35. Search Insert Position

Problem Statement

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

Solution:

Python

class Solution:
    def searchInsert(self, nums: list[int], target: int) -> int:
        if target in nums:
            return nums.index(target)
        else:
            nums.append(target)
            nums.sort()
            return nums.index(target)

        # Using binary search
        # start = 0
        # end = len(nums) - 1
        # while start <= end:
        #     mid = (start + end) // 2
        #     if nums[mid] == target:
        #         return mid
        #     elif nums[mid] < target:
        #         start = mid + 1
        #     else:
        #         end = mid - 1
        # return start

Rust

impl Solution {
    pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
        for (idx, val) in nums.iter().enumerate() {
            if target <= *val {
                return idx as i32;
            }
        }

        nums.len() as i32
    }
}

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