Given an array, check that the pattern of even then odd is upheld

Question

I'm stumped on this problem:

Given an array, check that every number switches from even to odd. All numbers in the array are positive integers. If there is a number that breaks the pattern, return the index of that number. If there aren't any numbers that break the pattern, return a -1

Example arrays + solutions:

//[1, 4, 5, 7, 4] // 3

//[25, 25, 25] // 1

//[4, 5, 2, 7, 4, 9] // -1

I have solved for this problem, except when it comes to the [25, 25, 25] array. Looking for this to be solved in JS.

Really not pretty, but here's what I have so far:

` function parity(numbers) {
   let evenIndex = []
   let oddIndex = []
   let container = []

   for (var i = 0; i < numbers.length; i+=2) {
    evenIndex.push(numbers[i])
   }

   for (var i = 1; i < numbers.length; i+=2) {
    oddIndex.push(numbers[i])
   }

   if (numbers[0] % 2 === 0) {
    for (var i = 0; i < evenIndex.length; i++) {
      if (evenIndex[i] % 2 !== 0) {
        container.push(numbers.indexOf(evenIndex[i]))
      }
    }
    for (var i = 0; i < oddIndex.length; i++) {
      if (oddIndex[i] % 2 !== 1) {
        container.push(numbers.indexOf(oddIndex[i]))
      }
    }
   }
  

   if (numbers[0] % 2 === 1) {
    for (var i = 0; i < evenIndex.length; i++) {
      if (evenIndex[i] % 2 !== 1) {
        container.push(numbers.indexOf(evenIndex[i]))
      }
    }
    for (var i = 0; i < oddIndex.length; i++) {
      if (oddIndex[i] % 2 !== 0) {
        container.push(numbers.indexOf(oddIndex[i]))
      }
    }
   }

   if (container.length === 0) {
    return -1
   }
   return Math.min(...container)
}`

What's the source of that programming puzzle — can you provide a link to it? I don't find any Google search query results for several substrings from the text in the block quote. — jsejcksn, May 18 '23 at 02:53
@jsejcksn I was given this problem to solve, but couldn't solve it. Afterwards, I tried to look everywhere on Google for this problem, but couldn't find it anywhere either (which led me to stackoverflow. I wrote this problem from memory) — codingsea, May 19 '23 at 01:03

score 2 · Accepted Answer · edited May 19 '23 at 01:52

2

You can simply check if the parity of each element is equal to the parity of the previous element. The first index for which this is true should be returned. If the end of the array is reached, then return -1.

function solve(arr) {
  let x, y;
  for (let i = 0; i < arr.length; i++) {
    y = arr[i] % 2;
    if (x === y) return i;
    x = y;
  }
  return -1;
}

console.log(solve([1, 4, 5, 7, 4])); // 3
console.log(solve([25, 25, 25])); // 1
console.log(solve([4, 5, 2, 7, 4, 9])); // -1

TypeScript Playground

edited May 19 '23 at 01:52

jsejcksn

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answered May 18 '23 at 00:36

Unmitigated

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Sorry I can't upvote you because I just joined stackoverflow, but tysm. This one stumped me – codingsea May 18 '23 at 00:41
Performance can be improved by starting at index `0` and performing the modulo calculation only once instead of twice every iteration, memoizing the remainder result outside the loop for the next comparison. – jsejcksn May 18 '23 at 01:45

Given an array, check that the pattern of even then odd is upheld

1 Answers1