Solution - #167 - Hans - 07/04/2025

Arrays & Strings/Two Sum/Explanation.md

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+## 🧮 Two Sum II - Input Array Is Sorted
+
+Difficulty: Easy
+Category: Arrays, Two Pointers
+Leetcode Link: https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
+
+### 📝 Introduction
+
+You're given a sorted array of integers in non-decreasing order and a target value. The task is to find the indices of the two numbers such that they add up to the target value. The return should be in 1-based indexing, and you can assume exactly one solution exists and no element is reused.
+
+### 💡 Approach & Key Insights
+
+    Key Insight: Since the array is sorted, we can use the two-pointer technique instead of brute force.
+
+    Naive approach checks all pairs — this is simple but inefficient (O(n²)).
+
+    Optimized approach uses two pointers (start and end) to shrink the search space intelligently in O(n) time.
+
+    We return a dynamically allocated array in C to store the result as C functions can’t directly return multiple values.
+
+### 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+Explanation:
+Check every pair (i, j) where i < j. If numbers[i] + numbers[j] == target, return their 1-based indices.
+
+Time Complexity: O(n²)
+Each element is paired with every other element ahead of it.
+
+Space Complexity: O(1)
+No extra space is used apart from result.
+
+Example / Dry Run:
+
+Example input: [2, 7, 11, 15], target = 9
+Steps:
+
+    2+7=9 → match found
+    Output: [1, 2]
+
+### 2️⃣ Optimized Approach
+
+Explanation:
+Use two pointers:
+
+    Start l = 0
+
+    End r = n - 1
+
+Repeat:
+
+    If numbers[l] + numbers[r] == target → return [l+1, r+1]
+
+    If sum is greater → move right pointer left
+
+    If sum is less → move left pointer right
+
+Time Complexity: O(n)
+Only one pass through the array.
+
+Space Complexity: O(1)
+Only two pointers and result array of size 2.
+
+Example / Dry Run:
+
+Input: [2, 7, 11, 15], target = 9
+Steps:
+
+    l = 0, r = 3 → 2 + 15 = 17 → too big → r--
+
+    l = 0, r = 2 → 2 + 11 = 13 → still big → r--
+
+    l = 0, r = 1 → 2 + 7 = 9 → match!
+    Output: [1, 2]
+
+### 3️⃣ Best / Final Optimized Approach (Same as above)
+
+Since the optimized two-pointer approach is already the most efficient for this problem, no further optimization is needed.
+
+### 📊 Complexity Analysis
+
+| Approach      | Time Complexity | Space Complexity |
+| ------------- | --------------- | ---------------- |
+| Brute Force   | O(n²)           | O(1)             |
+| Optimized     | O(n)            | O(1)             |
+| Best Approach | O(n)            | O(1)             |
+
+### 📉 Optimization Ideas
+
+    You can’t use hashing effectively here since the input is sorted, and space must be O(1).
+
+    No need to explore binary search as two-pointer provides linear time with constant space.
+
+### 📌 Example Walkthroughs & Dry Runs
+
+Example 1:
+
+Input: [2, 7, 11, 15], target = 9
+
+Initial pointers:
+l = 0 → 2
+r = 3 → 15
+
+Step 1: 2 + 15 = 17 → too big → r = 2
+Step 2: 2 + 11 = 13 → still big → r = 1
+Step 3: 2 + 7 = 9 → match found → return [1, 2]
+
+### 🔗 Additional Resources
+
+- [Understanding Two Pointer Technique - GeeksForGeeks](https://www.geeksforgeeks.org/two-pointers-technique/)
+
+
+Author: hanzel-sc
+Date: 07/07/2025

Arrays & Strings/Two Sum/Two Sum 2.c

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+
+int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
+    int l = 0; 
+    int r = numbersSize-1;
+
+    int* answer = (int*)malloc(2*sizeof(int)); //dynamic memory allocation
+    *returnSize = 2; //we're returning two values
+
+    while (l < r)
+    {
+        int sum = (numbers[l] + numbers[r]);
+        if (sum == target)
+        {
+            answer[0] = l+1; //facilitating 1-based indexing as required by the problem.
+            answer[1] = r+1;
+            return answer;
+        }
+        else if (sum > target)
+        {
+            r--; //point to a smaller value to reduce the sum
+        }
+        else
+        {
+            l++; //point to a larger value to increase the sum
+        }
+
+    }
+    return 0;
+
+}