Search Insert Position Leetcode Solution

In this post, we are going to solve the Search Insert Position Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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 = 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 <= 10⁴
-10⁴ <= nums[i] <= 10⁴

nums contains distinct values sorted in ascending order.
-10⁴ <= target <= 10⁴

Now, let’s see the leetcode solution of Search Insert Position Leetcode Solution.

Search Insert Position Leetcode Solution in Python

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

Search Insert Position Leetcode Solution in CPP

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) / 2;
      if (nums[m] == target)
        return m;
      if (nums[m] < target)
        l = m + 1;
      else
        r = m;
    }

    return l;
  }
};

Search Insert Position Leetcode Solution in Java

class Solution {
  public int searchInsert(int[] nums, int target) {
    int l = 0;
    int r = nums.length;

    while (l < r) {
      final int m = (l + r) / 2;
      if (nums[m] == target)
        return m;
      if (nums[m] < target)
        l = m + 1;
      else
        r = m;
    }

    return l;
  }
}

Note: This problem Search Insert Position is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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