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72 lines
1.7 KiB
Plaintext
72 lines
1.7 KiB
Plaintext
---
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title: '35. Search Insert Position'
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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
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Binary Search", slug: "binary-search" }
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video: "K-RYzDZkzCI"
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issue: "https://github.com/prdlk/leetcode/issues/63"
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sidebar:
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label: 'Search Insert Position'
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badge: 'Easy'
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---
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::::warning
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You must write an algorithm with O(log n) runtime complexity.
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::::
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### Example 1:
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- Input: `nums = [1,3,5,6], target = 5`
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- Output: `2`
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### Example 2:
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- Input: `nums = [1,3,5,6], target = 2`
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- Output: `1`
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### Example 3:
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- Input: `nums = [1,3,5,6], target = 7`
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- Output: `4`
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### Constraints:
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- `1 <= nums.length <= 10^4`
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- `-10^4 <= nums[i] <= 10^4`
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- `nums` contains distinct values sorted in ascending order.
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- `-10^4 <= target <= 10^4`
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## Approach
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```mermaid
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flowchart TD
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S(["searchInsert(nums, target)"]) --> I["l = 0, r = n - 1"]
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I --> W{"l <= r?"}
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W -- yes --> M["m = (l + r) // 2"]
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M --> Q{"nums[m] vs target"}
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Q -- equal --> R(["return m"])
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Q -- "nums[m] < target" --> L["l = m + 1"]
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L --> W
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Q -- "nums[m] > target" --> H["r = m - 1"]
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H --> W
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W -- no --> E(["return (l + r) // 2 + 1 — the insert slot"])
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```
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## Solution
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```py
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class Solution:
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def searchInsert(self, nums: List[int], target: int) -> int:
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l, r = 0, len(nums) - 1
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while l <= r:
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m = (l + r) // 2
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if nums[m] == target:
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return m
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elif nums[m] < target:
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l = m + 1
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elif nums[m] > target:
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r = m - 1
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return (l + r) // 2 + 1
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```
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