C++ Program for Binary Search Algorithm

Hello Everyone!

In this tutorial, we will learn how to Implement the Binary Search Algorithm, in the C++ programming language.

To understand the concept of Binary Search in detail, we will recommend you to visit https://www.studytonight.com/data-structures/binary-search-algorithm, where we have explained these concepts in detail.

For better understanding, refer to the well-commented CPP code given below.

Code:

#include <iostream>
#include <vector>
using namespace std;
//Program to return the index of b in the vector a
int BS(int a[], int l, int h, int b)
{
 int res = -1;
 while (l <= h)
 {
 int m = (l + h) / 2;
 if (a[m] == b)
 {
 return m;
 }
 else if (a[m] > b)
 {
 h = m - 1;
 }
 else
 {
 l = m + 1;
 }
 }
 return -1;
}
int main()
{
 cout << "\n\nWelcome to Studytonight :-)\n\n\n";
 cout << " ===== Program to demonstrate the Binary Search Algorithm, in CPP ===== \n\n";
 int i, n;
 int a[] = {2, 3, 5, 8, 9, 10};
 n = sizeof(a) / sizeof(a[0]);
 cout << "\n\nThe elements of the input sorted array are :\n\n";
 for (i = 0; i < n; i++)
 {
 cout << a[i] << " ";
 }
 int k1 = 8; //the element to find the index of
 //Calling BS() method to return the index of the element k1 if present, else -1.
 int f1 = BS(a, 0, n - 1, k1);
 if (f1 == -1)
 {
 cout << "\n\nThe element " << k1 << " is not present in the given array. ";
 }
 else
 {
 cout << "\n\nThe index of the element " << k1 << " (starting from 0) is: " << f1;
 }
 int k2 = 4; //the element to find the index of
 //Calling BS() method to return the index of the element k2 if present, else -1.
 int f2 = BS(a, 0, n - 1, k2);
 if (f2 == -1)
 {
 cout << "\n\nThe element " << k2 << " is not present in the given array. ";
 }
 else
 {
 cout << "\n\nThe index of the element " << k2 << " (starting from 0) is: " << f2;
 }
 cout << "\n\n\n";
 return 0;
}

Output:

C++ BS algo

We hope that this post helped you develop a better understanding of the concept of the Binary Search Algorithm and its implementation in CPP. For any query, feel free to reach out to us via the comments section down below.

Keep Learning : )