Rotate Array - LeetCode

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Java Solution :

class Solution {
    public void rotate(int[] nums, int k) {
        k = k%nums.length;
        reverse(nums,0,nums.length-1);
        reverse(nums,0,k-1);
        reverse(nums,k,nums.length-1);
    }
    
    public void reverse(int[] nums,int start , int end){
        while(start < end){
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

Rotate Image - LeetCode

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

Java Solution :

class Solution {
    public void rotate(int[][] matrix) {
        Collections.reverse(Arrays.asList(matrix));
        for(int i=0; i < matrix.length;  i++){
            for(int j =0 ; j < i ; j++){
                int temp = matrix[j][i];
                matrix[j][i] = matrix[i][j];
                matrix[i][j] = temp;
            }
        }
    }
}

Number of Islands - LeetCode

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

Java Solution :

class Solution {
    
   public void dfs(char[][] grid,int i ,int j){
        
    if(i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j]== '0')
       return;
       
            grid[i][j] = '0';
            dfs(grid,i,j+1);
            dfs(grid,i,j-1);
            dfs(grid,i+1,j);
            dfs(grid,i-1,j);
        }
    
    public int numIslands(char[][] grid) {
        if(grid == null || grid.length == 0)
            return 0;
         
    
        int numOfIslands = 0;
        
        for(int i=0 ; i < grid.length ; i++){
            for(int j=0; j < grid[i].length ; j++){
                if(grid[i][j] == '1'){
                    ++numOfIslands;
                 dfs(grid, i, j);
                }
            }
        }
       
        return numOfIslands;
    }
}

Partition Labels - LeetCode

A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.

Example 1:

Input: S = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.

Note:

1. S will have length in range [1, 500].
2. S will consist of lowercase letters ('a' to 'z') only

Java Solution:

class Solution {
    public List<Integer> partitionLabels(String S) {
        List<Integer> partitionlengths = new ArrayList<>();
        int[] lastIndexes = new int[26];
        for(int i=0 ; i < S.length() ; i++){
            lastIndexes[S.charAt(i)-'a'] = i;
        }
        
        int i = 0;
        while(i < S.length()){
            int end = lastIndexes[S.charAt(i) - 'a'];
            int j = i;
            while(j !=end){
                end = Math.max(end,lastIndexes[S.charAt(j++)-'a']);
            }
            partitionlengths.add(j-i+1);
            i = j+1;
        }
        return partitionlengths;
    }
}

Merge Two Sorted Lists - LeetCode

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Java Solution :

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode node = new ListNode(-1);
        ListNode head = node;
        while(l1 != null && l2 !=null){
            if(l1.val < l2.val){
                node.next = l1;
                l1 = l1.next;
            }else {
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
        }
        
        if(l1 !=null){
            node.next = l1;
        }else if(l2 != null){
            node.next = l2;
        }
        
        return head.next;
        
    }
}

Merge k Sorted Lists - LeetCode

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

Java Solution : 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        for(ListNode head : lists){
            while(head != null){
                minHeap.add(head.val);
                head = head.next;
            }
        }
        
        ListNode dummy = new ListNode(-1);
        ListNode head = dummy;
        
        while(!minHeap.isEmpty()){
            head.next = new ListNode(minHeap.remove());
            head = head.next;
        }
        return dummy.next;
    }
}

Valid Parentheses - LeetCode

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

Java Solution:

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<>();
        
        for(char c : s.toCharArray()){
            if(c == '[' || c == '{' || c == '('){
                stack.push(c);
            }else if(c == ']' && !stack.isEmpty() && stack.peek() == '['){
                stack.pop();
            }else if(c == '}' && !stack.isEmpty() && stack.peek() == '{'){
                stack.pop();
            }else if(c == ')' && !stack.isEmpty() && stack.peek() == '('){
                stack.pop();
            }else{
                return false;
            }           
        }
        return stack.isEmpty();
    }
}