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 possible pairs or not optimizing the solution to meet the required time complexity.
To solve this problem, we need to think about how to efficiently find the pair with the smallest absolute difference. A naive solution would involve checking all possible pairs, which would result in a time complexity of O(n*m). This is not optimal.
Instead, we can use sorting to reduce the complexity. By sorting both arrays, we can then use a two-pointer technique to find the closest pair in O(n log n) time.
1. Sort both arrays.
2. Use two pointers to traverse the arrays and find the pair with the smallest absolute difference.
1. Sort both arrays.
2. Initialize two pointers, one for each array.
3. Compare the elements pointed by the two pointers and update the minimum difference and the corresponding pair if a smaller difference is found.
4. Move the pointer that points to the smaller element to the next position.
5. Repeat steps 3 and 4 until one of the pointers reaches the end of its array.
import java.util.Arrays;
public class ArrayDiff {
public static int[] findClosestPair(int[] first, int[] second) {
// Sort both arrays
Arrays.sort(first);
Arrays.sort(second);
// Initialize pointers for both arrays
int i = 0, j = 0;
int minDiff = Integer.MAX_VALUE;
int[] result = new int[2];
// Traverse both arrays
while (i < first.length && j < second.length) {
int diff = Math.abs(first[i] - second[j]);
// Update the result if a smaller difference is found
if (diff < minDiff) {
minDiff = diff;
result[0] = first[i];
result[1] = second[j];
}
// Move the pointer that points to the smaller element
if (first[i] < second[j]) {
i++;
} else {
j++;
}
}
return result;
}
public static void main(String[] args) {
int[] first = {4, 1, 8};
int[] second = {12, 3, 17};
int[] result = findClosestPair(first, second);
System.out.println("Closest pair: [" + result[0] + ", " + result[1] + "]");
}
}
The time complexity of this approach is O(n log n) due to the sorting step. The space complexity is O(1) as we are using a constant amount of extra space.
1. Arrays of different lengths.
2. Arrays with negative numbers.
3. Arrays with duplicate numbers.
4. Arrays with only one element each.
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's essential to:
In this blog post, we discussed how to find a pair of integers from two arrays with the smallest absolute difference. We explored a naive solution and then optimized it using sorting and a two-pointer technique. Understanding and solving such problems is crucial for improving algorithmic thinking and problem-solving skills.
For further reading and practice, consider the following resources: