links for Data Structure

 

1) 𝐁𝐞𝐜𝐨𝐦𝐞 𝐌𝐚𝐬𝐭𝐞𝐫 𝐢𝐧 𝐋𝐢𝐧𝐤𝐞𝐝 𝐋𝐢𝐬𝐭: https://lnkd.in/gXQux4zj

2) 𝐀𝐥𝐥 𝐭𝐲𝐩𝐞𝐬 𝐨𝐟 𝐓𝐫𝐞𝐞 𝐓𝐫𝐚𝐯𝐞𝐫𝐬𝐚𝐥𝐬: https://lnkd.in/gKja_D5H

3) 𝐁𝐞𝐜𝐨𝐦𝐞 𝐌𝐚𝐬𝐭𝐞𝐫 𝐢𝐧 𝐑𝐞𝐜𝐮𝐫𝐬𝐢𝐨𝐧: https://lnkd.in/gQiasy8H

4) 𝐀 𝐆𝐞𝐧𝐞𝐫𝐚𝐥 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝐭𝐨 𝐁𝐚𝐜𝐤𝐭𝐫𝐚𝐜𝐤𝐢𝐧𝐠 𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬: https://lnkd.in/gVkQX5vA

5) 𝐈𝐦𝐩𝐨𝐫𝐭𝐚𝐧𝐭 𝐒𝐭𝐫𝐢𝐧𝐠 𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬 𝐏𝐚𝐭𝐭𝐞𝐫𝐧: https://lnkd.in/gkNvEi8j

6) 10-𝐥𝐢𝐧𝐞 𝐓𝐞𝐦𝐩𝐥𝐚𝐭𝐞 𝐭𝐡𝐚𝐭 𝐜𝐚𝐧 𝐬𝐨𝐥𝐯𝐞 𝐦𝐨𝐬𝐭 '𝐬𝐮𝐛𝐬𝐭𝐫𝐢𝐧𝐠' 𝐩𝐫𝐨𝐛𝐥𝐞𝐦𝐬: https://lnkd.in/giASrwds

7) 𝐒𝐥𝐢𝐝𝐢𝐧𝐠 𝐖𝐢𝐧𝐝𝐨𝐰 𝐓𝐞𝐦𝐩𝐥𝐚𝐭𝐞: https://lnkd.in/gjatQ5pK

8) 𝐓𝐰𝐨 𝐏𝐨𝐢𝐧𝐭𝐞𝐫𝐬 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬: https://lnkd.in/gBfWgHYe

9) 𝐏𝐨𝐰𝐞𝐫𝐟𝐮𝐥 𝐔𝐥𝐭𝐢𝐦𝐚𝐭𝐞 𝐁𝐢𝐧𝐚𝐫𝐲 𝐒𝐞𝐚𝐫𝐜𝐡 𝐓𝐞𝐦𝐩𝐥𝐚𝐭𝐞: https://lnkd.in/gKEm_qUK

10) 𝐓𝐞𝐦𝐩𝐥𝐚𝐭𝐞 𝐟𝐨𝐫 𝐌𝐨𝐧𝐨𝐭𝐨𝐧𝐢𝐜 𝐒𝐭𝐚𝐜𝐤 𝐏𝐫𝐨𝐛𝐥𝐞𝐦𝐬: https://lnkd.in/gdYahWVN

11) 𝐆𝐫𝐞𝐞𝐝𝐲 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬: https://lnkd.in/gw8CgMkC

12) 𝐀𝐥𝐥 𝐓𝐲𝐩𝐞𝐬 𝐨𝐟 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬 𝐟𝐨𝐫 𝐁𝐢𝐭𝐬 𝐌𝐚𝐧𝐢𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧𝐬: https://lnkd.in/gXzegWuU

13) 𝐆𝐫𝐚𝐩𝐡 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬: https://lnkd.in/gKE6w7Jb

14) 𝐃𝐲𝐧𝐚𝐦𝐢𝐜 𝐏𝐫𝐨𝐠𝐫𝐚𝐦𝐦𝐢𝐧𝐠 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬: https://lnkd.in/gbpRU46g

15) 14 𝐏𝐚𝐭𝐭𝐞𝐫𝐧𝐬 𝐭𝐨 𝐀𝐜𝐞 𝐂𝐨𝐝𝐢𝐧𝐠 𝐈𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰 𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬: https://lnkd.in/gMZJVkFf

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LeetCode 75 problems

 

Array

• Two Sum
• Best Time to Buy and Sell Stock
• Contains Duplicate
• Product of Array Except Self
• Maximum Subarray
• Maximum Product Subarray
• Find Minimum in Rotated Sorted Array
• Search in Rotated Sorted Array
• 3 Sum
• Container With Most Water

---

Binary

• Sum of Two Integers
• Number of 1 Bits
• Counting Bits
• Missing Number
• Reverse Bits

---

Dynamic Programming

• Climbing Stairs
• Coin Change
• Longest Increasing Subsequence
• Longest Common Subsequence
• Word Break Problem
• Combination Sum
• House Robber
• House Robber II
• Decode Ways
• Unique Paths
• Jump Game

