Search Insert Position
An easy Binary Search problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Binary Search
- Sheets
- 1
- Core for
- 7 roles
- Platform
- LeetCode
The problem
Given a sorted array of distinct integers and a target value, return the index where the target should be inserted to maintain sorted order. If the target is already present, return its index.
Example 1
- Input
- nums = [1,3,5,6], target = 5
- Output
- 2
- Why
- 5 is found at index 2.
Example 2
- Input
- nums = [1,3,5,6], target = 2
- Output
- 1
- Why
- 2 is not in the array but should be inserted at index 1 to maintain sorted order.
Example 3
- Input
- nums = [1,3,5,6], target = 7
- Output
- 4
- Why
- 7 should be inserted at the end, at index 4.
Constraints
- 1 <= nums.length <= 10^4
- -10^4 <= nums[i] <= 10^4
- nums contains distinct values sorted in ascending order
- -10^4 <= target <= 10^4
How to think about it
Updated 2026-09-09The problem is finding the first index whose value is at least the target. When the target is missing, the binary search termination condition itself parks the lower boundary pointer at precisely the first position where the new value can sit without violating order.
Approaches, worst first
Linear search for threshold
time O(n) · space O(1)
Step sequentially through the array and stop at the first element greater than or equal to target, or return array length at the end. Predictable and correct, but ignores the ordering during navigation.
Lower-bound binary searchWrite this one
time O(log n) · space O(1)
Divide the interval repeatedly. When the middle value is strictly smaller than target, the answer must lie strictly to the right, advancing low to mid + 1. When the search exits, low holds the exact insertion slot.
Where people lose marks · 3
- Returning mid inside the loop handles exact matches but leaves the missed case ambiguous unless low is consistently returned when the loop terminates.
- Target exceeding all elements requires inserting at index n, which is an out-of-bounds index for reading; code that indexes nums[low] after the loop crashes.
- Using high as the fallback insertion index instead of low is off by one because high finishes to the left of the insertion point when pointers cross.
The theory behind it
Binary Search — the ground this problem stands on. All Binary Search problems
What Binary Search is
Binary search is the guessing strategy used when searching a thick telephone directory or guessing a secret number between one and a hundred. Rather than inspecting names one by one from the first page, the search opens straight to the middle page. If the target precedes that middle entry, the entire back half is discarded; if it follows, the front half is eliminated. Repeating this halved split finds the item with remarkable speed.
When to reach for it
Reach for binary search when queries target sorted collections, rotated sorted arrays, or monotonic answer spaces. Strong hints include logarithmic time constraints like O(log n) or search ranges exceeding one billion where stepping one value at a time times out. It also applies when validating whether a guessed solution is possible via a monotonic boolean check, known as binary search on answer.
How the pattern works
Define the search territory with two inclusive pointers, low and high. Compute the midpoint using low plus half the difference to high, avoiding integer overflow. Formulate an exact boolean condition that divides the range into true and false halves. Decide whether the boundary condition includes the midpoint or shifts strictly past it. The loop invariant states that the sought target, if it exists, remains trapped inside the interval throughout every iteration.
What each operation costs
| Operation | Time |
|---|---|
| find target in sorted array | O(log n) |
| find boundary in monotonic range | O(log n) |
What usually goes wrong with Binary Search
- Triggering integer overflow by computing middle using low plus high divided by two instead of low plus half of high minus low in fixed-width numeric types.
- Creating an infinite loop when the interval shrinks to two elements by setting low equal to mid when mid was rounded down.
- Mismatched loop condition and bounds updates, such as pairing low less than or equal to high with non-advancing pointer assignments that never terminate.
Which roles need this problem
Binary Search is a core topic for these 7 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 17 more roles, including Full-Stack Developer, Data Engineer, Android Developer.
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