Filter out Strings from an Array

function filterOutStringfromArr(array) {
    var arr = []
    for (var i = 0; i < array.length; i++) {
        if (typeof array[i] === 'number') {
          arr.push(array[i])
        }
    }
    console.log(arr)
}
filterOutStringfromArr([1,2,'A','B',123])

MDN link that is pretty good

You should use push that will append to the end of an array. By inserting the found element into spot i you are going from index 3 to 5 thus the empty elements.

I suggest to use the Array.prototype.push() method.

function filterOutStringsfromArray(array) {
    var arr = [];
    for (let i = 0; i < array.length; i++) {
        if (typeof array[i] === 'string') {
          arr.push(array[i]);
        }
    }
    return arr;
}

filterOutStringfromArr([5, 2, 'h', 2, 's', 2]) // Expected Output: ['h', 's']

Filter method will solve your issue.

function filterOutStringfromArr(array) {
    const newArray = array.filter(item => {
        if (typeof item === 'number') {
            return item;
        }
    })    
    console.log(newArray);
}

filterOutStringfromArr([1,2,'A','B',123])

Personally i would just use a normal array filter function and test the condition. let native array.filter function do the work for you. The below will filter the array and return a new array (not modify the existing array).

function filterToArrayOfNumbers(arrayElement) {
    return typeof arrayElement === 'number';
}
console.log([1,2,'A','B',123].filter(filterToArrayOfNumbers));

arr[i] = array[i]  // <------------  your issue

i is still being incremented, so when you are looking at your last number 123, (index 4), it will set the new array at index 4 (leaving 2 and 3 empty)

rather than setting it explicitly, you can do arr.push(array[i])

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push