Count Strictly Increasing Subarrays in C++
We are given an array containing integer elements and the task is to firstly calculate the subarray out of the given array and then check whether the elements in a subarray are in increasing order or not. If yes, then we will consider the subarray else it will be discarded.
The approach here is to stop further checking the subarray if the elements in 0th and 1th position are not in increasing order.
For Example- in C++
Input: int a[] = {1, 7, 5}
Output: Count of strictly increasing subarrays is 1
Explanation - The possible subarrays include {1,7,5}, {1,7}, {7,5} where {1,7} is the only array which is in strictly increasing order.
Input: int a[] = {1, 2, 7, 10}
Output: Count of strictly increasing subarrays is 6
Explanation - The possible subarrays include {{1, 2}, {1, 2, 7}, {1, 2, 7, 10},{2, 7}, {2, 7, 10} and {7, 10} and all are in strictly increasing order.
Approach used in the below program is as follows
- An array is declared, the elements are inputted and data is then passed to the function named countIncSubarrays(a,a.length) along with the length of the array for further processing.
- Inside the function initialize a count flag.
- Start Loop FOR i from 0 till the length of an array
- Inside the loop, start another Loop FOR j from i+1 till the length of an array
- Inside the loop, check if a[j] is greater than a[j-1] and increment the count.
- Else break the loop as the increasing order check fails.
- In the main function the result is captured from the function call and printed as the output.
Example
import java.util.*;
class testqwe {
static int MAX = 4;
static int a[] = new int[MAX];
static Scanner scan = new Scanner(System.in);
static int countIncSubarrays(int a[], int n) {
int count = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (a[j] > a[j - 1])
count++;
else
break;
}
}
return count;
}
public static void main(String[] args) {
for (int i = 0; i < a.length; i++) {
a[i] = scan.nextInt(); //1,2,7,0
}
int result = countIncSubarrays(a, a.length);
System.out.println("Count of strictly increasing subarrays is " +
result);
}
}If we run the above code it will generate the following output −
Output
Count of strictly increasing subarrays is 6
Published on 29-Jan-2021 08:18:33
