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Find the 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
#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
Updated on: 19-Dec-2019
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