LeetCode - Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

Java Solution :

class Solution {
    public int trap(int[] height) {
        int[] leftMax = new int[height.length];
        int[] rightMax = new int[height.length];
        int waterCollected = 0;
        if(height == null || height.length == 0)
            return 0;
        // scan from left to right
        leftMax[0] = height[0];
        int max = leftMax[0];
        for(int i =1 ;i < height.length; i++){
            if(height[i] >= max){
                max = height[i];
                leftMax[i] = max;
            } else{
                leftMax[i] = max;
            }
        }
        
        // scan from right to left
        rightMax[height.length-1] = height[height.length-1];
        int mr = rightMax[height.length-1];
        for(int i = height.length-2; i >= 0 ; i--){
            if(height[i] >= mr){
                mr = height[i];
                rightMax[i] = mr;
            } else{
                rightMax[i] = mr;
            }
        }
        
        for(int i = 0 ;i < height.length; i++){
            waterCollected += Math.min(leftMax[i],rightMax[i]) - height[i];
        }
        
        return waterCollected;
    }
}

Design a Tic-tac-toe game Leetcode

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves is allowed.
3. A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Example:

Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

Follow up:
Could you do better than O(n2) per move() operation?

Java Solution :

class TicTacToe {
        int[] rowCounter;
        int[] colCounter;
        int diagLeft;
        int diagRight;
        int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
         rowCounter = new int[n];
         colCounter = new int[n];
         diagLeft = 0;
         diagRight = 0;
    }
    
    /** Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins. */
    public int move(int row, int col, int player) {
        int move = (player == 1) ? 1 : -1;
        rowCounter[row] +=move;
        colCounter[col] +=move;
        if(row == col)
            diagLeft +=move;
        if(row == n-col-1)
            diagRight +=move;
        if(rowCounter[row] == n || diagLeft == n || diagRight== n || colCounter[col] == n)
            return 1;
        else if(rowCounter[row] == -n || diagLeft == -n || diagRight== -n || colCounter[col] == -n)return 2;
        else
            return 0; 
    }
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */

LeetCode - Design Tic-Tac-Toe

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves is allowed.
3. A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Example:

Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

Follow up:
Could you do better than O(n2) per move() operation?

Java Solution :

class TicTacToe {
        int[] rowCounter;
        int[] colCounter;
        int diagLeft;
        int diagRight;
        int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
         rowCounter = new int[n];
         colCounter = new int[n];
         diagLeft = 0;
         diagRight = 0;
    }
    
    /** Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins. */
    public int move(int row, int col, int player) {
        int move = (player == 1) ? 1 : -1;
        rowCounter[row] +=move;
        colCounter[col] +=move;
        if(row == col)
            diagLeft +=move;
        if(row == n-col-1)
            diagRight +=move;
        if(rowCounter[row] == n || diagLeft == n || diagRight== n || colCounter[col] == n)
            return 1;
        else if(rowCounter[row] == -n || diagLeft == -n || diagRight== -n || colCounter[col] == -n)return 2;
        else
            return 0; 
    }
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */

LeetCode - K Closest Points to Origin

We have a list of points on the plane.  Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation: 
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

1. 1 <= K <= points.length <= 10000
2. -10000 < points[i][0] < 10000
3. -10000 < points[i][1] < 10000

Java Solution :

class Solution {
    public int[][] kClosest(int[][] points, int K) {
 PriorityQueue<int[]> maxHeap = new PriorityQueue<>(Comparator.comparing(a -> -a[0] * a[0] - a[1] * a[1]));        
        for(int[] point : points){
            maxHeap.add(point);
            if(maxHeap.size() > K)
                maxHeap.remove();
        }
        
        int[][] result = new int[K][2];
       
        while(K-- > 0){
            result[K] = maxHeap.poll();
        }
        
        return result;
    }
}

LeetCode - Grouped Anagrams

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

• All inputs will be in lowercase.
• The order of your output does not matter.

