Count distinct XOR values among pairs using numbers in range 1 to N

Given an integer N, the objective is to determine the count of distinct XOR values that can be generated from all possible pairs of numbers ranging from 1 to N, inclusive.

Example 1

Input: N = 3

Output: 4

Explanation: The following are all possible pairs using elements from 1 to N inclusive.

1^1 = 0

1^2 = 3

1^3 = 2

2^2 = 0

2^3 = 1

3^3 = 0

Therefore, there are 4 distinct possible XOR values.

Example 2

Input: N = 4

Output: 5

Explanation: The following are all possible pairs using elements from 1 to 4 inclusive.

1^1 = 0

1^2 = 3

1^3 = 2

1^4 = 5

2^2 = 0

2^3 = 1

2^4 = 6

3^3 = 0

3^4 = 7

4^4 = 0

Therefore, there are 7 distinct possible XOR values.

Example 3

Input: N = 1

Output: 1

Explanation: There is only one value 1 ^ 1 = 0. Hence, the output is 1.

Approach: Naive approach

The naive approach to count distinct XOR values among pairs of numbers in the range 1 to N in Java involves using nested loops to generate all possible pairs of numbers and calculate their XOR values.

Algorithm

Step 1: Initialize a set (xorValues) to store distinct XOR values among pairs.

Step 2: Loop through all pairs of numbers in the range 1 to N using two nested loops.

Step 3: For each pair of numbers (i, j), calculate their XOR value and add it to the xorValues set.

Step 4: Since XOR values should be counted only once, using a set ensures that duplicates are automatically eliminated.

Step 5: Continue this process for all possible pairs of numbers in the range.

Step 6: After the loops, the size of the xorValues set represents the count of distinct XOR values among pairs.

Step 7: Return this count as the final result.

Implementation

Filename: CountDistinctXORPairs.java

import java.util.HashSet;
import java.util.Set;

public class CountDistinctXORPairs {
    public static void main(String[] args) {
        int N = 3; // Replace with the desired value of N

        // Count distinct XOR values among pairs
        int count = countDistinctXORPairs(N);

        // Print the result
        System.out.println("Count of distinct XOR values among pairs: " + count);
    }

    public static int countDistinctXORPairs(int N) {
        Set<Integer> xorValues = new HashSet<>();

        //Apply  Loop over all pairs of numbers from 1 to N.
        for (int i = 1; i <= N; i++) {
            for (int j = i; j <= N; j++) {
                int xorValue = i ^ j; // Calculate XOR of the pair
                xorValues.add(xorValue);
            }
        }

        return xorValues.size();
    }
}

Output:

Count of distinct XOR values among pairs: 4

Time Complexity: The code employs a nested loop configuration to traverse through all possible pairs of integers within the inclusive range from 1 to N. The outer loop executes N times, and within each iteration of the outer loop, the inner loop operates i times, where i corresponds to the current value of the outer loop. The method executes around (N * (N+1))/2 iterations in total since the inner Loop gets smaller as i rises. Therefore, the overall complexity of the program is O(N²), where N is the input number.

Auxiliary Space: The auxiliary space complexity for the provided code to count distinct XOR values among pairs in the range 1 to N is indeed O(d), where d represents the number of distinct XOR values that can be generated for a given input N.

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