Search Insert Position 15 XP Easy

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Given a sorted array of distinct integers and a target, return the index if found. If not, return the index where it would be inserted to keep the array sorted.

nums = [1,3,5,6], target = 5  →  2  (found at index 2)
nums = [1,3,5,6], target = 2  →  1  (insert between 1 and 3)
nums = [1,3,5,6], target = 7  →  4  (insert at end)

This is binary search with a twist — when the target isn't found, left ends up at the correct insertion position. Think about why!

💡 Pro tip: Understand this problem deeply — don't just memorize the code. Try explaining the approach out loud as if teaching a friend. If you can explain it simply, you truly understand it!

📋 Instructions

Implement `search_insert(nums, target)` using binary search. Test with: - `search_insert([1,3,5,6], 5)` → 2 - `search_insert([1,3,5,6], 2)` → 1 - `search_insert([1,3,5,6], 7)` → 4 Print all three results.

Standard binary search. When the loop ends without finding target, return `left` — it's exactly where the target should be inserted.

🧪 Test Cases

Input

search_insert([1,3,5,6], 5)

Expected

2

Test case 1

Input

search_insert([1,3,5,6], 2)

Expected

1

Test case 2

Input

search_insert([1,3,5,6], 7)

Expected

4

Test case 3

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main.py

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Output

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