There are n friends in a group. When they meet, everybody shakes hands with everybody. How many hand shakes happen?
Example:
Input: n = 3 Output: 3 Explanation: Considering A, B, and C are the 3 friends. The hand shakes are: (A, B), (A, C) and (B, C)
The core challenge of this problem is to determine the number of unique handshakes that occur when each person in a group of n friends shakes hands with every other person exactly once. This problem is significant in combinatorics and has applications in network theory, social networking, and more.
Potential pitfalls include misunderstanding that each handshake is unique and not double-counting handshakes.
To solve this problem, we need to think about the nature of handshakes. Each person shakes hands with every other person exactly once. This can be visualized as a complete graph where each node represents a person and each edge represents a handshake.
A naive solution would involve iterating through each pair of friends and counting the handshakes. However, this approach is not optimal as it involves nested loops, leading to a time complexity of O(n^2).
We can derive a formula to calculate the number of handshakes directly. For n friends, each person shakes hands with n-1 others. However, this counts each handshake twice (once for each participant), so we divide by 2. The formula is:
Number of handshakes = (n * (n - 1)) / 2
This formula is derived from the combination formula C(n, 2), which counts the number of ways to choose 2 items from n items without regard to order.
Here is a step-by-step breakdown of the algorithm:
/**
* Function to calculate the number of handshakes
* @param {number} n - Number of friends
* @return {number} - Number of handshakes
*/
function countHandshakes(n) {
// Calculate the number of handshakes using the formula
return (n * (n - 1)) / 2;
}
// Example usage:
const n = 3;
console.log(countHandshakes(n)); // Output: 3
The time complexity of this solution is O(1) because it involves a constant number of arithmetic operations regardless of the input size. The space complexity is also O(1) as it uses a fixed amount of space.
Potential edge cases include:
These cases are handled correctly by the formula as (0 * -1) / 2 and (1 * 0) / 2 both yield 0.
To test the solution comprehensively, consider the following test cases:
console.log(countHandshakes(0)); // Output: 0 console.log(countHandshakes(1)); // Output: 0 console.log(countHandshakes(2)); // Output: 1 console.log(countHandshakes(3)); // Output: 3 console.log(countHandshakes(4)); // Output: 6 console.log(countHandshakes(10)); // Output: 45
These test cases cover simple, edge, and larger cases to ensure the function works correctly.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to solve the problem of counting handshakes among n friends using a combinatorial approach. We derived an optimized formula, implemented it in JavaScript, and analyzed its complexity. Understanding such problems enhances your problem-solving skills and prepares you for more complex challenges.
For further reading and practice, consider the following resources: