Feel like I'm on the right track but not sure.

Question about the task

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I understand the nature of the boolean compare function. If true, do this, if false do second. What I'm not really understanding is why the array is not assigning a value to when true or false. I expect that is should be doing this but I'm seeing nothing when I run test output in between the loops.

public static void sort(String[] array) {
        String[] s = new String[array.length];
        int i;
        int j;
        for(i = 0; i <= array.length-1; i++) {
            for (j = 0; j < array.length - 1; j++) {
                System.out.println(isGreaterThan(array[i], array[1 + j]));
                if (isGreaterThan(array[i], array[1 + j])) {
                    s[i] = array[i];

                } else {
                    s[i] = array[j + 1];
                }

            }
            for(int k = 0; k < array.length; k++){
                array[i]=s[i];
            }

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Guadalupe Gagnon

Level 37 , Tampa, United States

3 June 2020, 14:24

The problem that I see with this algorithm is that it will constantly be overwriting s[i] to the max value of the two passed to isGreaterThan(). Furthermore, because the last check of each of the 'i' loops will be the index of i against the last value in 'array', the result of the sort is going to be either of those 2 value on each index. Here are some examples. For clarity I will use numbers, lets say you have an array that is {1, 2, 3}, this is what happens i = 0, j = 0 - Checks 1 against 2 (j+1) - sets s[0] to 2 i = 0, j = 1 - Checks 1 against 3 - sets s[0] to 3 (end of j so value will finally be set to one of these two) i = 1, j = 0 - Checks 2 against 2 (j+1) - sets s[1] to 2 i = 1, j = 1 - Checks 2 against 3 (j+1) - sets s[1] to 3 (end of j so value will finally be set to one of these two) i = 2, j = 0 - Checks 3 against 2 (j+1) - sets s[2] to 3 i = 2, j = 1 - Checks 3 against 3 (j+1) - sets s[2] to 3 (end of j so value will finally be set to one of these two) The final result of 's' would be {3, 3, 3} Now if array was this [3,2,1] here is what would happen i = 0, j = 0 - Checks 3 against 2 (j+1) - sets s[0] to 3 i = 0, j = 1 - Checks 3 against 1 - sets s[0] to 3 (end of j so value will finally be set to one of these two) i = 1, j = 0 - Checks 2 against 2 (j+1) - sets s[1] to 2 i = 1, j = 1 - Checks 2 against 1 (j+1) - sets s[1] to 2 (end of j so value will finally be set to one of these two) i = 2, j = 0 - Checks 1 against 2 (j+1) - sets s[2] to 2 i = 2, j = 1 - Checks 1 against 1 (j+1) - sets s[2] to 1 (end of j so value will finally be set to one of these two) The result of this sort on 's' would equal {3, 2, 1}

Lucas Hoage

Level 14 , Savannah, United States

3 June 2020, 14:52

I understand why my code is doing what it is doing now. I have been struggling with algorithms for a while now. It is definitely my weakest skill. I also seem to have a small grasp on how compare is operating. I'll go back to the drawing board on this on. Thanks for the insight.

Guadalupe Gagnon

Level 37 , Tampa, United States

3 June 2020, 15:05

Creating algorithms is a skill that can be built. When I was first learning how to code I would put a whole bunch of sys.out statements through the code so that I could read the output and figure out what was going on. Here is an example with how I modified your code to get the answer:

public class Solution {
    public static void sort(int[] array) {
        int[] s = new int[array.length];
        System.out.println("Passed in array = " + Arrays.toString(array));
        for (int i = 0; i <= array.length - 1; i++) {
            System.out.println("New 'i' Loop - 'i' equals: " + i);
            for (int j = 0; j < array.length - 1; j++) {
                System.out.println("\tNew 'j' loop - 'j' equals: " + j);
                boolean testResult = isGreaterThan(array[i], array[1 + j]);
                System.out.println("\t\tTesting if array["+i+"]("+array[i]+") is greater than array["+(j + 1)+"]("+ array[1 + j]+"). result: " +
                        testResult);
                int intResult = testResult ? array[i] : array[j +1];
                System.out.println("\t\tSetting s["+i+"] to "+intResult);
                s[i] = intResult;
                System.out.println("\t\tArray s no equals: " + Arrays.toString(s));
            }
        }
        for (int k = 0; k < array.length; k++) {
            array[k] = s[k];
        }
    }

    private static boolean isGreaterThan(int s, int s1) {
        if(s > s1) return true;
        return false;
    }

    public static void main(String[] args) {
        int[] temp = {3,2,1};
        sort(temp);
        System.out.println(Arrays.toString(temp));
    }
}

Guadalupe Gagnon

Level 37 , Tampa, United States

3 June 2020, 15:08

feel free to copy + paste that and run it. Change line 29 to whatever you want to test for different things. You could do this instead:

int[] temp = {3,2,1,5,9,-3,1}; // more values including positive and negative numbers

or this:

int[] temp = {5}; // only 1 value in array

When you start to understand the algorithms and what they are actually doing you can modify them to get the results that you want.

Lucas Hoage

Level 14 , Savannah, United States

3 June 2020, 16:38

The problem that I'm seeing is that the shift is never happening from the original array. The original array is still evaluating the number that gets stored into the s array. Once the for loop jumps to the next iteration, the evaluation starts from (j+1) - end again. That's why I keep getting issues with the repeating numbers when the array gets piped into the for loop pre sorted. It sounds like the resolution to this problem, is to remove the entry that is assigned in s(i) from the original array until it has nothing left to sort against. In doing so, I would have to reset the "counter" for j wouldn't I?

Lucas Hoage

Level 14 , Savannah, United States

3 June 2020, 18:42

I think I'm super close. Just have to resolve the out of bounds issue.

public static void sort(String[] array) {
        String[] s = new String[array.length];
        int j;

        for (int i = 0; i <= array.length - 1;) {
            for (j = 0; j < array.length -1;j++){
                boolean result = isGreaterThan(array[i], array[j+1]);
                System.out.println("The value of arrayi before changes: " + i + array[i]);
                System.out.println("The value of arrayj before changes: " + (j+1) +array[j+1]);
                if(result == false){
                    s[i] = array[i];
                    array[i] = array[j+1];
                    System.out.println("s is: " + i + s[i]);
                    System.out.println("The value of arrayi after changes: " + i + array[i]);
                    System.out.println("The value of arrayj after changes: " + (j+1) + array[j+1]);
                    i++;

                }
                else{
                    s[i] = array[j+1];
                    array[j+1] = array[i];
                    System.out.println("s is: " + i + s[i]);
                    System.out.println("The value of arrayi2 after changes: " + i + array[i]);
                    System.out.println("The value of arrayj2 after changes: " + (j+1) + array[j+1]);
                    i++;

                }

I think... Don't give me the answer!

Guadalupe Gagnon

Level 37 , Tampa, United States

3 June 2020, 19:52

I didn't fully understand the post starting with "The problem that I'm seeing is that the shift is never happening from the original array", but that is ok if you are on to something. Let me explain a sorting algorithm that I know works and is fairly simple to implement, I will let you figure out how to implement it if you choose to do so. So you will want to start with a loop that will loop as long as the list is not sorted, we will call this the outer loop. You will then want to make a loop inside that makes comparisons to each value of the array with its neighbor (such as i compared to i +1). When it is found that i+1 is greater than i then swap those two values in the array. As soon as the inner loop finishes checking all the values then the outerloop checks if the list is sorted. If it is sorted then the method can end, if not then the inner loop starts all over. Some important info: - Requires only 2 loops, an outer and an inner as described above - Make sure that i + 1 does not result in an index out of range exception - Make sure that you do not lose data when swapping strings - a good way to tell if the list is sorted or not is if a swap happens during the inner loop. If the loop goes all the way through without swapping a single value then the list is sorted.

Lucas Hoage

Level 14 , Savannah, United States

3 June 2020, 21:02

I was going to type a long winded self destructive rant. But I'll make it short(er). I don't know if it's my way of thinking or not. I really have a bias that the building blocks needed to come to these conclusions are intentionally left out. I read what you wrote. I get it. I understand the purpose of index swapping to force the next iteration not to become screwy. I read a few other codes in the help section and I just don't understand how that conclusion was drawn based on our education material presented to us. Maybe I'm just looking for someone to blame for my inability to solve it. I'm too tired. I'll try again later. Thanks for the input.

Lucas Hoage

Level 14 , Savannah, United States

3 June 2020, 22:55

I solved it after watching the sorting method you mentioned in anothrr post. Two things that could be done during this chapter that act as an intermediary between this algorithm: Practice with swapping variables inside data structures using pre existing data only found within the data structure itself. Practice creating collapsing iterations within a nested for loop to emphasize how to prevent an OOB error. A great example of this would be to use collapsing iterations to display the remaining matrix that has to be displaerd as it dwindles in size. Additionally, although optional as the nested for loop is pretty popular: More practice with nested for loops and how they affect data structures like matrix. i believe with those fundamentals being expetienced by the student before even getting to a problem like that, they would have a great inderstanding of how to solve that problem.