Count Total Set Bits in First Natural Numbers in Java5 May 2025 | 4 min read Problem StatementThe task is to calculate the total number of bit sets (or 1 second) in the binary representation of the first natural number (n). Binary representations are the cornerstone of digital systems. And understanding bit sets is central to applications such as failure detection, cryptography, and machine design. The set bits in a binary number are the digits that are 1. For example:
If (n = 5) we calculate the combination of bits for the numbers (1) to (5): ( 1 (1), 1 (10), 2 (11), 1 (100), 2 (101) ), result i ( 1 + 1 + 2 + 1 + 2 = 7 ) Approaches to Solve the Problem1. Naive ApproachThe most straightforward way is to iterate through all numbers from ( 1 ) to ( n ) and calculate the set bits for each using bitwise operations. Complexity AnalysisThe naive approach iterates over each number from (1) to (n) and counts the set bits for each number using the Integer.bitCount() method. Here's the breakdown: 1. Counting Set Bits for Each Number:
2. Total Iterations:
Thus, the time complexity is: [O(n times log(n))] Space Complexity:
2. Optimized ApproachThis involves leveraging patterns in the binary system to count set bits without iterating through every number. Binary numbers follow repetitive patterns in bit positions, and mathematical tricks can speed up the computation significantly. File Name: CountSetBits.java Output: Naive Approach Total Set Bits: 25 Optimized Approach Total Set Bits: 25 Explanation The code calculates the total number of set bits in the binary representation of the first (n) natural numbers using two methods: a naive approach and an optimized approach. The naive approach iterates through each number from (1) to (n), using Integer.bitCount() to count set bits for each, then sums them up, with a time complexity of (O(n times log(n))). The optimized approach, with (O(log(n))) time complexity, uses bitwise operations and recursion to calculate set bits efficiently by leveraging patterns in binary numbers. It identifies the largest power of 2 ((2^x)) less than or equal to (n), calculates the set bits in all numbers up to (2^x - 1), adds the set bits in the numbers from (2^x) to (n), and recursively handles the remaining portion ((n - 2^x)). A base case ensures the recursion terminates when (n = 0). Complexity AnalysisOptimized ApproachThe optimized approach avoids iterating overall numbers. Instead, it leverages patterns in the binary representation and recursively computes the total set bits. The breakdown is as follows: 1. Bit Patterns:
2. Recursive Calls:
Thus, the time complexity is: [O(log(n))] Space Complexity:
ConclusionThe problem of counting total set bits highlights the utility of binary arithmetic and bitwise operations in solving real-world problems efficiently. Mastery of these techniques equips programmers to handle diverse computational challenges with elegance and performance in mind. Next TopicBouncy Number in Java |
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