FAQ: Selection Sort - Finding the Smallest Unsorted Element

This community-built FAQ covers the “Finding the Smallest Unsorted Element” exercise from the lesson “Selection Sort”.

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This exercise can be found in the following Codecademy content:

Study for the AP Computer Science A Exam (Java)

FAQs on the exercise Finding the Smallest Unsorted Element

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Inside of our for loop but after currentMinimumIndex create a nested for loop with a counter j set equal to i + 1, that loops through the remainder of our input array, checking the condition that i is less than size.

Is incorrect.

i + i

should be

j + 1

@catower, is this something in your wheelhouse?

Is

int arr[] = {30, 50, 10, 70};
 
for (int i = 0; i < arr.length; i++) {
  System.out.println(“Outer loop unsorted position “ + i);
  for (int j = i + 1; j < arr.length; j++) {
    System.out.println(“i = “ + i + “; j = “ + j);
  }
}

Supposed to be?

int arr[] = {30, 50, 10, 70};
 
for (int i = 0; i < arr.length; i++) {
  System.out.println(“Outer loop unsorted position “ + i);
  for (int j = i + 1; j < arr.length; j++) {
    System.out.println(“i = “ + arr[i] + “; j = “ + arr[j]);
  }
}

The difference is the print statement in the nested for-loop. If not why not, why not?