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Table of Contents
Basic Structure of Binary Search
Handling Duplicate Values
Common Mistakes and How to Avoid Them
Home Java javaTutorial How to implement a binary search in Java?

How to implement a binary search in Java?

Jul 15, 2025 am 03:08 AM

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

How to implement a binary search in Java?

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).

How to implement a binary search in Java?

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:

How to implement a binary search in Java?
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 or right), always move past the current mid.

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.

How to implement a binary search in Java?

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, set left = 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 if left and right 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 or right = 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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