Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input will have at most one solution, and you may not use the same index twice.
In case no solution exists, return [-1, -1]
Example:
Input: nums = [2, 7, 11, 15]
, target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9
For this lesson, your algorithm should run in O(n^2) time and use O(1) extra space.
(There exist faster solutions which we will discuss in future lessons)
The core challenge of the "Two Sum" problem is to find two distinct indices in an array such that the numbers at those indices add up to a given target. This problem is significant in various applications, such as financial transactions, where you need to find pairs of transactions that sum up to a specific amount.
Potential pitfalls include assuming that there are multiple solutions or using the same index twice. It's crucial to ensure that the indices are distinct and that the solution is unique.
To solve this problem, we can start with a naive approach and then discuss more optimized solutions.
The naive solution involves using two nested loops to check every possible pair of numbers in the array. This approach is straightforward but not optimal, as it has a time complexity of O(n^2).
We can optimize the solution using a hash map to store the indices of the numbers we have seen so far. This allows us to check in constant time if the complement of the current number (i.e., target - current number) exists in the hash map.
public int[] twoSum(int[] nums, int target) {
// Iterate through each element in the array
for (int i = 0; i < nums.length - 1; i++) {
// For each element, iterate through the remaining elements
for (int j = i + 1; j < nums.length; j++) {
// Check if the sum of the two elements equals the target
if (nums[i] + nums[j] == target) {
// Return the indices of the two elements
return new int[] { i, j };
}
}
}
// If no solution is found, return [-1, -1]
return new int[] { -1, -1 };
}
import java.util.HashMap;
import java.util.Map;
public int[] twoSumOptimized(int[] nums, int target) {
// Create a hash map to store the indices of the numbers
Map<Integer, Integer> map = new HashMap<>();
// Iterate through each element in the array
for (int i = 0; i < nums.length; i++) {
// Calculate the complement of the current number
int complement = target - nums[i];
// Check if the complement exists in the hash map
if (map.containsKey(complement)) {
// Return the indices of the two elements
return new int[] { map.get(complement), i };
}
// Add the current number and its index to the hash map
map.put(nums[i], i);
}
// If no solution is found, return [-1, -1]
return new int[] { -1, -1 };
}
The naive approach has a time complexity of O(n^2) and a space complexity of O(1). The optimized approach has a time complexity of O(n) and a space complexity of O(n) due to the hash map.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it's essential to:
The "Two Sum" problem is a classic example of using brute force and then optimizing the solution using a hash map. Understanding and solving such problems is crucial for developing problem-solving skills and preparing for coding interviews.
For further reading and practice problems, consider the following resources: