Add two sum

Author: Sanajit Jana

Easy/0001-two-sum.java

@@ -0,0 +1,55 @@
+# [Two Sum](https://leetcode.com/problems/two-sum/)
+
+## Question Description
+Given an array of integers `nums` and an integer `target`, return indices of the two numbers such that they add up to `target`. You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+---
+
+## Constraints
+- `2 <= nums.length <= 10^4`
+- `-10^9 <= nums[i] <= 10^9`
+- `-10^9 <= target <= 10^9`
+- Only one valid answer exists.
+
+---
+
+## Approach
+Use a HashMap to store visited numbers and their indices. For each number, check if `target - current` exists in the map. If it does, return the stored index and current index. Otherwise, store the current number with its index.
+
+This approach works because we need to find two numbers that sum to target, and using a hashmap allows O(1) lookups for the complement. Alternative approaches like brute force (O(n²)) would be too slow for large inputs, and sorting would require additional complexity to track original indices.
+
+---
+
+## Dry Run
+Example Input: `nums = [2,7,11,15], target = 9`
+
+Step-by-step execution:
+- i=0, nums[0]=2, complement=7, map is empty, put (2,0) in map
+- i=1, nums[1]=7, complement=2, map contains 2, return [0,1]
+
+Final Answer = `[0,1]`
+
+---
+
+## Solution
+```java
+class Solution {
+    public 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);
+        }
+        return new int[] {};
+    }
+}
+```
+
+---
+
+## Time and Space Complexity
+- **Time Complexity:** O(n) - single pass through the array
+- **Space Complexity:** O(n) - hashmap stores up to n elements in worst case