Given two positive integers n and sum, count the number of different pairs of integers (x, y) such that 1 <= x, y <= n and x + y equals sum
Example:
Input: n = 7, sum = 6 Output: 3 Explanation: There are 3 valid pairs: (1, 5), (2, 4), (3, 3). Note that pairs such as (1, 5) and (5, 1) are not considered different.
Your algorithm should run in O(n^2) time and use O(1) space.
The core challenge of this problem is to find all unique pairs (x, y) such that their sum equals a given value, and both x and y are within the range from 1 to n. This problem is significant in various applications such as finding pairs in arrays that sum up to a specific value, which is a common task in data analysis and algorithm design.
Potential pitfalls include counting pairs like (1, 5) and (5, 1) as different, which is incorrect as per the problem statement.
To solve this problem, we can start with a brute force approach and then discuss its optimization:
The naive solution involves using two nested loops to iterate through all possible pairs (x, y) and checking if their sum equals the given value. This approach is straightforward but not optimal in terms of time complexity.
public class PairCount {
public static int countPairs(int n, int sum) {
int result = 0;
for (int x = 1; x <= n; x++) {
for (int y = x; y <= n; y++) {
if (x + y == sum) {
result++;
}
}
}
return result;
}
public static void main(String[] args) {
int n = 7;
int sum = 6;
System.out.println(countPairs(n, sum)); // Output: 3
}
}
In this code, we use two nested loops to iterate through all possible pairs (x, y) and check if their sum equals the given value. If it does, we increment the result counter.
Given the constraints, the naive solution is already optimal in terms of time complexity (O(n^2)). However, we can make the code cleaner and more readable by avoiding unnecessary checks.
public class PairCount {
public static int countPairs(int n, int sum) {
int result = 0;
for (int x = 1; x <= n; x++) {
int y = sum - x;
if (y >= x && y <= n) {
result++;
}
}
return result;
}
public static void main(String[] args) {
int n = 7;
int sum = 6;
System.out.println(countPairs(n, sum)); // Output: 3
}
}
In this optimized version, we reduce the number of iterations by calculating y directly and checking if it falls within the valid range. This approach still has a time complexity of O(n^2) but is more efficient in practice.
The time complexity of both the naive and optimized solutions is O(n^2) because we are using nested loops to iterate through all possible pairs. The space complexity is O(1) as we are using a constant amount of extra space.
Potential edge cases include:
To test the solution comprehensively, we should include a variety of test cases:
When approaching such problems, it's essential to start with a brute force solution to understand the problem better. Then, look for patterns or mathematical properties that can help optimize the solution. Practice solving similar problems and studying algorithms to improve problem-solving skills.
In this blog post, we discussed how to solve the problem of counting pairs of integers that sum up to a given value. We explored both naive and optimized solutions, analyzed their complexities, and discussed edge cases and testing strategies. Understanding and solving such problems is crucial for developing strong algorithmic thinking and problem-solving skills.
For further reading and practice, consider the following resources: