"dichotomic search c++" Code Answer's
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binary search in c++
Helpful_Heroin
on Jul 20, 2020
#include<iostream>
using namespace std;
int binarySearch(int arr[], int p, int r, int num) {
if (p <= r) {
int mid = (p + r)/2;
if (arr[mid] == num)
return mid ;
if (arr[mid] > num)
return binarySearch(arr, p, mid-1, num);
if (arr[mid] < num)
return binarySearch(arr, mid+1, r, num);
}
return -1;
}
int main(void) {
int arr[] = {1, 3, 7, 15, 18, 20, 25, 33, 36, 40};
int n = sizeof(arr)/ sizeof(arr[0]);
int num = 33;
int index = binarySearch (arr, 0, n-1, num);
if(index == -1)
cout<< num <<" is not present in the array";
else
cout<< num <<" is present at index "<< index <<" in the array";
return 0;
}dichotomic search c++
Phil the ice cream man
on Dec 09, 2020
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;
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