Java Solution :

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> groupedAnagrams = new ArrayList<>();
        HashMap<String , List<String>> map = new HashMap<>();
        
        for(String s : strs){
            char[] charArray = s.toCharArray();
            Arrays.sort(charArray);
            String sortedString = new String(charArray);
            if(!map.containsKey(sortedString)){
                map.put(sortedString,new ArrayList<String>());
            }
            map.get(sortedString).add(s);
        }
         groupedAnagrams.addAll(map.values());
        return groupedAnagrams;
    }
}

LeetCode - Meeting Rooms II

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required.

Example 1:

Input: [[0, 30],[5, 10],[15, 20]]
Output: 2

Example 2:

Input: [[7,10],[2,4]]
Output: 1

Java Solution :

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        Arrays.sort(intervals,(a,b)-> a[0] - b[0]);
        int numOfMeetingRooms = 0;
PriorityQueue<Integer> minHeap = new PriorityQueue<>();        
        for(int[] interval : intervals){
            if(minHeap.isEmpty()){
                minHeap.offer(interval[1]);
                numOfMeetingRooms++;
            }else if(minHeap.peek() <= interval[0]){
                minHeap.poll();
                minHeap.offer(interval[1]);
            }else{
                minHeap.offer(interval[1]);
                numOfMeetingRooms++;
            }
        }
       return numOfMeetingRooms;
    }
}

LeetCode - Merge K Sorted Lists

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 result = dummy;
            while(!minHeap.isEmpty()){
                result.next = new ListNode(minHeap.remove());
                result = result.next;
            }
            return dummy.next;
    }
}

LeetCode - Number of Islands

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

LeetCode - Add Two Numbers

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 temp = new ListNode(0);
        int carry = 0;
        ListNode headA =l1 , headB = l2 , curr = temp;
     
           while (headA != null || headB !=null){
                int x = (headA != null) ? headA.val : 0;
                int y = (headB != null) ? headB.val : 0;
                int sum = x + y + carry;
                carry = sum / 10;
                curr.next = new ListNode(sum % 10);     
                curr = curr.next;
               if(headA != null) headA = headA.next;
               if(headB != null) headB = headB.next;
           }
            if(carry > 0)
                curr.next = new ListNode(carry);
            return temp.next;
        
    }
}

Time Complexity : O(max(m,n))

Space Complexity : O(max(m,n))+1

LeetCode - Valid Parentheses

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<Character>();
        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();
    }
}

LeetCode - Most Common Word

Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words.  It is guaranteed there is at least one word that isn't banned, and that the answer is unique.

Words in the list of banned words are given in lowercase, and free of punctuation.  Words in the paragraph are not case sensitive.  The answer is in lowercase.

Example:

Input: 
paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
banned = ["hit"]
Output: "ball"
Explanation: 
"hit" occurs 3 times, but it is a banned word.
"ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
Note that words in the paragraph are not case sensitive,
that punctuation is ignored (even if adjacent to words, such as "ball,"), 
and that "hit" isn't the answer even though it occurs more because it is banned.

Note:

• 1 <= paragraph.length <= 1000.
• 0 <= banned.length <= 100.
• 1 <= banned[i].length <= 10.
• The answer is unique, and written in lowercase (even if its occurrences in paragraph may have uppercase symbols, and even if it is a proper noun.)
• paragraph only consists of letters, spaces, or the punctuation symbols !?',;.
• There are no hyphens or hyphenated words.
• Words only consist of letters, never apostrophes or other punctuation symbols.

Java Solution:

class Solution {

    public String mostCommonWord(String paragraph, String[] banned) {

        String result =" ";

        Set<String> bannedset = new HashSet<String>();

        for(String word : banned) bannedset.add(word);

        Map<String,Integer> map = new HashMap<String,Integer>();

        String[] parawords = paragraph.replaceAll("[^a-zA-Z]"," ").toLowerCase().split(" ");

        for(String pword : parawords){

            if(!bannedset.contains(pword) && !pword.isEmpty())

                map.put(pword,map.getOrDefault(pword,0)+1);

        }

        

        int max =0;

        for(String word1 : map.keySet()){

            if(map.get(word1)>max){

                max = map.get(word1);

                result = word1;

            }

        }

        return result;

    }

}

LeetCode - Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Java Solution :

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        for(int i=0; i< nums.length;i++){
            int complement = target-nums[i];
            if(map.containsKey(complement)){
                return new int[] { map.get(complement),i};
            }
            map.put(nums[i],i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

Time complexity : O(n) Space Complexity O(n)

LeetCode-Merge Two Sorted Lists

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

Iterative Approach :

/**
 * 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 preNode = new ListNode(-1);
        ListNode prev = preNode;
        
        while (l1!=null && l2!=null){
            if(l1.val <= l2.val){
                prev.next = l1;
                l1 = l1.next;
            }else{
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }
        prev.next = l1 == null ? l2 : l1;
        return preNode.next;
    }
}

Time Complexity : O(n+m) Space Complexity O(1)

Recursive Approach:

/**
 * 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) {
        if(l1 == null){
            return l2;
        }else if(l2 == null){
            return l1;
        }else if(l1.val <l2.val){
                l1.next = mergeTwoLists(l1.next,l2);
            return l1;
            }else{
                l2.next = mergeTwoLists(l1,l2.next);
            return l2;
            }
        }
}

Time Complexity : O(n+m) Space Complexity : O(n+m)

LeetCode - Two Sum Less Than K

Given an array A of integers and integer K, return the maximum S such that there exists i < j with A[i] + A[j] = S and S < K. If no i, j exist satisfying this equation, return -1.

Example 1:

Input: A = [34,23,1,24,75,33,54,8], K = 60
Output: 58
Explanation: 
We can use 34 and 24 to sum 58 which is less than 60.

Example 2:

Input: A = [10,20,30], K = 15
Output: -1
Explanation: 
In this case it's not possible to get a pair sum less that 15.

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i] <= 1000
3. 1 <= K <= 2000

Java Solution ;

class Solution {

    public int twoSumLessThanK(int[] A, int K) {

        Arrays.sort(A);

        int maxSum = -1;

        int i = 0;

        int j = A.length-1;

        while(i < j){

            int sum = A[i]+A[j];

            if(sum<K){

                maxSum = Math.max(maxSum,sum);

                i++;

            }else{

                j--;

            }

        }

     return maxSum;

    }

}

LeetCode - Reorder Data in Log Files

You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier.  Then, either:

• Each word after the identifier will consist only of lowercase letters, or;
• Each word after the identifier will consist only of digits.

We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

Return the final order of the logs.

Example 1:

Input: logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"]
Output: ["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]

Constraints:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

Solution :

Java Solution :

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        Arrays.sort(logs,(log1,log2)->{
            String[] set1 = log1.split(" ",2);
            String[] set2 = log2.split(" ",2);
            boolean isCharacter1 = Character.isLetter(set1[1].charAt(0));
            boolean isCharacter2 = Character.isLetter(set2[1].charAt(0));
            if(isCharacter1 && isCharacter2){
                int cmp = set1[1].compareTo(set2[1]);
                if(cmp !=0) return cmp;
                return set1[0].compareTo(set2[0]);
            }
            return !isCharacter1 ? (!isCharacter2 ? 0 : 1) : -1;
        });
        return logs;
    }
}

Amazon Onsite Interview Questions

1.  Two Sum
2. Most Common Word
3. Reorder Log files
4.Trapping Rain Water
5.Copy List With Random Pointer
6.Number of Islands
7. Lowest common ancestor of Binary tree
8.Binary Tree Zig zag level order traversal
9. Word Ladder 1 , 2
10. word search 1  , 2
11. Meetings rooms 1 , 2
12. Kth largest element in array.
13. K closest points to Origin.
14.Longest Palindromic Substring.
15. LRC Cache
16. Tic Toc Toe
17. Leetcode all design questions.
18. Prison cells after N days.
19. word break 2
20.  stream of characters (DNA Sequence )
21.

