# Count smaller elements in sorted array in C++

• Difficulty Level : Easy
• Last Updated : 30 Mar, 2018

Given a sorted array and a number x, count smaller elements than x in the given array.

Examples:

```Input : arr[] = {10, 20, 30, 40, 50}
x = 45
Output : 4
There are 4 elements smaller than 45.

Input : arr[] = {10, 20, 30, 40, 50}
x = 40
Output : 3
There are 3 elements smaller than 40.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can use upper_bound() in C++ to quickly find the result. It returns iterator (or pointer) to first element which is greater than given number. If all elements smaller, then it returns size of array. If all elements are greater than it returns 0.

 `// CPP program to count smaller elements``// in an array.``#include ``using` `namespace` `std;`` ` `int` `countSmaller(``int` `arr[], ``int` `n, ``int` `x)``{``   ``return` `upper_bound(arr, arr+n, x) - arr;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = { 10, 20, 30, 40, 50 };``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``cout << countSmaller(arr, n, 45) << endl; ``    ``cout << countSmaller(arr, n, 55) << endl;``    ``cout << countSmaller(arr, n, 4) << endl;``    ``return` `0;``}`
Output:
```4
5
0
```

Time Complexity : O(Log n)

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