LeetCode Problem

Problem Description

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.

Examples

Input: [1,3,5,6], 5
Output: 2

Input: [1,3,5,6], 2
Output: 1

Input: [1,3,5,6], 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

Follow-up

Approach to Solve

Use a two-pointer approach to find the target.

Code Implementation

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

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;
    }

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;
    }

Explanation

This solution uses a two-pointer approach to find the target. Here's a detailed explanation of the algorithm:

Initialize left and right pointers to the start and end of the array, respectively.
While left is less than or equal to right:
- Calculate the middle index mid.
- If nums[mid] is equal to the target, return mid.
- If nums[mid] is less than the target, move the left pointer to mid + 1.
- Otherwise, move the right pointer to mid - 1.
If the target is not found, return left.

Complexity

Time Complexity: O(log n)
Space Complexity: O(1)

LeetSolved

Search Insert Position