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?
Consider the following implementation of the Binary Search function:
C++14
#include <iostream>// 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 -1int 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 expression
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.
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;
In C and C++ (where you don’t have the >>> operator), you can do this:
mid = ((unsigned int)low + (unsigned int)high)) >> 1
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.
