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();
    }
}

Maximum Subarray - LeetCode

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

Java Solution :

class Solution {
    public int maxSubArray(int[] nums) {
        if(nums.length == 0)
            return 0;
        int prev = nums[0];
        int maxValue = nums[0];
        for(int i =1; i < nums.length ; i++){
            prev = prev > 0 ? prev+nums[i] : nums[i];
            if(maxValue < prev)
                maxValue = prev;
        }
        return maxValue;
    }
}

Coin Change - LeetCode

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:

Input: coins = [1, 2, 5], amount = 11
Output: 3 
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Note:
You may assume that you have an infinite number of each kind of coin.

Java Solution:

class Solution {
    public int coinChange(int[] coins, int amount) {
        Arrays.sort(coins);
        int [] dp = new int[amount+1];
        Arrays.fill(dp,amount+1);
        dp[0] = 0;
        for(int i = 0; i <= amount; i++){
            for(int j = 0; j < coins.length ; j++){
                if(coins[j]<=i)
                dp[i] = Math.min(dp[i], 1+dp[i-coins[j]]);
                else
                    break;
            }
        }
        return dp[amount] > amount ? -1:dp[amount] ;
    }
}