Created
February 27, 2023 19:57
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Hints when to Use Binary Search in an interview.. - When an array or matrix is sorted or partially sorted that might be a sign that you want to use binary search - When the O(N) solution is obvious or the interviewer is looking for faster than O(N). Key Takeaways: - Search Condition can be determined without comparing to the element's neighbors …
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| def binarySearch(nums, target): | |
| """ | |
| :type nums: List[int] | |
| :type target: int | |
| :rtype: int | |
| """ | |
| if len(nums) == 0: | |
| return -1 | |
| lo, hi= 0, len(nums) - 1 | |
| while lo <= hi: | |
| mid = (lo + hi) // 2 | |
| if nums[mid] == target: | |
| return mid | |
| elif nums[mid] < target: | |
| lo = mid + 1 | |
| else: | |
| hi = mid - 1 | |
| # End Condition: lo > hi | |
| return -1 |
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