---

Graph

• Clone Graph
• Course Schedule
• Pacific Atlantic Water Flow
• Number of Islands
• Longest Consecutive Sequence
• Alien Dictionary (Leetcode Premium)
• Graph Valid Tree (Leetcode Premium)
• Number of Connected Components in an Undirected Graph (Leetcode Premium)

---

Interval

• Insert Interval
• Merge Intervals
• Non-overlapping Intervals
• Meeting Rooms (Leetcode Premium)
• Meeting Rooms II (Leetcode Premium)

---

Linked List

• Reverse a Linked List
• Detect Cycle in a Linked List
• Merge Two Sorted Lists
• Merge K Sorted Lists
• Remove Nth Node From End Of List
• Reorder List

---

Matrix

• Set Matrix Zeroes
• Spiral Matrix
• Rotate Image
• Word Search

---

String

• Longest Substring Without Repeating Characters
• Longest Repeating Character Replacement
• Minimum Window Substring
• Valid Anagram
• Group Anagrams
• Valid Parentheses
• Valid Palindrome
• Longest Palindromic Substring
• Palindromic Substrings
• Encode and Decode Strings (Leetcode Premium)

---

Tree

• Maximum Depth of Binary Tree
• Same Tree
• Invert/Flip Binary Tree
• Binary Tree Maximum Path Sum
• Binary Tree Level Order Traversal
• Serialize and Deserialize Binary Tree
• Subtree of Another Tree
• Construct Binary Tree from Preorder and Inorder Traversal
• Validate Binary Search Tree
• Kth Smallest Element in a BST
• Lowest Common Ancestor of BST
• Implement Trie (Prefix Tree)
• Add and Search Word
• Word Search II

---

Heap

• Merge K Sorted Lists
• Top K Frequent Elements
• Find Median from Data Stream

Minimum Subarray Of Sumk

Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2 

Explanation: The subarray [4,3] has the minimal length under the problem constraint. 

public class MinimumSubarrayOfSumk {

    public static void main(String[] args) {

        int[] arr = {

                10,5,13,4,8,4,5,11,14,9,16,10,20,8};

        int target  =80;

        int i =-1;

        int j=-1;

        int sum = 0;

        int length = Integer.MAX_VALUE;

        while (true){

        boolean f1 = false,f2 = false;

            while (i<arr.length-1 && sum<target){

                f1 = true;

                i++;

                sum+=arr[i];

                if(sum>=target){

                    length = Math.min(length,i-j);

                    break;

                }

               // f1 = true;

            }

            while (j<i && sum>=target){

                f2 = true;

                j++;

                sum-=arr[j];

                if(sum>=target){

                    length = Math.min(length,i-j);

                }else{

                    break;

                }

               // f2 = true;

            }

            if(!f1 && !f2){

                break;

            }

        }

        System.out.println(length);

    }

}

Result: 6

Permutation in String

 Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words,

one of the first string’s permutations is the substring of the second string.

For example:

Input: s1 = "ab" s2 = "eidbaooo"

Output: True

Explanation: s2 contains one permutation of s1 ("ba").

public class PermutationInString {

    public static void main(String[] args) {

        boolean bool = checkInclusion();

        System.out.println(bool);

    }

    private static boolean checkInclusion(){

        String str1 = "eidbaooo";

        String str2 = "ab";

        Map<Character, Integer> map2 = new HashMap<>();

        for(int i = 0;i<str2.length();i++){

            char ch = str2.charAt(i);

            map2.put(ch,map2.getOrDefault(ch,0)+1);

        }

        Map<Character, Integer> map1 = new HashMap<>();

        int i = -1;

        int j=-1;

        int desireCount = str2.length();

        int mCount = 0;

        boolean resultFlag= false;

        while (true){

            boolean f1 = false,f2 = false;

            while (i<str1.length()-1 && mCount<desireCount){

                f1= true;

                i++;

                char ch = str1.charAt(i);

                map1.put(ch,map1.getOrDefault(ch,0)+1);

                if(map1.getOrDefault(ch,0)<=map2.getOrDefault(ch,0)){

                    mCount++;

                }

                if(mCount==desireCount){

                    break;

                }

            }

            while (j<i && mCount==desireCount){

                f2 = true;

                j++;

                if(i+1-j==desireCount){ //check if size of string is same as desired count

                    //System.out.println("result");

                    resultFlag = true;

                    break;

                }

                char ch = str1.charAt(j);

                Integer val =map1.get(ch);

                if(val>1){

                    map1.put(ch,map1.get(ch)-1);

                }else{

                    map1.remove(ch);

                }

                if(map1.getOrDefault(ch,0)<map2.getOrDefault(ch,0)){

                    mCount--;

                }

                if(mCount<desireCount){

                    break;

                }

            }

            if(resultFlag){

                System.out.println("result found");

                break;

            }

            if(!f1 && !f2){

                System.out.println("result not found");

                break;

            }

        }

        if(resultFlag){

            //System.out.println("result found");

            return true;

        }else{

            //System.out.println("not found");

            return false;

        }

     //   return true;

    }

}

Result:true