# How to calculate “mid” or Middle Element Index in Binary Search?

The most common method to calculate mid or middle element index in Binary Search Algorithm is to find the middle of the highest index and lowest index of the searchable space, using the formula mid = low + \frac{(high – low)}{2}

Finding the middle index “mid” in Binary Search Algorithm

## Is this method to find mid always correct in Binary Search?Â

Consider the following implementation of the Binary Search function:

## C++14

 #include  // Binary search functionint binarySearch(int arr[], int low, int high, int x) {    // Continue searching while the low index is less than or equal to high index    while (low <= high) {        // Find the middle index        int mid = (low + high) / 2;         // If the element is present at the middle        if (arr[mid] == x)            return mid;        // If x is greater, ignore left half        else if (arr[mid] < x)            low = mid + 1;        // If x is smaller, ignore right half        else            high = mid - 1;    }    // If the element is not present    return -1;} int main() {    // Sorted array    int arr[] = {2, 3, 4, 10, 40};    // Element to be searched    int x = 10;     // Perform binary search    int result = binarySearch(arr, 0, sizeof(arr) / sizeof(arr[0]) - 1, x);     // Display the result    if (result != -1)        std::cout << "Element " << x << " is present at index " << result << std::endl;    else        std::cout << "Element " << x << " is not present in the array" << std::endl;     return 0;}

## C

 // A iterative binary search function. It returns location// of x in given array arr[l..r] if present, otherwise -1 int binarySearch(int arr[], int low, int high, int x){    while (low <= high) {         // Find index of middle element        int mid = (low + high) / 2;         // Check if x is present at mid        if (arr[mid] == x)            return mid;         // If x greater, ignore left half        if (arr[mid] <= x)            low = mid + 1;         // If x is smaller, ignore right half        else            high = mid - 1;    }     // If we reach here, then element was not present    return -1;}

## Java

 // An iterative binary search function. It returns the location// of x in the given array arr[l..r] if present, otherwise -1.public static int binarySearch(int[] arr, int low, int high, int x) {    while (low <= high) {        // Find the index of the middle element        int mid = (low + high) / 2;         // Check if x is present at mid        if (arr[mid] == x)            return mid;         // If x is greater, ignore the left half        if (arr[mid] < x)            low = mid + 1;         // If x is smaller, ignore the right half        else            high = mid - 1;    }     // If we reach here, then the element was not present    return -1;}

## Python3

 # An iterative binary search function.# It returns the location of x in the given array arr[l..r] if present, otherwise -1.def binary_search(arr, low, high, x):    while low <= high:        # Find the index of the middle element        mid = (low + high) // 2         # Check if x is present at mid        if arr[mid] == x:            return mid         # If x is greater, ignore the left half        elif arr[mid] < x:            low = mid + 1         # If x is smaller, ignore the right half        else:            high = mid - 1     # If we reach here, then the element was not present    return -1 # Example usagearr = [2, 3, 4, 10, 40]x = 10 # Perform binary searchresult = binary_search(arr, 0, len(arr) - 1, x) # Display the resultif result != -1:    print(f"Element {x} is present at index {result}")else:    print(f"Element {x} is not present in the array")

## C#

 using System; class Program{    // Binary search function    static int BinarySearch(int[] arr, int low, int high, int x)    {        // Continue searching while the low index is less than or equal to high index        while (low <= high)        {            // Find the middle index            int mid = (low + high) / 2;             // If the element is present at the middle            if (arr[mid] == x)                return mid;            // If x is greater, ignore left half            else if (arr[mid] < x)                low = mid + 1;            // If x is smaller, ignore right half            else                high = mid - 1;        }        // If element is not present        return -1;    }     static void Main()    {        // Sorted array        int[] arr = { 2, 3, 4, 10, 40 };        // Element to be searched        int x = 10;         // Perform binary search        int result = BinarySearch(arr, 0, arr.Length - 1, x);         // Display the result        if (result != -1)            Console.WriteLine($"Element {x} is present at index {result}"); else Console.WriteLine($"Element {x} is not present in the array");    }}

## Javascript

 // An iterative binary search function. It returns the location// of x in the given array arr[l..r] if present, otherwise -1.function binarySearch(arr, low, high, x) {    while (low <= high) {         // Find the index of the middle element        let mid = Math.floor((low + high) / 2);         // Check if x is present at mid        if (arr[mid] === x)            return mid;         // If x is greater, ignore the left half        if (arr[mid] < x)            low = mid + 1;         // If x is smaller, ignore the right half        else            high = mid - 1;    }     // If we reach here, then the element was not present    return -1;}

The above code looks fine except for one subtle thing, the expressionÂ

mid = (low + high)/2.

It fails for large values of low and high. Specifically, it fails if the sum of low and high is greater than the maximum positive value of int data type (i.e., 231 – 1). The sum overflows to a negative value, and the value stays negative when divided by two. This causes an array index out of bounds with unpredictable results.Â

### What is the correct way to calculate “mid” in Binary Search Algorithm?

The following is one way:

int mid = low + ((high – low) / 2);

Probably faster, and arguably as clear is (works only in Java, refer this):

int mid = (low + high) >>> 1;Â

In C and C++ (where you don’t have the >>> operator), you can do this:

mid = ((unsigned int)low + (unsigned int)high)) >> 1Â

A similar problem appears in other similar types of divide and conquer algorithms like Merge Sort as well. The above problem occurs when values of low and high are such that their sum is greater than the permissible limit of the data type. Although, this much size of an array is not likely to appear most of the time.

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