Count of Strictly Decreasing Subarrays in C++

Suppose we have an array A. And we have to find the total number of strictly decreasing subarrays of length > 1. So if A = [100, 3, 1, 15]. So decreasing sequences are [100, 3], [100, 3, 1], [15] So output will be 3. as three subarrays are found.

The idea is find subarray of len l and adds l(l – 1)/2 to result.

Example

 Live Demo

#include<iostream>
using namespace std;
int countSubarrays(int array[], int n) {
   int count = 0;
   int l = 1;
   for (int i = 0; i < n - 1; ++i) {
      if (array[i + 1] < array[i])
         l++;
      else {
         count += (((l - 1) * l) / 2);
         l = 1;
      }
   }
   if (l > 1)
   count += (((l - 1) * l) / 2);
   return count;
}
int main() {
   int A[] = { 100, 3, 1, 13, 8};
   int n = sizeof(A) / sizeof(A[0]);
   cout << "Number of decreasing subarrys: " << countSubarrays(A, n);
}

Output

Number of decreasing subarrys: 4