Java Coding Test Course, Utilizing Time Complexity

Introduction

Coding tests are a process that modern programmers inevitably go through. Among them, understanding and utilizing time complexity is very important. In this article, we will explain in detail how to solve algorithm problems using Java, along with how to analyze time complexity.

Problem Description

Given an integer array nums and an integer target, the problem is to find a pair of indices such that the two numbers in the array sum to target. If there is none, return -1, -1.

Problem Requirements

• 1 ≤ nums.length ≤ 10⁴
• -10⁹ ≤ nums[i] ≤ 10⁹
• -10⁹ ≤ target ≤ 10⁹
• Assume that there is only one solution.

Example

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9, so return [0, 1].

Solution Approach

This problem can be solved in two ways. The first method is using a double loop, and the second is using a hashmap. We will analyze the time complexity of each method and implement the hashmap method first.

Method 1: Double Loop

This method checks all pairs in the array to see if their sum equals target. The code implementation is as follows:

public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                return new int[]{i, j};
            }
        }
    }
    return new int[]{-1, -1};
}

Time Complexity Analysis

In the double loop, each element is checked against all other elements. Therefore, the time complexity is O(n^2). This can be inefficient in the worst case.

Method 2: Using Hashmap

To reduce time complexity, we can use a hashmap to solve this problem. This method stores the indices of each element and checks if the value of target - nums[i] exists as a key, returning the index if it does.

import java.util.HashMap;

public int[] twoSum(int[] nums, int target) {
    HashMap map = new HashMap<>();
    
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{-1, -1};
}

Time Complexity Analysis

The time complexity of this method is O(n). Since each element is checked only once, it is efficient. The insertion and retrieval from the hashmap are performed in constant time, making this approach superior.

Complexity Analysis and Conclusion

This problem highlighted the importance of time complexity. While the double loop approach is simple, it can lead to performance degradation, so it is necessary to use optimized data structures like hashmaps in practice.

Additional Tips

• When you first encounter a problem, consider various approaches. Transitioning from a double loop to a hashmap can significantly reduce time complexity.
• Always review whether the given data structure or algorithm is suitable for solving the problem.
• When calculating time complexity, always consider the worst case. It’s also beneficial to consider the best and average cases.

Conclusion

In this article, we emphasized the importance of time complexity through solving coding problems using Java. We saw how crucial it is to choose the optimal approach in solving algorithm problems. In the next session, we will explore more tips and tricks through another algorithm problem.