Solution #35 - Sudhir - 11/11/25 - commit #1

Arrays & Strings/#35 - Search Insert Position - Easy/searchInsertPosition - Explanation.md

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+# Problem Title
+
+**Difficulty:** Easy  
+**Category:** Binary Search, Arrays  
+**Leetcode Link:** [35. Search Insert Position](https://leetcode.com/problems/search-insert-position/)
+
+---
+
+## 📝 Introduction
+
+Given a sorted array of **distinct integers** and a target value, the task is to return the **index** if the target is found. If not, return the index where it **would be inserted** in order.
+
+This is a **classic binary search** problem with a small twist — instead of just finding the element, we also handle where it would fit in if it’s not present.
+
+---
+
+## 💡 Approach & Key Insights
+
+- The array is sorted — so binary search is the best tool here (O(log n) time).  
+- We look for the **first position** where `nums[i] >= target`.
+- If the target is greater than all elements, it should be inserted at the **end** (index = `len(nums)`).
+- This ensures correct placement whether or not the target exists in the array.
+
+---
+
+## 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+- **Explanation:**
+  - Iterate through the array.
+  - Return the index where `nums[i] >= target`.
+  - If target is larger than all elements, return `len(nums)`.
+- **Time Complexity:** O(n) — linear scan through the array.
+- **Space Complexity:** O(1) — no extra memory.
+- **Example/Dry Run:**
+
+Example input: `nums = [1, 3, 5, 6], target = 2`
+
+```
+Index 0: 1 < 2 → continue
+Index 1: 3 >= 2 → return 1
+Output: 1
+```
+
+---
+
+### 2️⃣ Optimized Approach
+
+- **Explanation:**
+  - Use binary search to narrow down where the element should be.
+  - Maintain `left` and `right` pointers.
+  - If `nums[mid] == target`, return `mid`.
+  - If `nums[mid] < target`, move `left = mid + 1`.
+  - Else move `right = mid - 1`.
+  - When loop ends, `left` will represent the **insert position**.
+- **Time Complexity:** O(log n) — binary search.
+- **Space Complexity:** O(1) — constant space.
+- **Example/Dry Run:**
+
+Example input: `nums = [1, 3, 5, 6], target = 5`
+
+```
+left = 0, right = 3
+mid = 1 (value = 3) → 3 < 5 → left = 2
+mid = 2 (value = 5) → match → return 2
+```
+
+Output: 2
+
+---
+
+## 📊 Complexity Analysis
+
+| Approach      | Time Complexity | Space Complexity |
+| ------------- | --------------- | ---------------- |
+| Brute Force   | O(n)            | O(1)             |
+| Optimized     | O(log n)        | O(1)             |
+
+---
+
+## 📉 Optimization Ideas
+
+- Using **binary search built-ins** like `bisect_left` in Python can reduce boilerplate.
+- For small arrays (n < 10), linear scan may be equally efficient — but binary search is best for scalability.
+
+---
+
+## 📌 Example Walkthroughs & Dry Runs
+
+```plaintext
+Example 1:
+nums = [1,3,5,6], target = 5
+→ Found at index 2
+
+Example 2:
+nums = [1,3,5,6], target = 2
+→ Insert before 3 → index 1
+
+Example 3:
+nums = [1,3,5,6], target = 7
+→ Beyond all → insert at end → index 4
+```
+
+---
+
+## 🔗 Additional Resources
+
+- [Binary Search (GeeksforGeeks)](https://www.geeksforgeeks.org/binary-search/)
+- [Python bisect module](https://docs.python.org/3/library/bisect.html)
+- [Leetcode Binary Search Patterns](https://leetcode.com/explore/learn/card/binary-search/)
+
+---
+
+Author:   
+Date: 11/11/2025
+

Arrays & Strings/#35 - Search Insert Position - Easy/searchInsertPosition - Solution.cpp

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+class Solution {
+public:
+    int searchInsert(vector<int>& nums, int target) {
+        int left = 0, right = nums.size() - 1;
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            if (nums[mid] == target)
+                return mid;
+            else if (nums[mid] < target)
+                left = mid + 1;
+            else
+                right = mid - 1;
+        }
+        return left;
+    }
+};

Arrays & Strings/#35 - Search Insert Position - Easy/searchInsertPosition - Solution.java

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+class Solution {
+    public int searchInsert(int[] nums, int target) {
+        int left = 0, right = nums.length - 1;
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            if (nums[mid] == target)
+                return mid;
+            else if (nums[mid] < target)
+                left = mid + 1;
+            else
+                right = mid - 1;
+        }
+        return left;
+    }
+}

Arrays & Strings/#35 - Search Insert Position - Easy/searchInsertPosition - Solution.py

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+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        left, right = 0, len(nums) - 1
+
+        while left <= right:
+            mid = (left + right) // 2
+            if nums[mid] == target:
+                return mid
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+        return left