GFG POTD: Sort By Set Bit Count | Bit-Magic Sorting (Java) | Day 05

Welcome to today's explanation of the GFG POTD (Problem of the Day)! In this video, I break down the strategies, ideas, and core concepts you need to solve the "Sort By Set Bit Count" problem. Understanding these underlying data structures will boost your overall problem-solving skills and help you build the intuition to tackle similar questions on your own next time. If you found this video helpful, please show your gratitude by liking the video, and don't forget to subscribe to the channel for more daily coding content. Happy coding! Problem Name: Sort By Set Bit Count Difficulty: Easy Platform: GeeksforGeeks Question: Given an array arr[] of integers, sort the array (in descending order) according to count of set bits in binary representation of array elements. Note: For integers having same number of set bits in their binary representation, sort according to their position in the original array i.e., a stable sort. The Approach: 1. Iterate The Loop: I am iterating the loop and each and every time I get the element and find the number of bit using countbit() function and the position 2. Store and Sort : Store in temp array the pair .and sort after the loop iterating are finished 3. Store in arrayList : Store the answer in arrayList the answer is value of the sorted array of each element Complexity : Time Complexity: O(Nlog K), where N is the value of the input number. K is the maximum value in the array. Space Complexity: O(1), as we use only a few integer variables. #gfg #dsa #starters #coding #problemoftheday #java #bitmanipulation #bitwiseoperators #sorting #logNcomplexity #easyquestion #leetcode #competitiveprogramming #hike #flipkart #microsoft #adobeillustrator Solution : https://github.com/Arnab-Pachal1234/GFG-POTD-SOLUTION