
    }
}

Odd Even Linked List -LeetCode

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Note:

• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even and so on ...

Java Solution :

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null)
            return null;
        ListNode odd = head;
        ListNode even =head.next;
        ListNode evenhead = even;
        while(even!=null && even.next !=null){
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        odd.next = evenhead;
        return head;
    }
}




Intersection of Two Linked Lists - LeetCode

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1)

• Java Solution:

• /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode(int x) {
   *         val = x;
   *         next = null;
   *     }
   * }
   */
  public class Solution {
      public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
          ListNode p = headA;
          ListNode q = headB;
          int lenA = 0;
          int lenB = 0;
          while(p!=null){
              lenA++;
              p = p.next;
          }
          while(q!=null){
              lenB++;
              q=q.next;
          }
          p = headA;
          q= headB;
          
          if(lenA > lenB){
              int diff = lenA-lenB;
              for(int i=0;i<diff;i++)
                  p = p.next;
          }else if(lenA < lenB){
              int diff = lenB-lenA;
              for(int i=0;i<diff;i++)
                  q = q.next;
          }
          
          while (p!=null && q!=null){
              if(p.val == q.val)
                  return p;
              p = p.next;
              q = q.next;
          }
          return null;
      }
  }

Add Two Numbers - LeetCode

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Java Solution :

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
    }
}

Two Sum II - Input array is sorted - LeetCode

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

Java Solution :

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
        for(int i=0; i < numbers.length ; i++){
            if(map.containsKey(target-numbers[i]) && i!=map.get(target-numbers[i])){
                return new int [] {map.get(target-numbers[i])+1,i+1};
            }else{
                map.put(numbers[i],i);
            }
        }
        return null;
    }
}

Best Time to Buy and Sell Stock || -LeetCode

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Java Solution:

class Solution {
    public int maxProfit(int[] prices) {
        int maxProfit =0;
        for(int i=1 ; i < prices.length; i++){
            if(prices[i]>prices[i-1])
                maxProfit+=prices[i]-prices[i-1];
        }
        return maxProfit;
    }
}