Hello Everyone,
Given a sorted array of size n. Find a number of elements that are less than or equal to a given element.
Examples:
Input : arr[] = {1, 2, 4, 5, 8, 10} key = 9 Output : 5 Elements less than or equal to 9 are 1, 2, 4, 5, 8 therefore result will be 5. Input : arr[] = {1, 2, 2, 2, 5, 7, 9} key = 2 Output : 4 Elements less than or equal to 2 are 1, 2, 2, 2 therefore result will be 4.
Naive approach: Search whole array linearly and count elements that are less than or equal to the key.
Efficient approach: As the whole array is sorted we can use binary search to find result.
Case 1: When key is present in array, the last position of key is the result.
Case 2: When key is not present in array, we ignore left half if key is greater than mid. If key is smaller than mid, we ignore right half. We always end up with a case where key is present before middle element.
- C++
// C++ program to count smaller or equal
// elements in sorted array.
#include <bits/stdc++.h>
using namespace std;
// A binary search function. It returns
// number of elements less than of equal
// to given key
int binarySearchCount( int arr[], int n, int key)
{
int left = 0, right = n;
int mid;
while (left < right) {
mid = (right + left) >> 1;
// Check if key is present in array
if (arr[mid] == key) {
// If duplicates are present it returns
// the position of last element
while (mid + 1 < n && arr[mid + 1] == key)
mid++;
break ;
}
// If key is smaller, ignore right half
else if (arr[mid] > key)
right = mid;
// If key is greater, ignore left half
else
left = mid + 1;
}
// If key is not found
// in array then it will be
// before mid
while (mid > -1 && arr[mid] > key)
mid--;
// Return mid + 1 because of 0-based indexing
// of array
return mid + 1;
}
// Driver program to test binarySearchCount()
int main()
{
int arr[] = { 1, 2, 4, 5, 8, 10 };
int key = 11;
int n = sizeof (arr) / sizeof (arr[0]);
cout << binarySearchCount(arr, n, key);
return 0;
}
Output
6
Although this solution performs better on average, the worst-case time complexity of this solution is still O(n).
The above program can be implemented using a more simplified binary search. The idea is to check if the middle element is greater than the given element then update right index as mid – 1 but if the middle element is less than or equal to key update answer as mid + 1 and left index as mid + 1 .
Below is the implementation of the above approach:
- C++
// C++ program to count smaller or equal
// elements in sorted array
#include <bits/stdc++.h>
using namespace std;
// A binary search function to return
// the number of elements less than
// or equal to the given key
int binarySearchCount( int arr[], int n, int key)
{
int left = 0;
int right = n - 1;
int count = 0;
while (left <= right) {
int mid = (right + left) / 2;
// Check if middle element is
// less than or equal to key
if (arr[mid] <= key) {
// At least (mid + 1) elements are there
// whose values are less than
// or equal to key
count = mid + 1;
left = mid + 1;
}
// If key is smaller, ignore right half
else
right = mid - 1;
}
return count;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 4, 11, 11, 16 };
int key = 11;
int n = sizeof (arr) / sizeof (arr[0]);
cout << binarySearchCount(arr, n, key);
return 0;
}
Output
5