Question 328722 in Javascript Question posted in
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Why Does Using a Variable for the Index Work in This Selection Sort Algorithm in JavaScript?

I’m trying to understand why using a variable to assign the index works fine, instead of directly using the index itself in the following code snippet of the selection sort algorithm using JavaScript.

What I Tried:
I attempted to use the index of the minimum element directly (j):

let myArray = [7, 9, 4, 16, 2, 0, 4, -18];

function selectionSortAlgorithm(anArray) {
    for (let i = 0; i < anArray.length - 1; i++) {
        let min = anArray[i];

        for (let j = i + 1; j < anArray.length; j++) {
            if (min > anArray[j]) {
                min = anArray[j];
            }
        }
        let temp = anArray[i];
        anArray[i] = min;
        anArray[j] = temp;
    }
    return anArray;
}

console.log(selectionSortAlgorithm(myArray));

The Correct Code:
I found that using a new variable for the minimum element (minIndex) corrects the issue:

let myArray = [7, 9, 4, 16, 2, 0, 4, -18];

function selectionSortAlgorithm(anArray) {
    for (let i = 0; i < anArray.length - 1; i++) {
        let min = anArray[i];
        let minIndex = i;

        for (let j = i + 1; j < anArray.length; j++) {
            if (min > anArray[j]) {
                min = anArray[j];
                minIndex = j;
            }
        }
        let temp = anArray[i];
        anArray[i] = min;
        anArray[minIndex] = temp;
    }
    return anArray;
}

console.log(selectionSortAlgorithm(myArray));
Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    Credit to @AztecCodes

    What makes you think that j would contain the index of the Min element? –

    The index j in the inner loop of the first attempt is not the index of the minimum value. j is always anArray.length when that loop terminates. It finds the minimum value but forgets the index where it's found.

    If you put console.log(j); in the first version and console.log(minIndex) in the second version, you'll see the difference.


  2. 0

    The problem in your first code is that j is declared inside of for statement.

    for (let j = i + 1; j < anArray.length; j++) {
  if (min > anArray[j]) {
    min = anArray[j];
  }
}

    So when you try to access it out of the statement, it isn’t accessible

    ReferenceError: j is not define

    In the second example, you create let minIndex outside of the for statement. That’s why you can access it.

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