Updated explanation with necessary changes

Author: hanzel-sc

Arrays & Strings/Two Sum/Explanation.md

@@ -1,7 +1,33 @@
 Explanation (C solution):
 
-Two Sum II, a slightly modified version of Two Sum I where we get a sorted array. 
-The solution's straightforward with the utilization of two pointers - at the start and end of the array respectively, let's say l and r.
+## Two Sum II - Input Array is Sorted
+
+### Difficulty : Easy
+### Category : Arrays, Two pointer
+
+Leetcode Link : https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
+
+### Introduction 
+
+Two Sum II, a slightly modified version of Two Sum I where we get a sorted array and a target. We need to find two numbers in the array such that they add up to the target and return their indices in 1-based indexing. 
+
+### Intuition and Naive Approach/Brute Force
+
+At first glance it seems simple. We could use two loops and check all the pairs and return the pair whose sum matches the target.
+
+for(int i=0; i < n; i++) {
+    for(int j=i+1; j < n; j++){
+        if (num[i] + num[j] == target)
+        return [i+1, j+1]
+    }
+}
+
+While this would give us the result it would totally negate any advantage we have with the array being sorted and furthermore it would result in a time complexity of O(n^2)
+
+Space complexity: O(1)
+
+### Optimized Solution
+The solution's optimized with the utilization of two pointers - at the start and end of the array respectively, let's say l and r.
 We move inwards into the array until our start and end pointers don't cross over i.e left > right.
 We check if the value at array[l] + array[r] (which is our sum) == target.
 
@@ -16,4 +42,18 @@ If Sum is equal to our target:
 -We've increased the index by 1 to facilitate 1-based indexing as given in the problem.
 -Return the malloced array.
 
-Time Complexity: O(n)
+Time Complexity: O(n)
+Space complexity: O(1)
+
+### Example
+Numbers = [2, 7, 11, 15], target = 9
+
+The working :
+
+l = 0, r = 3 → 2 + 15 = 17 → sum larger than target, reduce right pointer
+
+l = 0, r = 2 → 2 + 11 = 13 → sum larger than target, reduce right pointer
+
+l = 0, r = 1 → 2 + 7 = 9 → sum equal → return [1, 2]
+
+Output: [1, 2]