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In this post, you will find the solution for the **Search Insert Position** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Search Insert Position– LeetCode Problem

Search Insert Position – LeetCode Problem

**Problem:**

Given a sorted array 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.

**Example 1:**

Input:nums = [1,3,5,6], target = 5Output:2

**Example 2:**

Input:nums = [1,3,5,6], target = 2Output:1

**Example 3:**

Input:nums = [1,3,5,6], target = 7Output: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}

Search Insert Position– LeetCode Solutions

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

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

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