Minimum Number of Pushes to Type Word I and II | Leetcode 3014 | Leetcode 3016 | Weekly Contest скачать в хорошем качестве

Minimum Number of Pushes to Type Word I and II | Leetcode 3014 | Leetcode 3016 | Weekly Contest

Whatsapp Community Link : https://www.whatsapp.com/channel/0029... This is the 11th Video of our Playlist "Hash Map/Set : Popular Interview Problems". In this video we will try to solve two very good map problems - Minimum Number of Pushes to Type Word I and II (Leetcode 3014 and Leetcode 3016) 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 : Minimum Number of Pushes to Type Word I and II (Leetcode 3014 and Leetcode 3016) Company Tags : will update soon My solutions on Github(C++ & JAVA) : Part-1 : https://github.com/MAZHARMIK/Intervie... Part-2 : https://github.com/MAZHARMIK/Intervie... Leetcode Link : https://leetcode.com/problems/minimum... https://leetcode.com/problems/minimum... 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   Part-1 Summary : The approach involves creating a vector mp of size 26, initialized with zeros. For each character in the input word, the corresponding index in the vector is set to 1, indicating the presence of that character. The vector is then sorted in descending order based on the frequency of characters. The variable ans is initialized to zero, and a loop iterates through the sorted vector. For each element, the contribution to the answer is calculated by multiplying the frequency of the character by the expression ((i/8) + 1), where i is the current index. This expression is used to determine the group to which the character belongs, considering that characters with higher frequencies need fewer pushes. The result is the total minimum number of pushes required to rearrange the characters in the word to form a palindrome, and it is returned by the function. Part-2 Summary : Same approach as Approach-1 above. In this case we just keep in mind the frequency of each character can be more than 1. ╔═╦╗╔╦╗╔═╦═╦╦╦╦╗╔═╗ ║╚╣║║║╚╣╚╣╔╣╔╣║╚╣═╣ ╠╗║╚╝║║╠╗║╚╣║║║║║═╣ ╚═╩══╩═╩═╩═╩╝╚╩═╩═╝ ✨ 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 #2024 #newyear