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Kadane’s technique works by going from left to right throughout the array and calculating the largest sum of all the subarrays that finish at each element. The highest of these values will be the outcome. The primary problem, however, is how to determine the largest sum of all the subarrays that finish at an element in O(1) time. We can utilize the maximum sum ending at the previous element to get the maximum sum of the subarray ending at the current element, let’s say maxEnding. Therefore, we have two options for any element: C++ C Java Python C# JavaScript Output 11 Time Complexity: O(n), since we are traversing the array only one time.Auxiliary Space: O(1)

Problem Statement: String to Integer (atoi) Implement the function atoi which converts a string to an integer. The function should follow these rules: Example: Example 1: Example 2: Example 3: Example 4: Example 5: Approach: Algorithm: Time Complexity: Code Implementation: C: #include <stdio.h>#include <limits.h>#include <ctype.h>int myAtoi(char* str) { long result = 0; int i = 0, sign = 1; // Skip leading whitespaces while (str[i] == ‘ ‘) i++; // Check for optional sign if (str[i] == ‘+’ || str[i] == ‘-‘) { sign = (str[i] == ‘-‘) ? -1 : 1; i++; } // Convert digits to integer while (isdigit(str[i])) { result = result * 10 + (str[i] – ‘0’); i++; // Check for overflow/underflow if (result * sign > INT_MAX) return INT_MAX; if (result * sign < INT_MIN) return INT_MIN; } return (int)(result * sign);}int main() { char str[] = ” -42″; printf(“%d\n”, myAtoi(str)); return 0;} C++: #include <iostream>#include <climits>#include <cctype>using namespace std;int myAtoi(string str) { long result = 0; int i = 0, sign = 1; // Skip leading whitespaces while (i < str.size() && str[i] == ‘ ‘) i++; // Handle optional sign if (i < str.size() && (str[i] == ‘+’ || str[i] == ‘-‘)) { sign = (str[i] == ‘-‘) ? -1 : 1; i++; } // Convert digits to integer while (i < str.size() && isdigit(str[i])) { result = result * 10 + (str[i] – ‘0’); i++; // Check for overflow/underflow if (result * sign > INT_MAX) return INT_MAX; if (result * sign < INT_MIN) return INT_MIN; } return (int)(result * sign);}int main() { string str = ” -42″; cout << myAtoi(str) << endl; return 0;} Java: public class Solution { public int myAtoi(String str) { long result = 0; int i = 0, sign = 1; // Skip leading whitespaces while (i < str.length() && str.charAt(i) == ‘ ‘) i++; // Handle optional sign if (i < str.length() && (str.charAt(i) == ‘+’ || str.charAt(i) == ‘-‘)) { sign = (str.charAt(i) == ‘-‘) ? -1 : 1; i++; } // Convert digits to integer while (i < str.length() && Character.isDigit(str.charAt(i))) { result = result * 10 + (str.charAt(i) – ‘0’); i++; // Check for overflow/underflow if (result * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE; if (result * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE; } return (int)(result * sign); } public static void main(String[] args) { Solution sol = new Solution(); String str = ” -42″; System.out.println(sol.myAtoi(str)); }} Python: def myAtoi(s: str) -> int: result = 0 sign = 1 i = 0 # Skip leading whitespaces while i < len(s) and s[i] == ‘ ‘: i += 1 # Handle optional sign if i < len(s) and (s[i] == ‘+’ or s[i] == ‘-‘): sign = -1 if s[i] == ‘-‘ else 1 i += 1 # Convert digits to integer while i < len(s) and s[i].isdigit(): result = result * 10 + int(s[i]) i += 1 # Check for overflow/underflow if result * sign > 2**31 – 1: return 2**31 – 1 if result * sign < -2**31: return -2**31 return result * sign# Example usages = ” -42″print(myAtoi(s)) C#: using System;public class Solution { public int MyAtoi(string str) { long result = 0; int i = 0, sign = 1; // Skip leading whitespaces while (i < str.Length && str[i] == ‘ ‘) i++; // Handle optional sign if (i < str.Length && (str[i] == ‘+’ || str[i] == ‘-‘)) { sign = (str[i] == ‘-‘) ? -1 : 1; i++; } // Convert digits to integer while (i < str.Length && Char.IsDigit(str[i])) { result = result * 10 + (str[i] – ‘0’); i++; // Check for overflow/underflow if (result * sign > Int32.MaxValue) return Int32.MaxValue; if (result * sign < Int32.MinValue) return Int32.MinValue; } return (int)(result * sign); } public static void Main() { Solution sol = new Solution(); string str = ” -42″; Console.WriteLine(sol.MyAtoi(str)); }} JavaScript: var myAtoi = function(s) { let result = 0; let sign = 1; let i = 0; // Skip leading whitespaces while (i < s.length && s[i] === ‘ ‘) { i++; } // Handle optional sign if (i < s.length && (s[i] === ‘+’ || s[i] === ‘-‘)) { sign = (s[i] === ‘-‘) ? -1 : 1; i++; } // Convert digits to integer while (i < s.length && /\d/.test(s[i])) { result = result * 10 + (s[i] – ‘0’); i++; // Check for overflow/underflow if (result * sign > 2**31 – 1) return 2**31 – 1; if (result * sign < -(2**31)) return -(2**31); } return result * sign;};// Example usagelet s = ” -42″;console.log(myAtoi(s));

