notes/OJ notes/pages/Leetcode Search-Insert-Position.md

Leetcode Search-Insert-Position

2022-07-09 10:25

Algorithms:

#algorithm #binary_search

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

---

Problem

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Examples

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1

Constraints

• 1 <= nums.length <= 104
• -104 < nums[i], target < 104
• All the integers in nums are unique.
• nums is sorted in ascending order.

Thoughts

[!summary] This is a #binary_search problem.

similar to Leetcode First-Bad-Version, this relies on the converging nature of Bi-search

if the value is found, simply return it

if not found, at the end, position l - 1 should be smaller than val.

and position l should be bigger, which is the position for the answer:

If not, return the index where it would be if it were inserted in order.

Solution

Iteration

class Solution {
public:
  int searchInsert(vector<int>& nums, int target) {
    // variant of bi-search
    int r = nums.size() - 1;
    int l = 0;
    int mid, val;

    do {
      mid = l + (r - l) / 2;
      val = nums[mid];

      if (val == target) {
        return mid;
      } else if (val < target) {
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    } while (l <= r);

    return l;
  }
};