["A", "b", "C"] should return --> ["1: a", "2: b", "3: C"] but is instead returning --> ["1: a", "1: b", "1: C"]

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1.00/5 (1 vote)

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/*
Write a function which takes a list of strings and returns each line prepended by the correct number.

The numbering starts at 1. The format is "num: string". Notice the colon and space in between.

Examples: (Input --> Output)

[] --> []
["a", "b", "c"] --> ["1: a", "2: b", "3: c"]
*/

function orderedArray(array) {
  if (array != []) {
    let num = 1;
    let prefix = `${num}: `;
    for (let i = 0; i < array.length; i++) {
      array[i] = prefix + array[i];
      num++;
    }
    return array;
  }
  return array;
}

What I have tried:

On codewars, it shows the return result as --> ["1: a", "1: b", "1: c"]. I have tried changing incrementing num before adding the prefix to array[i]. I have tried setting num to other values such as 0, and the result shows --> ["0: a", "0: b", "0: c"], basically whatever the num variable is, it will not increment each element.

Posted 3-Jun-23 16:50pm

Naveen Krishna1

Updated 3-Jun-23 19:55pm

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2 solutions

Solution 1

Increment prefix inside the loop instead of num ... or generate prefix inside the loop instead of before it.

Posted 3-Jun-23 18:23pm

OriginalGriff

Updated 3-Jun-23 18:25pm

v2

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Patrice T · Accepted Answer · 2023-06-03T19:55:00

You want the prefix to be the position in array +1, so.

JavaScript

/*
Write a function which takes a list of strings and returns each line prepended by the correct number.
The numbering starts at 1. The format is "num: string". Notice the colon and space in between.
Examples: (Input --> Output)

[] --> []
["a", "b", "c"] --> ["1: a", "2: b", "3: c"]
*/

function orderedArray(array) {
  if (array != []) {
    let num = 1;
    let prefix = `${num}: `;
    for (let i = 0; i < array.length; i++) {
      let prefix = `${i+1}: `;
      array[i] = prefix + array[i];
      num++;
    }
    return array;
  }
  return array;
}