Add a Solution for Search Insert Position (LeetCode Problem 35)

dsa-solutions/lc-solutions/0000-0099/0035-search-insert-position.md

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+---
+id: search-insert-position
+title: Search Insert Position
+difficulty: Easy
+sidebar_label: 0038-SearchInsertPosition
+tags:
+  - Array
+  - Binary Search
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [Merge Two Sorted Lists](https://leetcode.com/problems/search-insert-position/) | [Merge Two Sorted Lists Solution on LeetCode](https://leetcode.com/problems/search-insert-position/solutions/) |  [VijayShankerSharma](https://leetcode.com/u/darkknight648/) |
+
+## Problem Description
+
+Given a sorted array `nums` of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You must write an algorithm with O(log n) runtime complexity.
+
+### Examples
+
+#### Example 1:
+
+- **Input:** 
+  - `nums = [1,3,5,6]`
+  - `target = 5`
+- **Output:** `2`
+
+#### Example 2:
+
+- **Input:** 
+  - `nums = [1,3,5,6]`
+  - `target = 2`
+- **Output:** `1`
+
+#### Example 3:
+
+- **Input:** 
+  - `nums = [1,3,5,6]`
+  - `target = 7`
+- **Output:** `4`
+
+### Constraints
+
+- `1 <= `nums.length` <= 10^4`
+- `-10^4 <= `nums[i]` <= 10^4`
+- `nums` contains distinct values sorted in ascending order.
+- `-10^4 <= `target` <= 10^4`
+
+### Approach
+
+To solve the problem with O(log n) runtime complexity, we can use binary search to find the insertion position of the target value.
+
+1. **Binary Search:**
+   - Start with low = 0 and high = length of `nums` - 1.
+   - While `low <= high`, compute mid as (low + high) / 2.
+   - If the target value is equal to the value at index mid, return mid.
+   - If the target value is less than the value at index mid, set high = mid - 1.
+   - If the target value is greater than the value at index mid, set low = mid + 1.
+   - After the loop, if the target value is not found, return low (or high + 1).
+
+### Solution Code
+
+#### Python
+
+```
+class Solution(object):
+    def searchInsert(self, nums, target):
+        left, right = 0, len(nums) - 1
+
+        while left <= right:
+            mid = left + (right - left) // 2
+            if nums[mid] == target:
+                return mid
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        return left
+
+```
+
+#### Java
+
+```
+class Solution {
+    public int searchInsert(int[] nums, int target) {
+        int low = 0, high = nums.length - 1;
+
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (nums[mid] == target) {
+                return mid;
+            } else if (nums[mid] < target) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+
+        return low;
+    }
+}
+```
+
+#### C++
+
+```
+class Solution {
+public:
+    int searchInsert(vector<int>& nums, int target) {
+        int low = 0, high = nums.size() - 1;
+
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (nums[mid] == target) {
+                return mid;
+            } else if (nums[mid] < target) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+
+        return low;
+    }
+};
+```
+
+### Conclusion
+
+The above solution efficiently finds the insertion position of a target value in a sorted array using binary search. It achieves a runtime complexity of O(log n), providing an optimal solution to the problem.