Lower Bound and Upper Bound
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
- GeeksforGeeks
The problem
Given a sorted array and a target value, find the lower bound — the index of the first element that is greater than or equal to the target. If all elements are less than the target, return the array length.
Example 1
- Input
- arr = [1,2,3,4,5], target = 3
- Output
- 2
- Why
- The first element >= 3 is at index 2 (value 3).
Example 2
- Input
- arr = [1,2,3,4,5], target = 6
- Output
- 5
- Why
- All elements are less than 6, so return the array length 5.
Example 3
- Input
- arr = [1,2,2,3,3,5,6], target = 2
- Output
- 1
- Why
- The first element >= 2 is at index 1.
Constraints
- 1 <= arr.length <= 10^5
- -10^9 <= arr[i], target <= 10^9
- arr is sorted in non-decreasing order
How to think about it
Updated 2026-09-09The condition arr[i] >= target partitions the sorted array into false on the left and true on the right. Finding the lower bound is locating the very first index where the predicate becomes true, discarding half the candidates on each step.
Approaches, worst first
Linear threshold scan
time O(n) · space O(1)
Iterate from index 0 until finding the first item satisfying arr[i] >= target. Correct but fails to leverage sorted order, scanning up to 10^5 items.
Lower-bound binary searchWrite this one
time O(log n) · space O(1)
Maintain search interval [low, high] initialized to [0, n - 1] and keep ans = n. Whenever arr[mid] >= target, record mid in ans and contract search left to mid - 1; otherwise shift right to mid + 1.
Where people lose marks · 3
- Returning -1 when target exceeds all elements instead of the specified default value of array length n.
- Using arr[mid] > target instead of arr[mid] >= target locates the upper bound instead of the lower bound, skipping exact matches.
- Setting high = mid with loop condition low <= high causes an infinite loop when low equals high.
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.
Track this in your role's order
Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.
Start freeMore Binary Search problems
Problem set and role mapping as of .