Redistribute Characters to Make All Strings Equal | 3 Ways | C++ | JAVA | Leetcode 1897 скачать в хорошем качестве

Redistribute Characters to Make All Strings Equal | 3 Ways | C++ | JAVA | Leetcode 1897

Whatsapp Community Link : https://www.whatsapp.com/channel/0029... This is the 16th Video of our Playlist "Leetcode Easy". In this video we will try to solve an easy but good problem - Minimum Difficulty of a Job Schedule (Leetcode 1897). We will solve it in 3 ways. I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY. We will do live coding after explanation and see if we are able to pass all the test cases. Also, please note that my Github solution link below contains both C++ as well as JAVA code. Problem Name : Redistribute Characters to Make All Strings Equal Company Tags : Will update soon My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Intervie... Leetcode Link : https://leetcode.com/problems/redistr... My DP Concepts Playlist :    • Roadmap for DP | How to Start DP ? | Topic...   My Graph Concepts Playlist :    • Graph Concepts & Qns - 1 : Graph will no m...   My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Intervie... Subscribe to my channel :    / @codestorywithmik   Instagram :   / codestorywithmik   Facebook :   / 100090524295846   Twitter :   / cswithmik   Approach-1 Summary : The approach uses a map to store the frequency of each character. It iterates through each word in the input vector, updating the character frequencies in the array. After counting the characters, the method checks if the count of each character is divisible by the total number of words (n). If any character count is not divisible, the method returns false, indicating that equal distribution is not possible. If all character counts are divisible, the method returns true, signifying that the words can be rearranged to have equal character frequencies. Approach-2 Summary : The approach uses an array count of size 26 (representing the English alphabet) to store the frequency of each character. It iterates through each word in the input vector, updating the character frequencies in the array. After counting the characters, the method checks if the count of each character is divisible by the total number of words (n). If any character count is not divisible, the method returns false, indicating that equal distribution is not possible. If all character counts are divisible, the method returns true, signifying that the words can be rearranged to have equal character frequencies. Approach-3 Summary : The also uses an array count of size 26 (representing the English alphabet) to store the frequency of each character. It iterates through each word in the input vector, updating the character frequencies in the array. After counting the characters, the method checks if the count of each character is divisible by the total number of words (n) using STL/library along with the lambda function. ╔═╦╗╔╦╗╔═╦═╦╦╦╦╗╔═╗ ║╚╣║║║╚╣╚╣╔╣╔╣║╚╣═╣ ╠╗║╚╝║║╠╗║╚╣║║║║║═╣ ╚═╩══╩═╩═╩═╩╝╚╩═╩═╝ ✨ Timelines✨ 00:00 - Introduction #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge#leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview#interviewtips #interviewpreparation #interview_ds_algo #hinglish #github #design #data #google #video #instagram #facebook #leetcode #computerscience #leetcodesolutions #leetcodequestionandanswers #code #learning #dsalgo #dsa