Need help for case when num larger than values of elements in array

JavaScript
#1

Tell us what’s happening:
Hello there, I’m currently stuck in the part where if the value of num is the largest value out of all elements in the array then I will have to return the index of the length of the array but somehow I cannot seem to get it right. I’m very new to JS so I don’t know many methods to use hence the long way for sorting the array.


function getIndexToIns(arr, num) {
if (arr.length === 0) {
  return 0;
} else {
  //SORTING
  for (let i = 0; i < arr.length; i++) {
    let temp = 0;
    for (let j = i + 1; j < arr.length; j++) {
      if(arr[j] < arr[i]){
        temp = arr[j];
        arr[j] = arr[i];
        arr[i] = temp;
      }
    }
  }
  for (let j = 0; j < arr.length; j++) {
    if (num <= arr[j]) {
      return j;
    } 
  }
}
}

getIndexToIns([40, 60], 50);


  **Your browser information:**

User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/605.1.15 (KHTML, like Gecko) Version/14.1.1 Safari/605.1.15

Challenge: Where do I Belong

Link to the challenge:

#2

Honestly, you should use the built in sort. It is faster, won’t have bugs, and is way easier for others to read at a glance.

Also, it’s generally frowned upon to modify input arrays. Those changes have impacts outside of your function.

I would use a sorted copy of the array. It is worth Googling the syntax to do so, and I can help you if you can’t find it.

You problem passing this challenge is probably here though. Where should you insert the new element if you have sorted the array?

#3

there is an easier logic to handle this challenge. The following isnt exactly a spoiler, but it will guide you to a way how to solve it easier, write a simpler code.
You dont really need to sort the array in order to find the place for a specific number. You can simply count the numbers which are lower than that number, which will tell you how many numbers you need to place in front of it, indirectly telling you at what position that number would go, if we were to sort them. (if all numbers are greater then the number, it will be placed in position zero, if only one number is lesser than the question number, we put it at second position, only one number would go ahead of it and so on).

#4

Thanks for reminding me to not change the input arrays :slight_smile:

#5

Thanks for your tip, it took me way much longer to solve as my problem was that I wrote my return inside my for-loop (in the second attempt of code) where as it much a very simple condition using a counter.

Both my first and second attempt code works but the second attempt’s code is much simpler and cleaner. Thanks for your help again ^^

  let tempArr = [...arr];
  console.log(tempArr);
  /* 
 FIRST ATTEMPT
  if (tempArr.length === 0) {
    return 0;
  } else {
    //SORTING
    for (let i = 0; i < tempArr.length; i++) {
      let temp = 0;
      for (let j = i + 1; j < tempArr.length; j++) {
        if(tempArr[j] < tempArr[i]){
          temp = tempArr[j];
           tempArr[j] = tempArr[i];
          tempArr[i] = temp;
        }
      }
    }
    for (let j = 0; j < tempArr.length; j++) {
      if (num <= tempArr[j]) {
        return j;
      } else if (num > tempArr[j] && j === tempArr.length - 1) {
        return j + 1;
      }
    }
  }
  */

  // to count how many numbers is num bigger than
  let counter = 0;
  
  //if length of arr is 0 then return 0
  if (tempArr.length === 0) {
    return 0;
  } else {
  //check where to insert num
    for (let i = 0; i < tempArr.length; i++) {
      if (num > tempArr[i]) {
        counter += 1;
      }
      console.log("i: " + i);
      console.log("counter: " + counter);
    }
    return counter;
  }
}
getIndexToIns([10, 20, 30, 40, 50], 35);
getIndexToIns([10, 20, 30, 40, 50], 30);
getIndexToIns([40, 60], 50);
getIndexToIns([3, 10, 5], 3);
getIndexToIns([5, 3, 20, 3], 5);
getIndexToIns([2, 20, 10], 19);
getIndexToIns([2, 5, 10], 15);```
#6

I would really stop rolling your own sort. It is a bad habit to get into.

But good work getting it passing! That little flip in the condition makes all the difference!

#7

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