# Number of subarrays with maximum values in given range

Given an array of N elements and L and R, print the number of sub-arrays such that the value of the maximum array element in that subarray is at least L and at most R.

Examples:

```Input : arr[] = {2, 0, 11, 3, 0}
L = 1, R = 10
Output : 4
Explanation: the sub-arrays {2}, {2, 0}, {3}
and {3, 0} have maximum in range 1-10.

Input : arr[] = {3, 4, 1}
L = 2, R = 4
Output : 5
Explanation: the sub-arrays are {3}, {4},
{3, 4}, {4, 1} and {3, 4, 1}```

naive approach will be to iterate for every sub-array and find the number of sub-arrays with maximum in range L-R. Time complexity of this solution is O(n*n)

An efficient approach is based on below facts :

• Any element > R is never included in any subarray.
• Any number of elements smaller than L can be included in subarray as long as there is at least one single element between L and R inclusive.
• The number of all possible subarrays of an array of size N is N * (N + 1)/2. Let countSubarrays(N) = N * (N + 1)/2

We keep track of two counts in current subarray.
1) Count of all elements smaller than or equal to R. We call it inc.
2) Count of all elements smaller than L. We call it exc.

Our answer for current subarray is countSubarrays(inc) – countSubarrays(exc). We basically remove all those subarrays which are formed by only elements smaller than L.

Below is the cpp implementation of the above approach-

`// CPP program to count subarrays whose maximum`
`// elements are in given range.`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// function to calculate N*(N+1)/2`
`long` `countSubarrys(``long` `n)`
`{`
`    ``return` `n * (n + 1) / 2;`
`}`
`// function to count the number of sub-arrays with`
`// maximum greater then L and less then R.`
`long` `countSubarrays(``int` `a[], ``int` `n, ``int` `L, ``int` `R)`
`{`
`    ``long` `res = 0;`
`    ``// exc is going to store count of elements`
`    ``// smaller than L in current valid subarray.`
`    ``// inc is going to store count of elements`
`    ``// smaller than or equal to R.`
`    ``long` `exc = 0, inc = 0;`
`    ``// traverse through all elements of the array`
`    ``for` `(``int` `i = 0; i < n; i++) {`
`        ``// If the element is greater than R,`
`        ``// add current value to result and reset`
`        ``// values of exc and inc.`
`        ``if` `(a[i] > R) {`
`            ``res += (countSubarrys(inc) - countSubarrys(exc));`
`            ``inc = 0;`
`            ``exc = 0;`
`        ``}`
`        ``// if it is less than L, then it is included`
`        ``// in the sub-arrays`
`        ``else` `if` `(a[i] < L) {`
`            ``exc++;`
`            ``inc++;`
`        ``}`
`        ``// if >= L and <= R, then count of`
`        ``// subarrays formed by previous chunk`
`        ``// of elements formed by only smaller`
`        ``// elements is reduced from result.`
`        ``else` `{`
`            ``res -= countSubarrys(exc);`
`            ``exc = 0;`
`            ``inc++;`
`        ``}`
`    ``}`
`    ``// Update result.`
`    ``res += (countSubarrys(inc) - countSubarrys(exc));`
`    ``// returns the count of sub-arrays`
`    ``return` `res;`
`}`
`// driver program`
`int` `main()`
`{`
`    ``int` `a[] = { 2, 0, 11, 3, 0 };`
`    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);`
`    ``int` `l = 1, r = 10;`
`    ``cout << countSubarrays(a, n, l, r);`
`    ``return` `0;`
`}`

Output:

```4
```

Time Complexity: O(n)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/number-subarrays-maximum-value-given-range/
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