JavaScript - Issue with For Loop when Finding Symmetric Difference

Question

JavaScript - Issue with For Loop when Finding Symmetric Difference

Here is my solution to a coding challenge on FreeCodeCamp called "Symmetric Difference."

I'm puzzled as to why my code is returning 2, 3, 4, 6 instead of the expected 2, 3, 4, 6, 7.

 function sym(args) {
    args = Array.from(arguments);
    var newArr = [];

    function removeFirstIndex() {
        for (var i = 0; i <= args[1].length; i++) {
            if (args[1].indexOf(args[0][i]) === -1) {
                newArr.push(args[0][i]);
            }
            if (args[0].indexOf(args[1][i]) === -1) {
                newArr.push(args[1][i]);
            }
        }
        args = args.splice(2);
        args.unshift(newArr);
        newArr=[];
    }
    while (args.length>1){
        removeFirstIndex();
    }
    args=args.reduce(function(a,b){
        return a.concat(b);
    });
    args=args.filter(function(elem,pos,self){
        return self.indexOf(elem)===pos && elem!==undefined;
    });
    return args.sort();
}
sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]); // should return 2, 3, 4, 6, 7

javascript arrays for-loop

Answer 1

Answer №1

Your inner for loop runs i to the length of one of the two arrays. However, you are using the same index to access an element in both arrays. This could lead to skipping elements or accessing values beyond the length of the shorter array if the arrays are of different lengths.

To address this issue, you should split the loop into two separate loops:

    for (var i = 0; i <= args[0].length; i++) {
        if (args[1].indexOf(args[0][i]) === -1) {
            newArr.push(args[0][i]);
        }
    }
    for (var i = 0; i <= args[1].length; i++) {
        if (args[0].indexOf(args[1][i]) === -1) {
            newArr.push(args[1][i]);
        }
    }

This adjustment will resolve the issue.

Additionally, you can enhance this solution by utilizing hashing, which can be achieved with a Set.

Solution with Sets:

Below is an ES6 code snippet that performs the operation using hashing. This approach eliminates the need for the indexOf operation (with a time complexity of O(n)), and instead utilizes Set.prototype.has (with an O(1) time complexity in most implementations):

function sym(...args) {
  return Array.from(
    args.map( a => new Set(a) )
      .reduce( (a, b) => new Set([...a, ...b].filter( x => !a.has(x) || !b.has(x) )) )
    ).sort();
}
var result = sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]); 
// should return 2, 3, 4, 6, 7

console.log(result);

Answer 2

Your inner for loop runs i to the length of one of the two arrays. However, you are using the same index to access an element in both arrays. This could lead to skipping elements or accessing values beyond the length of the shorter array if the arrays are of different lengths.

To address this issue, you should split the loop into two separate loops:

    for (var i = 0; i <= args[0].length; i++) {
        if (args[1].indexOf(args[0][i]) === -1) {
            newArr.push(args[0][i]);
        }
    }
    for (var i = 0; i <= args[1].length; i++) {
        if (args[0].indexOf(args[1][i]) === -1) {
            newArr.push(args[1][i]);
        }
    }

This adjustment will resolve the issue.

Additionally, you can enhance this solution by utilizing hashing, which can be achieved with a Set.

Solution with Sets:

Below is an ES6 code snippet that performs the operation using hashing. This approach eliminates the need for the indexOf operation (with a time complexity of O(n)), and instead utilizes Set.prototype.has (with an O(1) time complexity in most implementations):

function sym(...args) {
  return Array.from(
    args.map( a => new Set(a) )
      .reduce( (a, b) => new Set([...a, ...b].filter( x => !a.has(x) || !b.has(x) )) )
    ).sort();
}
var result = sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]); 
// should return 2, 3, 4, 6, 7

console.log(result);

JavaScript - Issue with For Loop when Finding Symmetric Difference

Answer №1

Solution with Sets:

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