Binary Search in Java: Explanation & Practice

Problem summary

Given a sorted array and target value, return the index of target, or -1 if not found. Implement classic binary search.

Starter code

public class Main {
    public static void main(String[] args) {
        int[] nums = {-1, 0, 3, 5, 9, 12};
        int target = 9;
        
        // left = 0, right = n-1
        // mid = left + (right - left) / 2
        // Adjust bounds based on comparison
        // Print: Index: 4
    }
}

Expected output and test cases

• 9 found at index 4

  Index: 4

• 2 not found

  Index: -1

• First element

  Index: 0

Hints

1. Use left + (right - left) / 2 to avoid overflow
2. If nums[mid] == target, return mid
3. If nums[mid] < target, search right half: left = mid + 1
4. If nums[mid] > target, search left half: right = mid - 1

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• Practice Task Scheduler in Java
• Practice Search in Rotated Sorted Array in Java

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