Valid Sudoku - LeetCode

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

Example 1:

Input:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: true

Example 2:

Input:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.
• The given board contain only digits 1-9 and the character '.'.
• The given board size is always 9x9.

Java Solution :

class Solution {
    public boolean isValidSudoku(char[][] board) {
        if(board == null || board.length == 0)
            return false;
        HashMap<Integer,Integer> [] rows = new HashMap[9];
        HashMap<Integer,Integer> [] cols = new HashMap[9];
        HashMap<Integer,Integer> [] boxes = new HashMap[9];
        
        for(int i = 0; i < 9 ; i++){
            rows[i] = new HashMap<Integer,Integer>();
            cols[i] = new HashMap<Integer,Integer>();
            boxes[i] = new HashMap<Integer,Integer>();
        }
        int num =0;
        for(int i =0 ; i < 9 ; i++){
            for(int j=0; j < 9; j++){
                char n = board[i][j];
                if(board[i][j] != '.'){
                 num = (int)n;
                int box_index = (i/3)*3 + j/3;
                rows[i].put(num,rows[i].getOrDefault(num,0)+1);
                cols[j].put(num,cols[j].getOrDefault(num,0)+1);
                boxes[box_index].put(num,boxes[box_index].getOrDefault(num,0)+1);
                
                if(rows[i].get(num) > 1 || cols[j].get(num) > 1 || boxes[box_index].get(num) > 1)
                    return false;
            }
        }
        }
        
        return true;
    }
}

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;
    }
}