Search Insert Position
- mosesg1123
- Apr 18
- 3 min read
Updated: Apr 22
Problem Statement - Search Insert Position
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 inserted in order to maintain the sorted order.
Constraints:
Time complexity must be O(log n)
The array contains no duplicates
Input array is sorted in ascending order
1. Clarify Requirements Before Jumping Into Code
Let me confirm a few details:
Are the elements all integers? → Yes
Are duplicates allowed in the array? → No
Can the target be smaller than all elements or larger than all? → Yes
Do we care about in-place mutation or just the index? → Just the index
Must we implement the binary search manually, or can we use built-ins like bisect? → Let’s assume I need to implement it myself
2. Identify the Category of the Question
The Search Insert Position question is clearly a binary search problem. The array is sorted, and I’m looking to find a target or a boundary condition based on the target’s value. Anytime I see "O(log n)" and "sorted array", my brain immediately goes to binary search.
3. Consider a Brute-Force Approach to Understand the Problem
Let’s keep it simple at first. I could:
Iterate from left to right
If I find the target, return its index
If I find an element greater than the target, return that index
If I get to the end, return len(array)
Time complexity: O(n)Space complexity: O(1)
This works, but doesn’t meet the O(log n) constraint. Still, it’s useful for validating output.
4. Brainstorm More Solutions
The brute-force version gives us the basic logic—but we can optimize with binary search.
In a binary search, instead of looking at every element, I narrow my search window by half on each step.
Start with left = 0, right = len(nums) - 1
While left <= right:
Compute mid = (left + right) // 2
If nums[mid] == target → return mid
If nums[mid] < target → search right half
Else → search left half
But here’s the twist: if I don’t find the target, I need to return the position where it would go.
Here’s the insight: when the loop ends, left is pointing to the first position where the target could be inserted while maintaining sorted order.
So I’ll return left as the final answer, whether or not the target is found.
5. Discuss Trade-Offs Between Your Solutions
Approach | Time Complexity | Space Complexity | Pros | Cons |
Brute Force | O(n) | O(1) | Easy to write and debug | Too slow for large input |
Binary Search | O(log n) | O(1) | Meets performance constraints | Slightly trickier to write correctly |
Binary search is clearly the winner here.
6. Write Pseudocode to Structure Your Thoughts
function searchInsert(nums, target):
left = 0
right = length of nums - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
else if nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return left
7. Consider Edge Cases
Here are a few that come to mind:
Target is smaller than all elements → Insert at index 0
Target is greater than all elements → Insert at the end
Target exists in the array → Return its index
Array has only one element
Array is empty (though problem usually assumes non-empty input)
8. Write Full Code Syntax (in Python)
def searchInsert(nums, target):
left = 0
right = 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
9. Test Your Code (include several assert-style examples)
assert searchInsert([1, 3, 5, 6], 5) == 2
assert searchInsert([1, 3, 5, 6], 2) == 1
assert searchInsert([1, 3, 5, 6], 7) == 4
assert searchInsert([1, 3, 5, 6], 0) == 0
assert searchInsert([1], 0) == 0
assert searchInsert([1], 1) == 0
assert searchInsert([1], 2) == 1
All of these help verify correctness for typical and boundary scenarios.
10. Key Lessons to Remember for Future Questions
When you see "sorted" and "find position", think binary search
If you don’t find the target, remember: left ends up at the right insertion point
In binary search, off-by-one errors are common—be careful with left, right, and loop conditions
Always validate both success and failure scenarios when returning an index
Writing a brute-force solution first helps clarify logic, even if you optimize later
This problem is a perfect warm-up—short, efficient, and full of useful edge-case thinking. It’s a great reminder of how much nuance a simple binary search can contain when it’s just slightly modified.
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