feat(problems): add medium array problem solutions for 3Sum, Two Sum II, and Subarray Sum Equals K

work/Medium/Array/167.two-sum-ii-input-array-is-sorted.js

@@ -0,0 +1,80 @@
+/*
+ * 167. Two Sum II - Input Array Is Sorted
+ * Difficulty: Medium
+ * https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given a 1-indexed array of integers numbers that is already sorted in
+ * non-decreasing order, find two numbers such that they add up to a
+ * specific target number. Let these two numbers be numbers[index1] and
+ * numbers[index2] where 1 <= index1 < index2 <= numbers.length.
+ * 
+ * Return the indices of the two numbers index1 and index2, each
+ * incremented by one, as an integer array [index1, index2] of length 2.
+ * 
+ * The tests are generated such that there is exactly one solution. You
+ * may not use the same element twice.
+ * 
+ * Your solution must use only constant extra space.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: numbers = [2,7,11,15], target = 9
+ * Output: [1,2]
+ * Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 =
+ * 2. We return [1, 2].
+ * 
+ * Example 2:
+ * 
+ * Input: numbers = [2,3,4], target = 6
+ * Output: [1,3]
+ * Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 =
+ * 3. We return [1, 3].
+ * 
+ * Example 3:
+ * 
+ * Input: numbers = [-1,0], target = -1
+ * Output: [1,2]
+ * Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2
+ * = 2. We return [1, 2].
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 2 <= numbers.length <= 3 * 10^4
+ * 
+ * 	• -1000 <= numbers[i] <= 1000
+ * 
+ * 	• numbers is sorted in non-decreasing order.
+ * 
+ * 	• -1000 <= target <= 1000
+ * 
+ * 	• The tests are generated such that there is exactly one solution.
+*/
+
+/**
+ * @param {number[]} numbers
+ * @param {number} target
+ * @return {number[]}
+ */
+var twoSum = function(numbers, target) {
+    let i = 0, j = numbers.length - 1;
+
+    while (i < j) {
+        const curr = numbers[i] + numbers[j];
+        if (curr === target){
+            return [i + 1, j + 1];
+        }else{
+            if (curr < target){
+                i++;
+            }else{
+                j--;
+            }
+        }
+    }
+    return [];
+};