Binary search in Java requires careful handling of boundaries and conditions to ensure correctness and efficiency. 1. Use left (right - left) / 2 to prevent integer overflow when calculating the midpoint. 2. Maintain the loop condition while (left
Binary search is a fundamental algorithm that efficiently finds a target value in a sorted array by repeatedly dividing the search interval in half. In Java, implementing binary search correctly means avoiding off-by-one errors, properly handling edge cases, and ensuring performance stays at O(log n).

Basic Structure of Binary Search
At its core, binary search works like this: compare the middle element of the current range with the target. If it matches, return the index. If the target is smaller, search the left half; if it’s larger, search the right half.
Here’s a basic implementation:

public static int binarySearch(int[] arr, int target) { int left = 0; int right = arr.length - 1; while (left <= right) { int mid = left (right - left) / 2; if (arr[mid] == target) { return mid; } else if (arr[mid] < target) { left = mid 1; } else { right = mid - 1; } } return -1; // Not found }
A few key points:
- Always use
left (right - left) / 2
to avoid integer overflow. - The loop condition is
while (left <= right)
— this makes sure the last single element is still checked. - When adjusting boundaries (
left
orright
), always move past the currentmid
.
Handling Duplicate Values
If your array contains duplicates and you want to find the first occurrence of the target, standard binary search won’t cut it — it stops at any match.

To find the first occurrence, modify the logic:
public static int findFirst(int[] arr, int target) { int index = -1; int left = 0; int right = arr.length - 1; while (left <= right) { int mid = left (right - left) / 2; if (arr[mid] == target) { index = mid; right = mid - 1; // keep searching left for earlier occurrence } else if (arr[mid] < target) { left = mid 1; } else { right = mid - 1; } } return index; }
This approach keeps moving left even after finding a match until it finds the earliest one.
Similarly, to find the last occurrence, you’d do the opposite:
- When
arr[mid] == target
, setleft = mid 1
and continue. - Keep updating the result until the loop ends.
Common Mistakes and How to Avoid Them
Even experienced developers can make subtle mistakes when writing binary search. Here are some common ones:
- ? Using
mid = (left right) / 2
: This can cause integer overflow ifleft
andright
are large. - ? Incorrect loop condition: Forgetting
in <code>while (left can skip the last possible match.
- ? Off-by-one errors in boundary updates: Setting
left = mid
orright = mid
without adding/subtracting 1 can create infinite loops. - ? Returning early without checking all possibilities: Especially in duplicate scenarios, not continuing the search may miss the correct index.
To prevent these issues:
- Always test with edge cases: empty arrays, single-element arrays, targets at beginning/end.
- Use print statements or a debugger to trace mid values and boundaries.
- Prefer using built-in libraries like
Arrays.binarySearch()
unless you need custom behavior.
That's basically how you implement binary search in Java — straightforward but easy to mess up if you're not careful with bounds and conditions.
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