Finding the mode without an array?

JavaScript
1

Finding the mean and median were pretty easy, I was happy! This one is seeming to be a bit more difficult: finding the mode. The answer cannot be [10], it must be 10.

function mode(numbers) {
    //declare variables
  //array to be returned as answer
  let modes = []
  //count how many numbers in array
  let count = []
  let i;
  let num;
  //keep track of most consistent number
  let max = 0;
  
  for (i = 0; i < numbers.length; i++) {
      num = numbers[i];
      count[num] = (count[num] || 0) + 1;
      if (count[num] > max) {
          max = count[num];
      }
  }
    for (i in count)
        if (count.hasOwnProperty(i)) {
            if (count[i] === max) {
                modes.push(Number(i));
            }
        }
    
    return modes;
3

I’m not sure. That’s why I thought an array would be perfect for this situation. They only link this: https://www.khanacademy.org/math/statistics-probability/summarizing-quantitative-data/mean-median-basics/a/mean-median-and-mode-review

5

Of course I broke something…

 function mode(numbers) {
    //declare variables
  //array to be returned as answer
  let modes = []
  //count how many numbers in array
  let count = []
  let i;
  let num;
  //keep track of most consistent number
  let max = 0;
  
  for (i = 0; i < numbers.length; i++) {
      num = numbers[i];
      count[num] = (count[num] || 0) + 1;
      if (count[num] > max) {
          max = count[num];
      }
  }
    for (i in count)
        if (count.hasOwnProperty(i)) {
            if (count[i] === max) 
                modes.push(Number(i));
            else if (count[i].length === 1){
                return count[i]
            }
                
            }
    }
    
    return modes
7

Thanks…I currently have this but it still gives me back an array if it is one number.

function mode(numbers) {
  let modes = []
  let count = []
  let i;
  let num;
  let max = 0;
  
  for (i = 0; i < numbers.length; i += 1) {
      num = numbers[i];
      count[num] = (count[num] || 0) + 1;
      if (count[num] > max) {
          max = count[num];
      }
  }
    for (i in count){
        if (count.hasOwnProperty(i)) {
            if (count[i] === max) {
                modes.push(Number(i));
                return modes;
            }
        }else{
            return count[i];
        }
  
}
}
9

It isn’t there yet. I’m not sure if I need the else if statement or not.

10

Gonna just give this one up for the day and call it early. I guess I’ll come back tomorrow.