Q4: How do you solve the Two Sum problem in O(n) time?
Given an array of integers and a target sum, return the indices of two numbers that add up to target. Use a HashMap<Integer, Integer> storing (value -> index). Iterate through the array once, computing complement = target - nums[i]. If complement exists in map, return current index and stored complement index.
Optimal Two Sum Solution
import java.util.HashMap;
import java.util.Map;
public class TwoSum {
public static int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> 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);
}
throw new IllegalArgumentException("No two sum solution");
}
}💡 Maps element values to array indices to resolve complement lookups instantly.
Key Takeaways
- HashMap provides $O(1)$ lookup time for target complements.
- Achieves $O(n)$ time complexity in a single pass.
Knowledge Check
What is the time complexity of finding a Two Sum pair using a Hash Table?