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

1 Answer

0 votes
public class MyClass {
    public static int countStrictlyIncreasingSubarrays(int[] arr) {
    	int size = arr.length;
    	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
    				 System.out.print(arr[k] + " "); // for print only
    			}                                    // for print only
    			System.out.println();                // for print only
    		}
    	}
    	return count;
    }
    public static void main(String args[]) {
        int[] arr = {1, 3, 4, 0, 7, 5, 9};
        // {1, 3}, {1, 3, 4}, {3, 4}, {0, 7}, {5, 9}

        System.out.print(countStrictlyIncreasingSubarrays(arr));
        System.out.print(" = Count of strictly increasing subarrays");

    }
}






/*
run:
 
1 3 
1 3 4 
3 4 
0 7 
5 9 
5 = Count of strictly increasing subarrays
 
*/

 



answered Aug 21, 2022 by avibootz
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