Given a 2d array in which each row is sorted and rotated, you need to come up with an algorithm which efficiently sort the entire 2d matrix in descenting order.

eg:
input: arr[][] = {
{41, 45, 20, 21},
{1 ,2, 3, 4},
{30, 42, 43, 29 },
{16, 17, 19, 10}
}

output: {
{ 45, 43, 42, 41},
{30, 29, 21, 20},
{19, 17, 16, 10},
{4, 3, 2, 1}
}

Interviewer was expecting the solution to run with a complexity < O(n^3) solution.

22.  https://leetcode.com/problems/cut-off-trees-for-golf-event/

23. https://leetcode.com/problems/top-k-frequent-elements/

nput is in <productId, timeStamp> format. So assume you have a list of productIDs and their timestamps which they were accessed:

[<product1, timestamp1>, <product2, timestamp2>, <product3, timestamp3>, ...]

Find the top K products purchased in the last one hour.

24. We have two Queues where each queue contains timestamp price pair. We have to return list of list[[price1, price 2]] for all those timestamps where abs(ts1-ts2) <= 1 second where ts1 and price1 are the from the first queue and ts2 and price2 are from the second queue.

Follow up:- one queue is slow

25.Deep copy of an N-ary tree.

public TreeNode clone(TreeNode root) {
  if (root == null) {
    return null;
  }

  TreeNode rootClone = new TreeNode(root.val);
  for (TreeNode child : root.children) {
    TreeNode childClone = clone(child);
    rootClone.children.add(childClone);
  }

  return rootClone;
}

private static class TreeNode {

  int val;
  List<TreeNode> children = new ArrayList<>();

  public TreeNode() {
  }

  public TreeNode(int val) {
    this.val = val;
  }
}

26.https://leetcode.com/problems/subsets/

Now the follow-up question is to remove that for loop from backtrack funtion and pass the indexes. I tried to keep i index by passing it to the function by couldn't figure out the index incrementing logic.

ay we have a machine which scans a printed word and assigns a probablity to each character of the word.
For example, if we scan LBC, the machine produces the following map for the word (first list will be for the first char, second list for the second char etc.) :





{
 ['L': 90, 'I': 50, 'N': 10],      // for first character (L)
 ['B': 95, '8': 80, 'S': 15],      // for secodn character (B)
 ['C': 90, 'O': 90, 'D': 70]       // for third character (C)
}




Meaning for the first character there is a 90 percent chance that it is L, 50 percent that it is I etc.
Given a list of Strings, return the word with the highest probablity based on this map. For example, if the list is {LBD, IBC, LSD}, LBD should be returned (with the probablity of 90 + 95 + 70).



27. Consider that there are 26 tables, listed from A to Z. Each table is connected with table next to it. A connected to B, B to C so forth and so on till Z.



How will you connect table A and Z with minimum number of joins?

28. https://leetcode.com/problems/k-similar-strings/

There are two anagram strings, what is the minimum number of swaps required to match one string to another? You can swap any 2 characters. What is time complexity?



Example 1:


Input: s1 = "AABC", s2 = "AACB"
Output: 1
Explanation: swap 'B' and 'C' AABC => AACB
29.Position: SDE2






Design a configuration management system



User should be able to add configuration
User should be able to delete configuration
User should be able to search for configuration
User should be able to subscribe to Configuration So that any updates in configuration will gets notfied to user


30.I have my amazon interview coming up for SDE in seattle. One of my friends recently went on-site in Seattle and was asked this question in OOD. I have not been able to find a good approach to this question. Any suggestions would be helpful



implemnet linux find command as an api ,the api willl support finding files that has given size requirements and a file with a certain format like



find all file >5mb
find all xml
Assume file class
{
get name()
directorylistfile()
getFile()
create a library flexible that is flexible
Design clases,interfaces.


31. https://leetcode.com/problems/analyze-user-website-visit-pattern




32. Given an array of strings, you need to group isomorphic strings together.



Example:


Input: ["apple", "apply", "dog", "cog", "romi"]
Output: [["dog", "cog"], ["romi"], ["apple", "apply"]]
33. Load Packages