Mastering Binary Search in Java: A Comprehensive Guide with Examples

If you've ever worked with sorted arrays, you’ve probably encountered binary search—a highly efficient way to find elements in a sorted list. Today, we’ll break down how binary search works in Java, using a simple example. By the end of this post, you’ll not only understand the logic behind binary search but also how it can be applied in real-world scenarios to solve problems efficiently.

Let’s dive in!

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What is Binary Search?

Binary search is an algorithm used to search for a specific element in a sorted array. It works by repeatedly dividing the search interval in half. If the value of the target is less than the value in the middle of the interval, the search continues in the lower half. Otherwise, it continues in the upper half. This process continues until the target value is found or the search interval is empty.

The key advantages of binary search are its efficiency and speed, which makes it much faster than linear search, especially for large datasets.

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The Problem: Finding an Element in a Sorted Array

Suppose we have a sorted array of integers:

javaCopy codeint arr[] = {1, 2, 3, 4, 5, 6, 7};

Our task is to find the index of the target value 7 in the array. We’ll use binary search to achieve this.

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Code Walkthrough

Here’s a Java implementation of the binary search algorithm:

javaCopy codepublic class BinarySearchExample {
    public static void main(String[] args) {
        int arr[] = {1, 2, 3, 4, 5, 6, 7};
        int target = 7;
        int start = 0;
        int end = arr.length - 1;

        while (start <= end) {
            int mid = start + (end - start) / 2;

            if (arr[mid] > target) {
                end = mid - 1;
            } else if (arr[mid] < target) {
                start = mid + 1;
            } else {
                System.out.println("Element found at index: " + mid);
                return;
            }
        }

        System.out.println("Element not found");
    }
}

Let’s break down how this code works step by step.

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Step-by-Step Execution

We’ll start by initializing the array and the target value we’re searching for:

javaCopy codeint arr[] = {1, 2, 3, 4, 5, 6, 7};
int target = 7;
int start = 0;
int end = arr.length - 1;  // end = 6

We then enter a while loop that continues as long as start <= end. In each iteration, we calculate the middle index (mid) of the current sub-array being considered.

Iteration 1:

• Mid Calculation:

    javaCopy codemid = start + (end - start) / 2
    mid = 0 + (6 - 0) / 2
    mid = 3
  

  The value at arr[3] is 4.

• Compare arr[mid] with the target:

  • arr[mid] = 4 is less than target = 7.

  • Since arr[mid] < target, we adjust the start index to mid + 1:

    javaCopy codestart = mid + 1
    start = 3 + 1
    start = 4

it proceeds to the else if block arr[mid] < target

Iteration 2:

• Mid Calculation:

    javaCopy codemid = start + (end - start) / 2
    mid = 4 + (6 - 4) / 2
    mid = 5
  

  The value at arr[5] is 6.

• Compare arr[mid] with the target:

  • arr[mid] = 6 is still less than target = 7.

  • Since arr[mid] < target, we adjust the start index to mid + 1:

    javaCopy codestart = mid + 1
    start = 5 + 1
    start = 6

it proceeds to the else if block arr[mid] < target again.

Iteration 3:

• Mid Calculation:

    javaCopy codemid = start + (end - start) / 2
    mid = 6 + (6 - 6) / 2
    mid = 6
  

  The value at arr[6] is 7.

• Compare arr[mid] with the target:

  • arr[mid] = 7 is equal to target = 7.

  • Since arr[mid] == target, we print the index of the target:

    javaCopy codeSystem.out.println("Element found at index: " + mid);  // Output: 6

At this point, the search is complete, and we have found the target at index 6.

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Key Takeaways from Each Step

1. Iteration 1:

   • mid = 3 → arr[mid] = 4 (less than 7).

   • Adjust start = 4.

2. Iteration 2:

   • mid = 5 → arr[mid] = 6 (less than 7).

   • Adjust start = 6.

3. Iteration 3:

   • mid = 6 → arr[mid] = 7 (equal to 7).

   • Target found at index 6.

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Why Binary Search is Efficient

One of the most significant advantages of binary search is its time complexity. Unlike linear search, which has a time complexity of O(n) (where n is the number of elements in the array), binary search operates in O(log n) time. This means that with each step, the search space is reduced by half, making it exponentially faster as the size of the array increases.

For example:

• With a list of 1,000 elements, binary search would only need 10 steps to find the target (since log⁡2(1000)≈10\log_2(1000) \approx 10log2​(1000)≈10).

• For a list of 1,000,000 elements, binary search would need just 20 steps.

This efficiency makes binary search especially valuable for large datasets.

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Conclusion

Binary search is a powerful and efficient algorithm for finding elements in sorted arrays. By repeatedly halving the search space, it ensures that the search process remains fast, even for large datasets. In this post, we walked through a simple example in Java to help you understand how binary search works step by step. With its time complexity of O(log n), binary search is one of the most important algorithms to master for any programmer.

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