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The aim is to return all conceivable ways to break the sentence in individual dictionary words given a dictionary dict[] with a list of non-empty words and a non-empty sequence s.Note: When breaking, a dictionary word may be used more than once. Examples:  Input: s = “catsanddog” , dict = [“cat”, “cats”, “and”, “sand”, “dog”]Output: “cats and dog” “cat sand dog”Explanation: The string is split into above 2 ways, in each way all are valid dictionary words. Input: s = “pineapplepenapple” , dict = [“apple”, “pen”, “applepen”, “pine”, “pineapple”]Output: “pine apple pen apple” “pineapple pen apple” “pine applepen apple” Explanation: The string is split into above 3 ways, in each way all are valid dictionary words. How to put the aforementioned concept into practice: C++ Java Python JS Output i like sam sung mobile i like samsung mobile

Determine whether an input string can be divided into a space-separated sequence of dictionary terms given a dictionary of words. For further information, see the examples that follow.This is a well-known Google interview question that is being asked by many other businesses as well. C++ Java Python JS Output Yes Yes Yes Yes Yes No

How to use a random pointer and next to clone a linked list: C++ Java Python C# JS Output 1(3) -> 2(1) -> 3(5) -> 4(3) -> 5(2)

Each node in a linked list of size N contains two links: a random pointer to any random node in the list and a next pointer to the next node. Making a copy of this linked list is the work at hand. Table of Content How to use a random pointer and next to clone a linked list: To store the new nodes that correspond to their original nodes, create a hash table with the value mp.For each node in the original linked list, say curr,mp[curr] = new Node() creates a new node that corresponds to curr and pushes it into a hash table.To update the next and random pointer of every new node, iterate through the original linked list once more: mp[curr]->next = mp[curr->next] and mp[curr]->random = mp[curr->random].Return the cloned linked list’s head, mp[head].The application of the aforementioned strategy is seen below: C++14 Java Python JS Output Original linked list: 1(3) -> 2(1) -> 3(5) -> 4(3) -> 5(2) Cloned linked list: 1(3) -> 2(1) -> 3(5) -> 4(3) -> 5(2) Time Complexity: O(2n), as we are traversing the linked list twice. Auxiliary Space: O(2n), extra O(n) space as we are using a hash table to store the new nodes.

It has previously been detailed how to clone a Binary Tree and LinkedList using random pointers. The concept of graph copying is essentially the same. The goal is to perform a BFS traversal of the graph and create a clone node—a replica of the original node—while visiting a node. A clone node is already present if a node that has already been visited is encountered. How can I monitor the nodes that have been visited or cloned? Maintaining all of the previously generated nodes requires a hash map or map. Important stores: The original node’s address or reference Value shops: Address/Reference of the Cloned Node A duplicate of every node in the graph has been created. How are clone nodes connected? Visit all of the neighboring nodes for u, find the corresponding clone node for each neighbor (or create one if none exists), and then push into the neighboring vector of the cloneNodeU node. This process is done while visiting the neighboring vertices of a node u to obtain the corresponding cloned node for u, which we’ll call cloneNodeU. How can I tell whether the cloned graph is accurate? Before and after the graph is cloned, perform a BFS traversal. In BFS traversal, show a node’s value in addition to its address or reference. Examine the node order; if the values are the same but the address or reference varies between the two traversals, the cloned graph is accurate. C++ Java Python3 Javascript Output BFS Traversal before cloning Value of Node 1 Address of Node 0x1b6ce70 Value of Node 2 Address of Node 0x1b6cea0 Value of Node 4 Address of Node 0x1b6cf00 Value of Node 3 Address of Node 0x1b6ced0 BFS Traversal after cloning Value of Node 1 Address of Node 0x1b6e5a0 Value of Node 2 Address of Node 0x1b6e5d0 Value of Node 4 Address of Node 0x1b6e620 Value of Node 3 Address of Node 0x1b6e670

The aim is to make a duplicate of a node and place it between the original node and the next node, rather than putting it in a different hash table. New nodes will now be present at different locations. The duplicate of node X will now be X->next, and since it is the duplicate of X->random, the random pointer of the duplicate should point to X->random->next. In order to separate the original linked list from the cloned linked list, iterate over the complete linked list to update the random pointer of each cloned node. To put the concept into practice, take the actions listed below: Make a duplicate of node 1 and place it between nodes 1 and 2 in the original linked list. Then, make a duplicate of node 2 and place it between nodes 2 and 3, and so on. After the Nth node, add the copy of N.Update the random pointers to connect the clone node.Update the following pointers to distinguish the cloned linked list from the original list. C++ Java Python JS Output Original linked list: 1(3) -> 2(1) -> 3(5) -> 4(3) -> 5(2) Cloned linked list: 1(3) -> 2(1) -> 3(5) -> 4(3) -> 5(2) Time Complexity: O(3n), because we are traversing the linked list three times. Auxiliary Space: O(1), as we are storing all the cloned nodes in the original linked list itself, no extra space is required.

Determine the length of the largest subsequence in an integer array where the components are consecutive integers; the numbers can be in any order. Examples:   Input: arr[] = {1, 9, 3, 10, 4, 20, 2}Output: 4Explanation: The subsequence 1, 3, 4, 2 is the longest subsequence of consecutive elements Input: arr[] = {36, 41, 56, 35, 44, 33, 34, 92, 43, 32, 42}Output: 5Explanation: The subsequence 36, 35, 33, 34, 32 is the longest subsequence of consecutive elements. [Naive Approach] Using Sorting – O(N*logN) Time and O(N) Space: Finding the longest subarray with consecutive elements is the first step in sorting the array. Run a loop after sorting the array and eliminating entries that appear more than once, maintaining a count and max (both originally zero). Run a loop from beginning to end, setting the count to 1 if the current element is not equal to the previous (element+1); otherwise, increase the count. With a maximum of count and max, update max. To solve the issue, take the actions listed below: C++ Java Python JS Output Length of the Longest contiguous subsequence is 3 Time complexity: O(Nlog(N)), Time to sort the array is O(Nlog(N)).Auxiliary space: O(N). Extra space is needed for storing distinct elements.

Problem Statement: Interleaving String Given three strings, s1, s2, and s3, determine if s3 is formed by an interleaving of s1 and s2. An interleaving of two strings s1 and s2 is formed by inserting characters of s1 and s2 in such a way that the relative order of characters in both strings is preserved. For example: Here, s3 is an interleaving of s1 and s2, so the answer is True. Constraints: Dynamic Programming Approach We can solve this problem using a 2D DP table. Time Complexity: Code Implementation in Different Languages C: #include <stdio.h>#include <string.h>#include <stdbool.h>bool isInterleave(char *s1, char *s2, char *s3) { int m = strlen(s1); int n = strlen(s2); if (m + n != strlen(s3)) return false; bool dp[m + 1][n + 1]; memset(dp, false, sizeof(dp)); dp[0][0] = true; for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i > 0 && s1[i – 1] == s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i – 1][j]; } if (j > 0 && s2[j – 1] == s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i][j – 1]; } } } return dp[m][n];}int main() { char s1[] = “abc”; char s2[] = “def”; char s3[] = “adbcef”; if (isInterleave(s1, s2, s3)) { printf(“True\n”); } else { printf(“False\n”); } return 0;} C++: #include <iostream>#include <vector>#include <string>using namespace std;bool isInterleave(string s1, string s2, string s3) { int m = s1.size(), n = s2.size(); if (m + n != s3.size()) return false; vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false)); dp[0][0] = true; for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i > 0 && s1[i – 1] == s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i – 1][j]; } if (j > 0 && s2[j – 1] == s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i][j – 1]; } } } return dp[m][n];}int main() { string s1 = “abc”, s2 = “def”, s3 = “adbcef”; if (isInterleave(s1, s2, s3)) { cout << “True” << endl; } else { cout << “False” << endl; } return 0;} Java: public class Main { public static boolean isInterleave(String s1, String s2, String s3) { int m = s1.length(), n = s2.length(); if (m + n != s3.length()) return false; boolean[][] dp = new boolean[m + 1][n + 1]; dp[0][0] = true; for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i > 0 && s1.charAt(i – 1) == s3.charAt(i + j – 1)) { dp[i][j] = dp[i][j] || dp[i – 1][j]; } if (j > 0 && s2.charAt(j – 1) == s3.charAt(i + j – 1)) { dp[i][j] = dp[i][j] || dp[i][j – 1]; } } } return dp[m][n]; } public static void main(String[] args) { String s1 = “abc”, s2 = “def”, s3 = “adbcef”; System.out.println(isInterleave(s1, s2, s3) ? “True” : “False”); }} Python: def isInterleave(s1, s2, s3): m, n = len(s1), len(s2) if m + n != len(s3): return False dp = [[False] * (n + 1) for _ in range(m + 1)] dp[0][0] = True for i in range(m + 1): for j in range(n + 1): if i > 0 and s1[i – 1] == s3[i + j – 1]: dp[i][j] = dp[i][j] or dp[i – 1][j] if j > 0 and s2[j – 1] == s3[i + j – 1]: dp[i][j] = dp[i][j] or dp[i][j – 1] return dp[m][n]# Example Usages1 = “abc”s2 = “def”s3 = “adbcef”print(“True” if isInterleave(s1, s2, s3) else “False”) C#: using System;class Program { public static bool IsInterleave(string s1, string s2, string s3) { int m = s1.Length, n = s2.Length; if (m + n != s3.Length) return false; bool[,] dp = new bool[m + 1, n + 1]; dp[0, 0] = true; for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i > 0 && s1[i – 1] == s3[i + j – 1]) { dp[i, j] = dp[i, j] || dp[i – 1, j]; } if (j > 0 && s2[j – 1] == s3[i + j – 1]) { dp[i, j] = dp[i, j] || dp[i, j – 1]; } } } return dp[m, n]; } static void Main() { string s1 = “abc”, s2 = “def”, s3 = “adbcef”; Console.WriteLine(IsInterleave(s1, s2, s3) ? “True” : “False”); }} JavaScript: function isInterleave(s1, s2, s3) { const m = s1.length, n = s2.length; if (m + n !== s3.length) return false; const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(false)); dp[0][0] = true; for (let i = 0; i <= m; i++) { for (let j = 0; j <= n; j++) { if (i > 0 && s1[i – 1] === s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i – 1][j]; } if (j > 0 && s2[j – 1] === s3[i + j – 1]) { dp[i][j] = dp[i][j] || dp[i][j – 1]; } } } return dp[m][n];}// Example Usagelet s1 = “abc”, s2 = “def”, s3 = “adbcef”;console.log(isInterleave(s1, s2, s3) ? “True” : “False”); Summary:

Two terms of the same length, “start” and “target,” are provided in a dictionary. Determine the length of the smallest chain, if any, between “start” and “target,” such that each word in the chain is a legitimate word—that is, it appears in the dictionary—and that adjacent words in the chain differ by no more than one character. It is reasonable to assume that the “target” term is present in a dictionary and that all dictionary words have the same length. Example:  Input: Dictionary = {POON, PLEE, SAME, POIE, PLEA, PLIE, POIN}, start = TOON, target = PLEAOutput: 7Explanation: TOON – POON – POIN – POIE – PLIE – PLEE – PLEA Input: Dictionary = {ABCD, EBAD, EBCD, XYZA}, start = ABCV, target = EBADOutput: 4Explanation: ABCV – ABCD – EBCD – EBAD Approach: Using BFS is the suggested solution to the issue. Start with the start word and push it in a queue to determine the shortest path via BFS. Return that level of BFS traversal after the target is located for the first time. All of the phrases that can be created with that many stages are available in each BFS step. Thus, the length of the shortest word chain will be the first time the target word is discovered. C++ Java Python JS Output Length of shortest chain is: 7 Time Complexity: O(N * M), where N is the number of entries originally in the dictionary and M is the size of the string.Auxiliary Space: O(N+M)

Given a list indicating words and two separate words, startWord and targetWorda list of distinct words with comparable lengths. Determine which transformation sequences from startWord to targetWord are the shortest. They can be returned in any feasible order. The following requirements must be considered in this problem statement: Examples: Example 1: Input: startWord = “der”, targetWord = “dfs”, wordList = {“des”,”der”,”dfr”,”dgt”,”dfs”} Output: [ [ “der”, “dfr”, “dfs” ], [ “der”, “des”, “dfs”] ] Explanation: The length of the smallest transformation sequence here is 3. Following are the only two shortest ways to get to the targetWord from the startWord : “der” -> ( replace ‘r’ by ‘s’ ) -> “des” -> ( replace ‘e’ by ‘f’ ) -> “dfs”. “der” -> ( replace ‘e’ by ‘f’ ) -> “dfr” -> ( replace ‘r’ by ‘s’ ) -> “dfs”. Example 2: Input: startWord = “gedk”, targetWord= “geek” wordList = {“geek”, “gefk”} Output: [ [ “gedk”, “geek” ] ] Explanation: The length of the smallest transformation sequence here is 2. Following is the only shortest way to get to the targetWord from the startWord : “gedk” -> ( replace ‘d’ by ‘e’ ) -> “geek”. Intuition: The idea behind applying the BFS traversal strategy to these types of issues is that, if we pay close attention, we continue to replace the characters one by one, giving the impression that we are going level-wise to get to the targetWord. It is evident from the example below that there are two ways to get to the targetWord. Since we require all of the shortest sequences to get to the destination word, unlike the prior case, we continue the traversal for as many instances of the targetWord as possible rather than stopping it at the first occurrence. The only difficulty here is that, even if a word matches the changed word during character replacement, we do not have to remove it from the wordList right away. Rather, we remove it whenever the traversal for a specific level is over, enabling us to investigate every potential route. This enables us to find several sequences using related words that lead to the targetWord. We can infer from the preceding image that there are two shortest feasible sequences that lead to the word “cog.” C++ Java Output:  der des dfs  der dfr dfs