Computation Complexity - Double Loop

Given a sorted array in descending order, following code counts pairs of two elements with the difference d. It is easy to give an O(n²) bound for the time complexity of the function countingDifference in below codes; n times in outer for-loop and n times inner while-loop.

But a more careful analysis shows that in fact we can give an O(2n) = O(n) bound on the time complexity of this functions.  The command j++; is executed no more than n times totally, while outer for-loop is looping n times.  So, time complexity of this function is O(n+n) = O(2n) = O(n). 

The statement 'this function(algorithm) is O(n²)' is still right. 'this algorithm is O(n)' gives more information about the function.

public class CodeTest {
static int counter1 =0;
static int counter2 =0;

public static void main(String[] args) {
int input[] = {112, 39, 34, 32, 30, 27, 12, 11, 9, 8, 5, 2, 0};
int d = 2;

System.out.print("Input: ");
for(int i = 0; i<input.length;i++)
System.out.print(input[i] + "  ");
System.out.println();
System.out.println( "For " + input.length + " -length array: " + countDiffernce(input, d) + " pairs are " + d + "-differences " );
System.out.println ("Outer Loop: " + counter1 + " times, Inner Loop: " + counter2 + " times");
}

public static int countDiffernce(int input[], int d) {
int cnt =0, n = input.length;
int j = 1;
for (int i = 0; i < n && j < n; i++) {
counter1++;
while ( (j < n-1) && (input[i] - input[j] < d)) {
counter2++;
j++;
}
if(input[i] - input[j] == d) {
cnt++;
System.out.println("\t" + input[i] + " - " + input[j] + " = " + d);
} 
}
return cnt;
}
}

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