Question

How to count the number of strictly increasing subarrays in an array with C

Answer 1

#include <stdio.h>

int countStrictlyIncreasingSubarrays(int arr[], int size) {
    int count = 0;

    for (int i = 0; i < size; i++) {
        for (int j = i + 1; j < size; j++) {
            if (arr[j - 1] >= arr[j]) {
                break;
            }
            count++;
            for (int k = i; k < j + 1; k++) { // for print only 
                printf("%d ", arr[k]);        // for print only 
            }                                 // for print only 
            printf("\n");                     // for print only 
        }
    }
    return count;
}

int main()
{
    int arr[] = { 1, 3, 4, 0, 7, 5, 9 };
    // {1, 3}, {1, 3, 4}, {3, 4}, {0, 7}, {5, 9}

    int size = sizeof(arr) / sizeof(arr[0]);

    printf("%d = Count of strictly increasing subarrays", countStrictlyIncreasingSubarrays(arr, size));

}




/*
run:

1 3
1 3 4
3 4
0 7
5 9
5 = Count of strictly increasing subarrays

*/

 

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