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Quickly Find Bitwise OR Of A Range In C: A Simple Guide

10 min read 11-15- 2024
Quickly Find Bitwise OR Of A Range In C: A Simple Guide

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When working with binary numbers in C programming, bitwise operations can often simplify tasks and improve performance. One such operation is the Bitwise OR, which can be particularly useful when trying to combine numbers in a given range. In this guide, we will explore how to quickly find the Bitwise OR of a range of numbers in C, highlighting key concepts and providing practical examples. 🚀

What is Bitwise OR?

Before diving into the specifics of implementing the Bitwise OR in C, let's first clarify what the Bitwise OR operation entails. The Bitwise OR operator (|) compares two binary numbers bit by bit. If either bit of the operands is 1, the resulting bit is set to 1. Otherwise, it is set to 0.

Truth Table for Bitwise OR

Here’s a brief overview of the truth table for Bitwise OR:

A B A B
0 0 0
0 1 1
1 0 1
1 1 1

From this table, we see that:

  • 0 | 0 = 0
  • 0 | 1 = 1
  • 1 | 0 = 1
  • 1 | 1 = 1

This operation can be beneficial when working with ranges, allowing you to compute a composite result that reflects all numbers within that range.

Understanding the Problem Statement

The challenge here is to efficiently find the Bitwise OR of a range of integers [L, R], where L is the starting point and R is the endpoint of the range. For example, if we need to find the Bitwise OR of the numbers from 3 to 6, we want to compute:

3 | 4 | 5 | 6

This would yield 7 since:

  • 3 in binary is 011
  • 4 in binary is 100
  • 5 in binary is 101
  • 6 in binary is 110

When we perform the Bitwise OR across these numbers, the result is 111 (which is 7 in decimal).

Efficient Calculation of Bitwise OR

Brute Force Approach

The simplest way to calculate the Bitwise OR of a range is through a brute-force method. This involves iterating through each number in the range and accumulating the result. Here’s an implementation in C:

#include 

int rangeBitwiseOR(int L, int R) {
    int result = 0;
    for (int i = L; i <= R; i++) {
        result |= i;  // Perform Bitwise OR
    }
    return result;
}

int main() {
    int L = 3, R = 6;
    printf("Bitwise OR of range [%d, %d] is: %d\n", L, R, rangeBitwiseOR(L, R));
    return 0;
}

Time Complexity

The time complexity of this brute-force approach is O(N), where N is the number of integers in the range from L to R. While this method is straightforward, it may not be efficient for large ranges.

Optimal Approach: Bit Manipulation

An optimal approach involves using properties of Bitwise OR and binary manipulation. The result of L | (L + 1) | ... | R can be computed more efficiently:

  1. Observe that the Bitwise OR accumulates the highest bits across the range.
  2. We can start with the leftmost bit and shift until we identify the range.

Here’s how you can implement this approach in C:

#include 

int rangeBitwiseOR(int L, int R) {
    int shift = 0;
    
    // Find the common prefix of L and R
    while (L < R) {
        L >>= 1;
        R >>= 1;
        shift++;
    }
    // Construct the result with the common prefix and the remaining bits set to 1
    return (L << shift) | ((1 << shift) - 1);
}

int main() {
    int L = 3, R = 6;
    printf("Bitwise OR of range [%d, %d] is: %d\n", L, R, rangeBitwiseOR(L, R));
    return 0;
}

Explanation of the Code

  • The function first determines the position of the highest bit that varies between L and R. This is done by right-shifting both L and R until they are equal.
  • It counts how many shifts it took using the shift variable.
  • Finally, it constructs the result by left-shifting L back to its original position and filling the lower bits with 1s using the formula (1 << shift) - 1.

Time Complexity

This optimal solution runs in O(log R) time complexity, as each right shift effectively halves the value, leading to a logarithmic number of operations.

Practical Examples

Let’s explore some more examples to solidify your understanding of how to find the Bitwise OR of a range:

Example 1: Range [1, 5]

  1. Brute Force Calculation:

    1 | 2 | 3 | 4 | 5 = 7

  2. Optimal Calculation:

    • Right shift until both become equal:
    • Start with 1 and 5: (1 = 0001, 5 = 0101).
    • After 2 shifts, they become equal (0).
    • The result is 0 << 2 | (1 << 2) - 1 = 7.

Example 2: Range [10, 15]

  1. Brute Force Calculation:

    10 | 11 | 12 | 13 | 14 | 15 = 15

  2. Optimal Calculation:

    • Start with 10 and 15: (10 = 1010, 15 = 1111).
    • After 1 shift, they become 5 (both become equal).
    • The result is 5 << 1 | (1 << 1) - 1 = 15.

Summary of Outputs

Range Bitwise OR
[3, 6] 7
[1, 5] 7
[10, 15] 15

Important Considerations

  • Edge Cases: Always consider edge cases where L is equal to R. The Bitwise OR will simply yield the same number.

    if (L == R) return L;

  • Performance: For very large ranges, the optimal solution will significantly reduce the computation time and resources required.

By following this guide, you should now be well-equipped to quickly find the Bitwise OR of a range in C using both brute-force and optimized techniques. Whether you're building algorithms or tackling problems, mastering bitwise operations is a valuable skill in programming. Happy coding! ✨

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