Binary Search in C++

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Binary Search in C++

Solution 1

Using recursion


<predata-previewers=””>#include<iostream>
using namespace std;
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not
// present in array
return -1;
}
int main(void)
{
int arr[] = { 2, 3, 4, 10, 40 };
int x = 10;
int n = sizeof(arr) / sizeof(arr[0]);
int result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}

Solution 2

Simple Solution


<predata-previewers=””>#include<iostream>
using namespace std;
int binarySearch(int arr[], int num, int n)
{
int min = 0;
int max = n-1;
while (min <= max)
{
int mid = (max + (max - min)) / 2;
if (arr[mid] == num)
return mid;
if (arr[mid] > num)
{
max = mid - 1;
}
else
{
min = mid + 1;
}
}
return -1;
}
int main(void)
{
int arr[] = { 1, 2, 3, 4, 10, 40 };
int x = 10;
int n = sizeof(arr) / sizeof(arr[0]);
int result = binarySearch(arr, 10, n);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}

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