Introduction:
When preparing for coding interviews, optimizing your solutions is often the key to success. One such common interview problem is LeetCode's "Product of Array Except Self." In our previous article, we explored the brute-force solution with O(n²) time complexity. In this article, we will dive into the optimized approach, which solves the problem in O(n) time using Java. This solution is highly efficient and is often the preferred method during technical interviews.
Problem Overview:
You are given an array of integers, and the task is to return a new array where each element at index i is the product of all the elements of the original array except arr[i].
Example:
For An input: [1, 2, 3, 4], Output: [24, 12, 8, 6]
Optimized Approach: (O(n) Time Complexity)
In this solution, we avoid the inefficiencies of nested loops by leveraging two auxiliary arrays to store the products of elements to the left and right of each index. By combining these, we can achieve the desired result in just two passes through the array.
Code in Java:
import java.util.Arrays;
public class ProductOfArrayExceptSelf {
public static void main(String[] args) throws Exception {
int[] arr = {1, 2, 3, 4};
int[] effecientSolution = getProductOfArrayEfficiently(arr);
System.out.println("Array after product: " +Arrays.toString(effecientSolution));
}
//Solution with time complexity O(n)
private static int[] getProductOfArrayEfficiently(int[] arr) throws Exception {
if (arr.length == 0) {
throw new Exception("Empty Array!");
}
int n = arr.length;
int[] result = new int[n];
// Step 1: Fill the result array with left products
result[0] = 1; // There is no element to the left of the first element
for (int i = 1; i < n; i++) {
result[i] = result[i - 1] * arr[i - 1];
}
// Step 2: Multiply with the right products
int rightProduct = 1;
for (int i = n - 1; i >= 0; i--) {
result[i] *= rightProduct;
rightProduct *= arr[i];
}
return result;
}
}
Time Complexity Breakdown:
- Time Complexity: O(n)
- The array is traversed twice: once to calculate the left products and once to calculate the right products, making it linear time complexity.
- Space Complexity: O(1) (excluding the output array)
- No extra space is used apart from the result array.
Explanation:
In the first pass, we compute the product of all elements to the left of each index and store them in the result array. In the second pass, we multiply these left products with the products of all elements to the right. This efficient use of auxiliary variables ensures that we only traverse the array twice, optimizing both time and space usage.
Why the Optimized Solution is Important:
The O(n) solution is a prime example of how to approach array manipulation problems efficiently. For coding interviews, this is the preferred solution because it handles larger datasets better than the brute-force approach, making it a key problem for DSA and LeetCode preparation in Java.
When to Use the Optimized Approach:
The optimized approach is ideal for coding interviews and real-world scenarios where performance matters. It's especially useful when working with large datasets, where time complexity directly impacts the efficiency of your solution.
Conclusion:
Mastering the optimized O(n) solution for the Product of Array Except Self problem is a valuable asset for anyone preparing for coding interviews or practicing DSA. While the brute-force solution helps in understanding the problem, this efficient approach ensures that your solution scales well, making it a must-know for serious Java developers.
Tags: #LeetCode #Java #DSA #OptimizedSolution #ArrayManipulation #CodingChallenges #Programming #InterviewPrep #O(n) #TechInterview
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