📖 Scenario: You are working on a system that needs to count how many bits are set to 1 in a number's binary form. This is useful in many areas like error detection, cryptography, and network masks.
🎯 Goal: Build a program that counts the number of set bits (1s) in a given integer using Brian Kernighan's algorithm.
📋 What You'll Learn
Create an integer variable with a specific value
Create a counter variable to count set bits
Use Brian Kernighan's algorithm to count set bits
Print the count of set bits
💡 Why This Matters
🌍 Real World
Counting set bits is useful in network mask calculations, cryptography, and error detection systems.
💼 Career
Understanding bit manipulation and efficient algorithms is important for software engineers working in systems programming, embedded systems, and performance-critical applications.
Progress0 / 4 steps
1
Create the integer variable
Create an integer variable called number and set it to 29.
DSA C++
Hint
Use int number = 29; to create the variable.
2
Create the counter variable
Create an integer variable called count and set it to 0 to keep track of set bits.
DSA C++
Hint
Initialize count to zero before counting.
3
Apply Brian Kernighan's algorithm
Use a while loop with the condition number != 0. Inside the loop, update number to number & (number - 1) and increment count by 1.
DSA C++
Hint
Use Brian Kernighan's trick to clear the least significant set bit in each iteration.
4
Print the count of set bits
Use std::cout to print the value of count.
DSA C++
Hint
Print the count using std::cout << count << std::endl;.
Practice
(1/5)
1. What does the Brian Kernighan algorithm do when counting set bits in an integer?
easy
A. It repeatedly clears the least significant set bit until the number becomes zero.
B. It counts the total number of digits in the integer.
C. It shifts the bits to the right one by one and counts all bits.
D. It converts the number to a string and counts '1' characters.
Solution
Step 1: Understand the algorithm's main operation
The Brian Kernighan algorithm works by clearing the least significant set bit in each iteration.
Step 2: Effect of clearing bits
Each clearing reduces the number of set bits by one until the number becomes zero.
Final Answer:
It repeatedly clears the least significant set bit until the number becomes zero. -> Option A
Quick Check:
Clearing bits one by one counts set bits [OK]
Hint: Clear least significant set bit repeatedly to count bits [OK]
Common Mistakes:
Counting all bits instead of only set bits
Using string conversion unnecessarily
Shifting bits without checking if set
2. Which of the following is the correct syntax for the main loop in Brian Kernighan's algorithm in C++?
easy
A. while (n > 0) { count += n % 2; n /= 2; }
B. for (int i = 0; i < n; i++) { count += n & 1; n >>= 1; }
C. while (n != 0) { n = n & (n - 1); count++; }
D. for (int i = 0; i < 32; i++) { count += (n >> i) & 1; }
Solution
Step 1: Identify Brian Kernighan's loop
The algorithm uses while (n != 0) and clears the least significant set bit by n = n & (n - 1).
Step 2: Confirm increment of count
Each iteration increments count by one for each cleared set bit.
Final Answer:
while (n != 0) { n = n & (n - 1); count++; } -> Option C
Quick Check:
Loop clears bits and counts [OK]
Hint: Use while with n = n & (n-1) to clear bits [OK]
Common Mistakes:
Using for loop instead of while
Counting bits by shifting instead of clearing
Using modulo instead of bitwise operations
3. What is the output of the following C++ code snippet using Brian Kernighan's algorithm?
int countSetBits(int n) {
int count = 0;
while (n) {
n = n & (n - 1);
count++;
}
return count;
}
int main() {
int x = 13;
std::cout << countSetBits(x);
return 0;
}
medium
A. 5
B. 3
C. 2
D. 4
Solution
Step 1: Convert 13 to binary
13 in binary is 1101, which has three set bits (1s).
Step 2: Apply Brian Kernighan's algorithm
Each iteration clears one set bit, so count increments 3 times before n becomes 0.
Final Answer:
3 -> Option B
Quick Check:
Number of 1s in 1101 = 3 [OK]
Hint: Count 1s in binary of number for quick answer [OK]
Common Mistakes:
Counting total bits instead of set bits
Miscounting binary 1s
Confusing bitwise AND with OR
4. Identify the error in this Brian Kernighan algorithm implementation:
int countSetBits(int n) {
int count = 0;
while (n > 0) {
n = n & (n + 1);
count++;
}
return count;
}
medium
A. Missing return statement.
B. Not initializing count to zero.
C. Using while (n > 0) instead of while (n != 0).
D. Using n + 1 instead of n - 1 in bit clearing step.
Solution
Step 1: Check bit clearing operation
The correct operation to clear the least significant set bit is n = n & (n - 1), not n + 1.
Step 2: Effect of wrong operation
Using n + 1 does not clear bits properly and causes incorrect count or infinite loop.
Final Answer:
Using n + 1 instead of n - 1 in bit clearing step. -> Option D
Quick Check:
Correct bit clearing uses n & (n - 1) [OK]
Hint: Use n & (n - 1) to clear bits, not n + 1 [OK]
Common Mistakes:
Replacing n - 1 with n + 1
Confusing loop condition syntax
Forgetting to increment count
5. Given an array of integers, how can you use Brian Kernighan's algorithm to find the total number of set bits in all elements?
Choose the best approach.
hard
A. Iterate over each element, apply Brian Kernighan's algorithm to count set bits, and sum all counts.
B. Convert all integers to strings and count '1's in each string, then sum.
C. Use a single Brian Kernighan's algorithm call on the sum of all integers.
D. Count set bits only in the first element and multiply by array size.
Solution
Step 1: Understand counting set bits per element
Brian Kernighan's algorithm counts set bits in one integer at a time.
Step 2: Apply algorithm to each array element
Iterate through the array, count set bits for each element, and sum these counts for total.
Final Answer:
Iterate over each element, apply Brian Kernighan's algorithm to count set bits, and sum all counts. -> Option A
Quick Check:
Sum of individual counts gives total set bits [OK]
Hint: Count bits per element, then sum for total [OK]
Common Mistakes:
Counting bits of sum instead of each element
Using string conversion unnecessarily
Assuming uniform bit count per element
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