Added rust

Author: javadev

README.md

@@ -0,0 +1,63 @@
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+
+## 1\. Two Sum
+
+Easy
+
+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.
+
+You can return the answer in any order.
+
+**Example 1:**
+
+**Input:** nums = [2,7,11,15], target = 9
+
+**Output:** [0,1]
+
+**Output:** Because nums[0] + nums[1] == 9, we return [0, 1]. 
+
+**Example 2:**
+
+**Input:** nums = [3,2,4], target = 6
+
+**Output:** [1,2] 
+
+**Example 3:**
+
+**Input:** nums = [3,3], target = 6
+
+**Output:** [0,1] 
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+*   **Only one valid answer exists.**
+
+**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>) </code>time complexity?
+
+## Solution
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
+        let mut index_map: HashMap<i32, usize> = HashMap::new();
+        
+        for (i, &num) in numbers.iter().enumerate() {
+            let required_num = target - num;
+            if let Some(&index) = index_map.get(&required_num) {
+                return vec![index as i32, i as i32];
+            }
+            index_map.insert(num, i);
+        }
+        
+        vec![-1, -1]
+    }
+}
+```