I have a simple problem which should return an integer whose count is odd in an array

Question

function findOdd(A) {
  var distinctElements = [...new Set(A)];
  distinctElements.forEach(x => {
    var count = A.filter(y => y === x).length;
    if (count % 2 != 0) {
      return x;
    }
  });
}

I don't understand why program is still executing after return statement.

The `return` ends the *current* function, and the current function is the callback. `forEach` then just continues to the next element and executes the callback again. — VLAZ, Mar 28 '20 at 13:16
What is `x` and `y` set to? Also, can you provide some example input and outputs to your function? — Richie Bendall, Mar 28 '20 at 13:16
You need to understand function scopes in javascript first. You are returning inside a child function. — Berk Öztürk, Mar 28 '20 at 13:17

Richie Bendall · Answer 1 · 2020-03-28T13:18:25.047

1

Your return statement is being executed within a scope.

Try replacing the function with:

function findOdd(A) {
    var distinctElements = [...new Set(A)];
    for (x of distinctElements) {
        var count = A.filter(y => y === x).length;
        if (count % 2 != 0) {
            return x;
        }
    }
}

edited Mar 28 '20 at 13:18

answered Mar 28 '20 at 13:17

Richie Bendall

7,738
4
38
58

You should `return` out of the `forEach` after assigning a value to `out` to maintain the behavior of the original program. – Aplet123 Mar 28 '20 at 13:18

I have a simple problem which should return an integer whose count is odd in an array

1 Answers1