Toggle Boolean on 'flick'

Question

Have a next task

Create a function that always returns True/true for every item in a given list. However, if an element is the word 'flick', switch to always returning the opposite boolean value.

Examples

['codewars', 'flick', 'code', 'wars'] // ➞ [True, False, False, False]

['flick', 'chocolate', 'adventure', 'sunshine'] // ➞ [False, False, False, False]

['bicycle', 'jarmony', 'flick', 'sheep', 'flick'] // ➞ [True, True, False, False, True]

["flick", "flick", "flick", "flick", "flick"]) // ➞ [false, true, false, true, false]

How I understand problem statment.

If an element in the array is 'flick', it toggles a switch. 'flick' may turn the switch to false, and the elements after it will follow this state. If the next 'flick' is encountered, it toggles the switch again to true, and the following elements will follow that state, and so on.

Here is how I solved this task.

function flickSwitch(arr) {
    let flag = true
    let array = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i].toLowerCase() == 'flick') {
            flag = false;
            array.push(flag);
        } else {
            array.push(flag);
        }
    }
    return array;
}

console.log(flickSwitch(["codewars", "flick", "code", "wars"]));

And most of the tests passed successfully.

But if there are two 'flick' elements next to each other in the array, like ["codewars", "flick", "flick", "wars"], my solution breaks.

How can I solve this problem? Please help me!

"If the next 'flick' is encountered, it toggles the switch again to true": this is a correct understanding, but where in the loop do you switch flag back to true when it was false?* — trincot
– trincot, Commented Jun 7 at 11:46

sourcechange · Accepted Answer · 2025-06-07 10:30:03Z

1

The problem with the code you wrote is that when 'flick' is seen you set the flag to false but you want toggle so you need to replace `flag = !flag`
boolean = !boolean always stores the opposite of the previous state.

Modify your code

function flickSwitch(arr) {
    let flag = true
    let array = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i].toLowerCase() == 'flick') {
            flag = !flag;  // just here is changed
            array.push(flag);
        } else {
            array.push(flag);
        }
    }
    return array;
}

console.log(flickSwitch(["codewars", "flick", "code", "wars"]));

answered Jun 7 at 10:30

sourcechange

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1 Comment

Follow Magnus Jun 7 at 12:29

Thank you so much, that's right, it was very helpful — I really appreciate it

Roko C. Buljan · Accepted Answer · 2025-06-07 12:05:56Z

1

You can set flick as your second function's argument (that way you can also decide which initial flick state to use, in order to inverse the logic if necessary)
Toggle the flick if str === "flick"
You could use Array.prototype.map() in order to return a new Array with the boolean values

const flickSwitch = (arr, flick = true) => {
  return arr.map(str => {
    if (str === "flick") flick = !flick;
    return flick;
  });
};

console.log(flickSwitch(['codewars', 'flick', 'code', 'wars'])) // [True, False, False, False]
console.log(flickSwitch(['flick', 'chocolate', 'adventure', 'sunshine'])) // [False, False, False, False]
console.log(flickSwitch(['bicycle', 'jarmony', 'flick', 'sheep', 'flick'])) // [True, True, False, False, True]
console.log(flickSwitch(["flick", "flick", "flick", "flick", "flick"])) // [false, true, false, true, false]

Or the shorter, with Arrow function's implicit return:

const flickSwitch = (arr, f = true) => arr.map(s =>  s === "flick" ? f = !f : f);

console.log(flickSwitch(['codewars', 'flick', 'code', 'wars'])) // [True, False, False, False]
console.log(flickSwitch(['flick', 'chocolate', 'adventure', 'sunshine'])) // [False, False, False, False]
console.log(flickSwitch(['bicycle', 'jarmony', 'flick', 'sheep', 'flick'])) // [True, True, False, False, True]
console.log(flickSwitch(["flick", "flick", "flick", "flick", "flick"])) // [false, true, false, true, false]

(Fun offtopic) Code-golfing it would look like:

const flickSwitch=(a,f=0)=>a.map(s=>!(s=="flick"?f^=1:f));

edited Jun 7 at 12:05

answered Jun 7 at 11:50

Roko C. Buljan

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2 Comments

Follow Magnus Jun 7 at 12:32

Thank you so much, that's right, it was very helpful — I really appreciate it

Roko C. Buljan Jun 7 at 12:53

@FollowMagnus you're very welcome, happy coding

Solt · Accepted Answer · 2025-06-07 10:30:02Z

0

You have to invert the flag but not just change to false

And do not put same actions inside if() and else, the code will be shorter.

function flickSwitch(arr) {
    let flag = true
    let array = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i].toLowerCase() == 'flick')
            flag = !flag;
        array.push(flag);
    }
    return array;
}

console.log(flickSwitch(["codewars", "flick", "code", "flick"]));

answered Jun 7 at 10:30

Solt

1,5014 silver badges17 bronze badges

1 Comment

Follow Magnus Jun 7 at 12:29

Thank you so much, that's right, it was very helpful — I really appreciate it

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