What steps can be taken to identify the two highest numbers in an array, one being positive and the other

Seeking the optimal solution to resolve this issue

Issue: Develop a function called ArrayChallenge (Javascript) that takes a single argument "arr" representing an array of numbers. The function should return the string true if there exist two numbers in the array that can be multiplied to yield a product greater than double the sum of all elements in the array. If such numbers do not exist, the function should return the string false.

For instance: if the array "arr" is [2, 5, 6, -6, 16, 2, 3, 6, 5, 3], the sum of these elements is 42, and doubling it results in 84. In this case, the two elements 16 and 6 can be multiplied to get 96, which is greater than 84, so the function should return true. Conversely, an array like [1, 2, 4] should return false as the product of the two largest elements (4 * 2 = 8) is less than double the sum (14).

my approach was

function ArrayChallenge(arr) {
  if (arr.length < 2) return 'false'

  let maxNeg = 0
  let neg = 0

  let pos = 0
  let maxPos = 0

  const sum = arr.reduce((total, num) => {
    if (num < 0) {
      if (num < neg) maxNeg = num
      else neg = num
    } else {
      if (num >= maxPos) {
        pos = maxPos
        maxPos = num
      } else if (num > pos) pos = num
    }

    return total + num
  }, 0)

  if (maxPos * pos > sum * 2 || maxNeg * neg > sum * 2) return 'true'
  else return 'false'
}

https://codepen.io/hamodey85/pen/ExmrdgM

Answer №1

When tackling this issue, it is important to recognize that if the highest possible product of two numbers in the array is not greater than twice the sum of the array, then there are no feasible pairs available.

Methods to Resolve the Issue

Step 1: Calculate the sum of the array in advance. This can be easily achieved using a for loop or the reduce function.

Step 2: Obtain the two largest values from the array. Depending on the allowed complexity, you can either sort the array (O(nlogn)) or traverse it twice (O(2n), which is more efficient.

Step 3: Compare the product of the two maximum values with the precomputed sum and return true if the product is greater than the precomputed sum.

Sorting Approach

function ArrayChallenge(arr){  
  var precomputedSum = arr.reduce((a,c) => a+c,0); //Step 1   
  var sortedArray = arr.sort(function(a, b){return b-a});// Step 2 
  var product = sortedArray[0] * sortedArray[1];//part of step 2  
  return product > 2*precomputedSum ;
}

Looping Approach

function ArrayChallenge(arr){  
  var precomputedSum = arr.reduce((a,c) => a+c,0); //Step 1   
  int firstMax = -2147483648;// Step 2
  for(int i=0;i<arr.length;i++){
    if(arr[i]>firstMax)firstMax=arr[i];//step 2
  } 
  int secondMax = -2147483648;// Step 2
  for(int i=0;i<arr.length;i++){
    if(arr[i]>secondMax && arr[i]!=firstMax)secondMax=arr[i];//step 2
  } 
  var product = firstMax * secondMax;//part of step 2  
  return product > 2*precomputedSum ;
}

Answer №2

After experimenting with various approaches, I have stumbled upon what seems to be the most effective solution so far.

function ArrayChallenge(arr) {
  if (arr.length < 2) return 'false'

  let maxNeg = 0
  let neg = 0

  let pos = 0
  let maxPos = 0

  const sum = arr.reduce((total, num) => {
    if (num < 0) {
      if (num < neg) maxNeg = num
      else neg = num
    } else {
      if (num >= maxPos) {
        pos = maxPos
        maxPos = num
      } else if (num > pos) pos = num
    }

    return total + num
  }, 0)

  if (maxPos * pos > sum * 2 || maxNeg * neg > sum * 2) return 'true'
  else return 'false'
}

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