Strange issue with the splice function in JavaScript

Other Questions JavaScript
1

I have an array with multiple “Zeros” and I need to rearrange it so that all the “Zeros” are positioned at the end of the array.

The result that I am aiming for is:

[1,2,3,0,0,0,0]

However, the output I’m getting is:

[1,2,0,3,0,0,0]

Here’s my code snippet:

let numbers = [0, 1, 2, 0, 0, 3, 0];
let zeroCount = 0;
const totalLength = numbers.length;

for (let index = 0; index < totalLength; index++) {
  if (numbers[index] === 0) {
    zeroCount++;
    numbers.splice(index, 1);
    index--;
  }
}

for (let i = 0; i < zeroCount; i++) {
  numbers.push(0);
}

console.log(numbers);
2

Your splice operation is altering the array in place and causing shifts in index positions. This means each time you remove a zero, the remaining elements shift left, leading to skipped checks on subsequent elements. Instead of using splice within the loop, you can use a more straightforward approach: filter out the zeros and then add them back at the end of the array. An example implementation could be:

let numbers = [0, 1, 2, 0, 0, 3, 0];
let filteredNumbers = numbers.filter(number => number !== 0);
let zeroCount = numbers.length - filteredNumbers.length;
for (let i = 0; i < zeroCount; i++) {
  filteredNumbers.push(0);
}
console.log(filteredNumbers);

This will reliably place all zeros at the end, maintaining the intended order for other numbers.

3

From personal experience, you can benefit from using an extra array or counter approach to keep track of the non-zero and zero elements separately. Here’s a way I’d tackle it: go through the array once to collect all non-zeros, then push zeros based on the count collected. Here’s a simple illustration:

let numbers = [0, 1, 2, 0, 0, 3, 0];
let nonZeros = [];
let zeroCount = 0;

numbers.forEach(number => {
  if (number !== 0) {
    nonZeros.push(number);
  } else {
    zeroCount++;
  }
});

for (let i = 0; i < zeroCount; i++) {
  nonZeros.push(0);
}

console.log(nonZeros);

This method minimizes operations by ensuring you’re only iterating twice, which can be a bit more efficient in some scenarios.

4

you could try using the sort method too, funny enough it can work for this kind of task. something like numbers.sort((a, b) => a === 0 ? 1 : -1). it will move all zeros to the end by treating 0 as larger than any other number. give it a try!