Given two arrays of integers, find a pair of integers, one from each array, whose absolute difference is the closest to zero.
Example:
Input: first =[4, 1, 8], second = [12, 3, 17] Output: [4, 3] Explanation: abs(4-3) = 1 which is the closest to 0
Your algorithm should run in O(n log n) time and use O(1) extra space.
The core challenge of this problem is to find a pair of integers, one from each array, such that their absolute difference is minimized. This problem is significant in various applications such as minimizing error in measurements, finding closest points in computational geometry, and more.
Potential pitfalls include not considering all pairs or not optimizing the solution to meet the required time complexity.
To solve this problem efficiently, we can use the following approach:
Sorting the arrays helps in efficiently finding the closest pair by leveraging the sorted order to minimize the difference.
A naive solution would involve checking all possible pairs from both arrays, which would result in a time complexity of O(n * m), where n and m are the lengths of the two arrays. This is not optimal for large arrays.
The optimized solution involves sorting both arrays and then using a two-pointer technique to find the pair with the minimum absolute difference. This approach ensures a time complexity of O(n log n) due to the sorting step, and O(1) extra space.
Here is a step-by-step breakdown of the optimized algorithm:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
std::pair<int, int> findClosestPair(std::vector<int>& first, std::vector<int>& second) {
// Sort both arrays
std::sort(first.begin(), first.end());
std::sort(second.begin(), second.end());
// Initialize pointers for both arrays
int i = 0, j = 0;
int minDiff = INT_MAX;
std::pair<int, int> result;
// Traverse both arrays
while (i < first.size() && j < second.size()) {
int diff = std::abs(first[i] - second[j]);
// Update the minimum difference and result pair
if (diff < minDiff) {
minDiff = diff;
result = {first[i], second[j]};
}
// Move the pointer in the array with the smaller current element
if (first[i] < second[j]) {
i++;
} else {
j++;
}
}
return result;
}
int main() {
std::vector<int> first = {4, 1, 8};
std::vector<int> second = {12, 3, 17};
std::pair<int, int> result = findClosestPair(first, second);
std::cout << "Closest pair: [" << result.first << ", " << result.second << "]" << std::endl;
return 0;
}
The time complexity of the optimized solution is O(n log n) due to the sorting step. The space complexity is O(1) as we are using only a constant amount of extra space.
Potential edge cases include:
Each of these cases should be tested to ensure the algorithm handles them correctly.
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it is crucial to:
In this blog post, we discussed how to solve the problem of finding the pair of integers from two arrays with the closest absolute difference. We explored both naive and optimized solutions, provided a detailed algorithm, and implemented the solution in C++. Understanding and solving such problems is essential for improving algorithmic thinking and coding skills.
For further reading and practice, consider the following resources: