703236. Product Pair
Product Pair
Summary
Given an array arr and two integers k1 and k2, find all pairs (a, b) in the array such that a * b = k1 * k2. The problem is about finding product pairs in an array based on the tags "Arrays" and "Data Structures".
Detailed Explanation
The problem can be solved by iterating through the array and checking each pair of elements to see if their product equals k1 * k2. Since we need all such pairs, we'll store them in a result set. Here's a step-by-step breakdown of the solution:
- Initialize an empty list or vector
resultto store the product pairs. - Iterate through the array using two nested loops. The outer loop iterates over each element
ain the array, and the inner loop iterates over each elementbthat comes aftera. - For each pair
(a, b), calculate their productp. Ifpequalsk1 * k2, add the pair to the result set. - After iterating through the entire array, return the result set containing all product pairs.
Here's an ASCII art diagram of the algorithm:
+---------------+
| Iterate over |
| each element a |
+---------------+
|
v
+---------------+ +---------------+
| Inner loop: | | Calculate p = a * b |
| iterate over | | If p == k1 * k2, add (a, b) |
| elements after | | to the result set |
+---------------+ +---------------+
The time complexity of this algorithm is O(n^2), where n is the size of the array. This is because we're using two nested loops to iterate through the array. The space complexity is O(n), as in the worst-case scenario, we might need to store all pairs in the result set.
Optimized Solutions
Java
import java.util.*;
public class ProductPair {
public List<int[]> findPairs(int[] arr, int k1, int k2) {
List<int[]> result = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
if (arr[i] * arr[j] == k1 * k2) {
result.add(new int[] {arr[i], arr[j]});
}
}
}
return result;
}
}