Problem Statement: Given a string s and an integer numRows, arrange the characters of the string in a zigzag pattern on a number of rows and then read the string line by line. The number of rows is given by numRows. You should return the string after the zigzag conversion, which means reading the characters row by row. For example: The zigzag pattern for this example would look like: P A H NA P L S I I GY I R Approach: Algorithm: Time Complexity: Code Implementation: C: #include <stdio.h>#include <string.h>char* convert(char* s, int numRows) { if (numRows == 1 || numRows >= strlen(s)) { return s; } char* result = (char*)malloc(strlen(s) + 1); int index = 0; // Create an array to store rows char rows[numRows][strlen(s)]; for (int i = 0; i < numRows; i++) { memset(rows[i], 0, sizeof(rows[i])); } int row = 0, direction = 1; for (int i = 0; i < strlen(s); i++) { rows[row][i] = s[i]; if (row == 0) direction = 1; if (row == numRows – 1) direction = -1; row += direction; } // Concatenate all rows for (int i = 0; i < numRows; i++) { for (int j = 0; j < strlen(s); j++) { if (rows[i][j] != 0) { result[index++] = rows[i][j]; } } } result[index] = ‘\0’; return result;}int main() { char* s = “PAYPALISHIRING”; int numRows = 3; char* result = convert(s, numRows); printf(“%s\n”, result); free(result); return 0;} C++: #include <iostream>#include <vector>#include <string>using namespace std;string convert(string s, int numRows) { if (numRows == 1 || numRows >= s.size()) return s; vector<string> rows(numRows); int currentRow = 0; bool goingDown = false; for (char c : s) { rows[currentRow] += c; if (currentRow == 0 || currentRow == numRows – 1) goingDown = !goingDown; currentRow += goingDown ? 1 : -1; } string result = “”; for (string row : rows) { result += row; } return result;}int main() { string s = “PAYPALISHIRING”; int numRows = 3; cout << convert(s, numRows) << endl; return 0;} Java: public class ZigzagConversion { public static String convert(String s, int numRows) { if (numRows == 1 || numRows >= s.length()) return s; StringBuilder[] rows = new StringBuilder[numRows]; for (int i = 0; i < numRows; i++) { rows[i] = new StringBuilder(); } int currentRow = 0; boolean goingDown = false; for (char c : s.toCharArray()) { rows[currentRow].append(c); if (currentRow == 0 || currentRow == numRows – 1) goingDown = !goingDown; currentRow += goingDown ? 1 : -1; } StringBuilder result = new StringBuilder(); for (StringBuilder row : rows) { result.append(row); } return result.toString(); } public static void main(String[] args) { String s = “PAYPALISHIRING”; int numRows = 3; System.out.println(convert(s, numRows)); }} Python: def convert(s: str, numRows: int) -> str: if numRows == 1 or numRows >= len(s): return s rows = [”] * numRows currentRow, goingDown = 0, False for c in s: rows[currentRow] += c if currentRow == 0 or currentRow == numRows – 1: goingDown = not goingDown currentRow += 1 if goingDown else -1 return ”.join(rows)# Example usages = “PAYPALISHIRING”numRows = 3print(convert(s, numRows)) C#: using System;using System.Text;public class ZigzagConversion { public static string Convert(string s, int numRows) { if (numRows == 1 || numRows >= s.Length) return s; StringBuilder[] rows = new StringBuilder[numRows]; for (int i = 0; i < numRows; i++) { rows[i] = new StringBuilder(); } int currentRow = 0; bool goingDown = false; foreach (char c in s) { rows[currentRow].Append(c); if (currentRow == 0 || currentRow == numRows – 1) goingDown = !goingDown; currentRow += goingDown ? 1 : -1; } StringBuilder result = new StringBuilder(); foreach (var row in rows) { result.Append(row.ToString()); } return result.ToString(); } public static void Main() { string s = “PAYPALISHIRING”; int numRows = 3; Console.WriteLine(Convert(s, numRows)); }} JavaScript: var convert = function(s, numRows) { if (numRows === 1 || numRows >= s.length) return s; let rows = Array(numRows).fill(”); let currentRow = 0; let goingDown = false; for (let i = 0; i < s.length; i++) { rows[currentRow] += s[i]; if (currentRow === 0 || currentRow === numRows – 1) goingDown = !goingDown; currentRow += goingDown ? 1 : -1; } return rows.join(”);};// Example usagelet s = “PAYPALISHIRING”;let numRows = 3;console.log(convert(s, numRows));

What is this problem with N-Queen? Placing N chess queens on a N×N chessboard without any queens attacking one another is known as the N Queen problem. For example, the following is a solution for the 4 Queen problem. The anticipated result takes the shape of a matrix with “Qs” for the blocks containing queens and “.” for the empty spaces. For instance, the output matrix for the 4-Queen solution mentioned above looks like this. N Queen Problem using Backtracking: Starting with the leftmost column, the queens are supposed to be arranged one after the other in various columns. We look for conflicts with other queens before positioning a queen in a column. We mark a row and column in the current column as part of the solution if we locate a row for which there is no collision. We go back and return false if we are unable to locate such a row because of clashes. To put the concept into practice, take the actions listed below: C++ C Java Python C# Time Complexity: O(N!)Auxiliary Space: O(N2)

Given an array of words, print all anagrams together. Making a hash table is an easy way. Determine the hash value of every word so that the hash value of every anagram is the same. Add these hash values to the Hash Table. Lastly, use the same hash values to print those words together. The modulo sum of all characters’ ASCII values can serve as a basic hashing technique. Two non-anagram words may have the same hash value when modulo sum is used. Individual characters must be matched in order to manage this. Here’s another way to print all the anagrams at once. Consider the word array, index array, and two auxiliary arrays. Add the specified word sequence to the word array. The word array’s individual words are sorted. Lastly, sort the word array while monitoring the associated indices. All of the anagrams group together after sorting. To print the strings from the initial array of strings, use the index array. Let’s use the following word sequence as input to better understand the steps: “cat”, “dog”, “tac”, “god”, “act” 1) Create two auxiliary arrays index[] and words[]. Copy all given words to words[] and store the original indexes in index[]  index[]: 0 1 2 3 4 words[]: cat dog tac god act 2) Sort individual words in words[]. Index array doesn’t change. index[]: 0 1 2 3 4 words[]: act dgo act dgo act 3) Sort the words array. Compare individual words using strcmp() to sort index: 0 2 4 1 3 words[]: act act act dgo dgo 4) All anagrams come together. But words are changed in the words array. To print the original words, take the index from the index array and use it in the original array. We get  “cat tac act dog god” The aforesaid algorithm’s implementations are shown below. Both index and word arrays are stored in an array of structure “Word” in the program that follows. Another structure that holds an array of “Word” structures is called Dupray. C C++ Java Python C# JavaScript Results: Cat tac act dog god Time Complexity: Assume that there are N words with a maximum of M characters per word. O(NMLogM + MNLogN) is the upper bound.O(NMLogM) time is required for step 2. Word sorting requires a maximum of O(MLogM) time. Thus, it takes O(NMLogM) time to sort N-words. O(MNLogN) is taken in step 3. Sorting a word array requires NLogN comparisons. Maximum O(M) time may be required for a comparison. Thus, O(MNLogN) will be the time required to sort a word array. Complexity of space: O(N*M)

Let’s look at how to use Python to rotate an image in this tutorial. The image rotated by a predetermined number of degrees around its center using Image Rotation. An image is transformed geometrically when it rotates. Either forward transformation or inverse transformation can be used. Here, inverse transformation is used by the Image Processing Library with Pillow. Some pixel values are beyond the image boundaries, or outside the image’s dimensions, if the number of degrees specified for image rotation is not an integer multiple of 90 degrees. The generated image will not display such values. Technique 1: Making use of the Image Processing Library Pillow Python3 The rotate() method of Python Image Processing Library Pillow Takes the number of degrees as a parameter and rotates the image in Counter Clockwise Direction to the number of degrees specified. Method 2: Using Open-CV to rotate an image by an angle in Python This is common that everyone knows that Python Open-CV is a module that will handle real-time applications related to computer vision. Open-CV works with image processing library imutils which deals with images. The imutils.rotate() function iused to rotate an image by an angle in Python. Python3 Even with this Open-CV, the image rotates counterclockwise to the designated number of degrees. Are you looking to hone your Python skills or explore the world of programming? Your best resource for learning Python is our Master Python: Complete Beginner to Advanced Course. From basic programming concepts to sophisticated approaches, this course covers all you need to establish a strong foundation. With practical projects, real-world examples, and knowledgeable advice, you’ll develop the self-assurance to take on challenging coding tasks. This training is ideal whether you want to improve your skills or start from fresh. Learn Python, the language of the future, by enrolling today!

Write a function that, given two identically sized unsorted arrays, returns true if the arrays are permutations of one another and false otherwise. Examples We strongly advise you to try this yourself first by minimizing your browser.Sorting both arrays and comparing the sorted arrays is a straightforward solution. This solution’s temporal complexity is O(nLogn). Using Hashing is a better option. The application of this strategy is seen below. C++ Java Python3 C# JavaScript Output Arrays are permutations of each other

The goal at hand is to publish every conceivable permutation of an array arr[]. For instance: C++ C Java Python C# JavaScript Output

Whether you manage a company or wish to produce and profit from your own material, you need a thorough action plan in place. Let us so dissect the ten processes required to transform your interactive ideas into excellent material. Specify your intended audience first. You are creating material for whom? A blog post targeted at IT experts, for instance, will look and be written differently than one intended at single mothers with small children. It’s far wiser to choose a particular niche and write to them than try to produce material for “everyone.” Not only is it more likely to catch on, but unique material can keep you concentrated. Plan your objectives. With your material, what results you expect for? If you run a company, you could wish to boost brand awareness or raise conversion rates. If you create material, your main objective could be to build an audience so you might make money off of it. Quantifying these objectives will help you to determine if you are reaching them or not. For a single video, one could decide to aim for 100 comments and 1,000 shares. Set another once you have that one. Recognize content constraints. Kind of manufacturing infrastructure you have? Do you own lights, microphones, and video gear? Alternatively are you aiming to start with a blog heavy in text and grow fresh material over time? Start with your strongest suit and then add additional content kinds as you develop your audience. You can also get comments on the next projects your followers wish to see. produce excellent, interesting material. Content marketing is where quality always beats quantity. Yes, with a constant publishing schedule you should be able to keep momentum; two highly valuable posts will outperform ten average ones. Create material overall not only to have for the feed. Make sure every article complements your more general marketing objectives and has a purpose. Find which platforms fit best. You are not obliged to use all the several social media sites just because they exist. Your material might be excellent for Instagram and TikHub, for instance, but not so much for Facebook or Twitter. Concentrate on those platforms that will move your needle rather than waste time on ones unlikely to provide results. Create a library of contents. Multiple parts should ideally be ready for use as soon as your page launches. This helps you to create a buffer and relieves pressure to create fresh content every week or month. Start with central material that covers the pillars of your brand. From there, you can investigate other points of view and widen out into your niche. Create a content agenda. While some content producers choose once a week, others want to create a fresh piece every day. Though you want to avoid waiting too long between new releases, there is no “right” response. Otherwise, keeping pace and growing an audience becomes challenging. Starting small—that is, twice a month—then adding more material as you optimize the process is also ideal. Sort material for every social media network. Social media marketing is so successful since search engines and each platform allow you rapidly to strengthen your brand. Sometimes the Facebook page of a company may show higher than its website. Thus, make sure you maximize all of your material using transcripts, photos, caption keywords, and more. Every element is worth something, thus don’t just come up with a great title and assume that is enough. Get fresh ideas from keyword research. Coming up with ideas over the long run is one of the main difficulties of producing social media content. By knowing what your audience is looking for, keyword research helps you to craft works to answer those questions. Then you can create a content schedule in line. Focus on social developments. Trending issues change depending on several factors and can move on a dime. You can still grab on them as soon as they start even if you never know what will be hot at any one moment. In this situation, you will require a simplified content creation process to profit on trends before they go out. Forms of Content to Produced Effective content marketing depends on knowing exactly the kinds of materials to produce. Content can be arranged under blogs, videos, infographics, interviews, podcasts, webinars, and articles. Your content strategy’s intended use will determine whether you decide to produce several kinds of materials or concentrate on one type at a time. These are some of the best kinds of material you should use into your approach to article development. These let you interact with possible clients and offer your knowledge. Blog Entry Articles One of the most often used kinds of material on blogs is article. Usually they are shared, easy to read, and quite instructive. Make sure your blog article’s images are interesting, headlines are appealing, and pertinent keywords abound. Another excellent strategy to spread your message and network other thought leaders in your field is guest blogging on other websites. Several of the several advantages of including articles consist in: Using keywords in your writings will drive more people to your website in search engine optimization.Interesting and instructive articles can help readers to be involved.Sharing your skills and experience will help you establish your brand as an authority in your field of business.VideoOne great approach to involve and instruct your viewers is with videos. Company updates, product features, or expert field-based interviews can all be posted using them. Showcasing client success stories and proving how your solutions might address issues is another wonderful approach to use videos. Make sure your movies feature interesting images, crisp audio, and a gripping story. Furthermore keep in mind optimizing meta descriptions, tags, and titles to enable the videos to be quickly searchable and improve the ranking of your website. InfographicsMost businesses agree that visiting websites should last 52 seconds. People search websites for less than a minute, thus grabbing the users’ interest right away is crucial. For this