javascript - Special Array: An array is special if every even index contains an even number and every odd index contains an odd

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what is the problem of this code? it's showing false. this should be true.

function isSpecialArray(arr) {
    for(i=0; i<arr.length; i++){
        
        return ((arr[i % 2 == 0]) % 2 == 0) && ((arr[i % 2 !==0]) % 2 !== 0)
    }   
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) // false??

what is the problem of this code? it's showing false. this should be true.

function isSpecialArray(arr) {
    for(i=0; i<arr.length; i++){
        
        return ((arr[i % 2 == 0]) % 2 == 0) && ((arr[i % 2 !==0]) % 2 !== 0)
    }   
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) // false??

• javascript
• arrays
• loops
• logical-operators

Share Improve this question Follow edited Aug 19, 2021 at 14:53 Pointy 414k6262 gold badges595595 silver badges629629 bronze badges asked Aug 19, 2021 at 14:52 Surya YadavSurya Yadav 5311 silver badge55 bronze badges 4

• 6 Your return statement in the for loop means that your function will return after examining only the first element of the array; return exits the function immediately. – Pointy Commented Aug 19, 2021 at 14:54
• Any suggestion to make the loop worthy to examine each and every element , apart from every method of array ? – Surya Yadav Commented Aug 19, 2021 at 15:00
• Use an if and return false on an invalid entry in the loop. After the loop return true. – Andreas Commented Aug 19, 2021 at 15:00
• 1 You could do it in one line and save unnecessary loops if you use Array.prototype.every(): function isSpecial(arr) { return arr.every((item, index) => item % 2 === index % 2); } – secan Commented Aug 19, 2021 at 15:01

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7 Answers 7

Sorted by: Reset to default 3

You can simplify your code and avoid unnecessary looping as soon as the first element breaking the rule is found.

Using a for loop

function isSpecial(arr) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] % 2 !== i % 2) return false
  }
  return true;
}

Using Array.prototype.every

function isSpecial(arr) {
  return arr.every((item, index) => item % 2 === index % 2);
}

Your return ((arr[i % 2 == 0]) % 2 == 0) && ((arr[i % 2 !==0]) % 2 !== 0) syntax is wrong. Hope the below function meets your requirement.

Logic

• Loop through the array, check each node for "special" condition and store it in isNodeSpecial
• "special" condition means, even index has even number and odd index have odd number.
• Initialize a variable outside the loop which holds the isSpecial status of array.
• Update the isSpecial varaible as logical and of isSpecial and isNodeSpecial.
• Whenever a single node violates the condition the isSpecial is set to false and loop exits.

function isSpecialArray(arr) {
  let isSpecial = true;
  for (i = 0; i < arr.length && isSpecial; i++) {
    const isNodeSpecial = (i % 2 === 0) ? arr[i] % 2 === 0 : arr[i] % 2 === 1;
    isSpecial = isSpecial && isNodeSpecial;
  }
  return isSpecial;
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])); // true
console.log(isSpecialArray([2, 7, 4, 10, 6, 1, 6, 3]));// false

You can check if any element does not satisfy the condition and immediately return false. If all elements satisfy the condition, then return true. That is best solution when it es to performance. You can change your code like this:

function isSpecialArray(arr) {
  for (i = 0; i < arr.length; i++) {
    if (arr[i] % 2 !== i % 2) return false;
  }
  return true;
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) //true
console.log(isSpecialArray([1, 7, 4, 9, 6, 1, 6, 3])) //false

what you are doing is wrong because your code will return right after first execution of loop, the easiest thing you can do is to put an if condition inside your loop and check whether your odd index element is odd or even index element is even and return false if your element is wrong and keep checking for the whole array and return true outside the loop if array is special.

function isSpecialArray(arr) {
    for(i=0; i<arr.length; i++){
        if(!((i%2==0 && arr[i]%2==0) || (i%2==1 && arr[i]%2==1))) {
            return false;
        }
    }
    return true;
}

Thankyou

arr[i % 2 == 0] //returns undefined, b/c you are trying to do arr[true]
undefined % 2 //returns NaN
NaN == 0 //returns false, so you always get false

It is also true what the others are saying about the loop breaking right away, but that is not the main problem.

Try this instead:

function isSpecial(element, index) {
    if(index%2 == 0) {
        return element%2 == 0
    }
    else {
        return element%2 != 0
    }
}
function isSpecialArray(arr) {
    return arr.every(isSpecial);
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) // true
console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 4])) // false

simply do direct way :

const isSpecialArray = arr => arr.every((v,i)=>i%2===v%2)

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) // true
console.log(isSpecialArray([1, 7, 4, 9, 6, 1, 6, 3])) // false

same idea with array.reduce():

const isSpecialArray = arr =>
  arr.reduce((r,v,i)=>r && ((i%2) === (v%2)), true)

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3]))
console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 4]))

Here's the sweet and simple way to solve this(python)

        def is_special_array(list):
        res=[]
        for i in range(0,len(list),2):
            if (list[i]%2==0) and (list[i+1]%2!=0):
                res.append(True)
            else:
                res.append(False)
        return all(res)

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