Determine the lowest amplitude possible by eliminating K consecutive elements from a given array

Question

Determine the lowest amplitude possible by eliminating K consecutive elements from a given array

I'm looking for a way to efficiently solve this challenge using JavaScript, aiming for O(N) time complexity.

Challenge: Given an array A consisting of N positive integers and an integer k. The objective is to eliminate k consecutive elements from A in order to minimize the amplitude of the remaining elements. Here, amplitude refers to the difference between the smallest and largest elements.

For example, if A[] = [8,7,4,1] and k=2. The expected output would be 1 as we should remove 4 and 1.

A straightforward solution exists, but is it feasible to achieve this in O(n) time? Appreciate any insights or suggestions. Thank you.

javascript arrays

Answer 1

Answer №1

Are you finding this information helpful? Initially, I identify the three largest numbers (choosing three because removing two consecutive numbers might eliminate the two largest ones, requiring us to consider the next largest). Then, a similar process is repeated for the three smallest numbers.

An array of the consecutive numbers that were removed is created without changing the original array.

Following that, various if/else statements are used to determine whether the maximum and minimum values are present in the consecutive numbers that were removed.

While it may not be the most elegant solution due to the frequent usage of if/else statements.

const arr = [8, 7, 4, 1]
const arr2 = [8, 7, 4, 1, 4, 6, 8]
const arr3 = [8, 7, 4, 1, 4, 6, 8, 3, 11, 4, 15]

function slice2Consecutive(arr) {
  let newArr = []
  for (let i = 0; i < arr.length - 1; i++) {
    let s1 = arr[i]
    let s2 = arr[i + 1]
    newArr.push([arr[i], arr[i + 1]])
  }
  return newArr
}

function maxThree(arr) {
  let one = -Infinity;
  let two = -Infinity;
  let three = -Infinity;
  for (let i = 0; i < arr.length; i += 1) {
    let num = arr[i];
    if (num > three) {
      if (num >= two) {
        three = two;
        if (num >= one) {
          two = one;
          one = num;
        } else {
          two = num;
        }
      } else {
        three = num;
      }
    }
  }
  return [one, two, three]
}

function minThree(arr) {
  let one = +Infinity;
  let two = +Infinity;
  let three = +Infinity;
  for (let i = 0; i < arr.length; i += 1) {
    let num = arr[i];
    if (num < three) {
      if (num <= two) {
        three = two;
        if (num <= one) {
          two = one;
          one = num;
        } else {
          two = num;
        }
      } else {
        three = num;
      }
    }
  }
  return [one, two, three]
}

function minAmplitude(arr) {
  const [max, secondMax, thirdMax] = maxThree(arr)
  const [min, secondMin, thirdMin] = minThree(arr)
  const slicedArr = slice2Consecutive(arr)
  const amplitudeArr = []
  for (let i = 0; i < slicedArr.length; i++) {
    let m = max
    let n = min
    if (slicedArr[i][0] === max || slicedArr[i][1] === max) {
      if (slicedArr[i][0] === secondMax || slicedArr[i][1] === secondMax) {
        m = thirdMax
      } else {
        m = secondMax
      }
    }
    if (slicedArr[i][0] === min || slicedArr[i][1] === min) {
      if (slicedArr[i][0] === secondMin || slicedArr[i][1] === secondMin) {
        n = thirdMin
      } else {
        n = secondMin
      }
    }
    amplitudeArr.push(m - n)

  }
  return Math.min(...amplitudeArr)
}

console.log(minAmplitude(arr))
console.log(minAmplitude(arr2))
console.log(minAmplitude(arr3))

Answer 2

Are you finding this information helpful? Initially, I identify the three largest numbers (choosing three because removing two consecutive numbers might eliminate the two largest ones, requiring us to consider the next largest). Then, a similar process is repeated for the three smallest numbers.

An array of the consecutive numbers that were removed is created without changing the original array.

Following that, various if/else statements are used to determine whether the maximum and minimum values are present in the consecutive numbers that were removed.

While it may not be the most elegant solution due to the frequent usage of if/else statements.

const arr = [8, 7, 4, 1]
const arr2 = [8, 7, 4, 1, 4, 6, 8]
const arr3 = [8, 7, 4, 1, 4, 6, 8, 3, 11, 4, 15]

function slice2Consecutive(arr) {
  let newArr = []
  for (let i = 0; i < arr.length - 1; i++) {
    let s1 = arr[i]
    let s2 = arr[i + 1]
    newArr.push([arr[i], arr[i + 1]])
  }
  return newArr
}

function maxThree(arr) {
  let one = -Infinity;
  let two = -Infinity;
  let three = -Infinity;
  for (let i = 0; i < arr.length; i += 1) {
    let num = arr[i];
    if (num > three) {
      if (num >= two) {
        three = two;
        if (num >= one) {
          two = one;
          one = num;
        } else {
          two = num;
        }
      } else {
        three = num;
      }
    }
  }
  return [one, two, three]
}

function minThree(arr) {
  let one = +Infinity;
  let two = +Infinity;
  let three = +Infinity;
  for (let i = 0; i < arr.length; i += 1) {
    let num = arr[i];
    if (num < three) {
      if (num <= two) {
        three = two;
        if (num <= one) {
          two = one;
          one = num;
        } else {
          two = num;
        }
      } else {
        three = num;
      }
    }
  }
  return [one, two, three]
}

function minAmplitude(arr) {
  const [max, secondMax, thirdMax] = maxThree(arr)
  const [min, secondMin, thirdMin] = minThree(arr)
  const slicedArr = slice2Consecutive(arr)
  const amplitudeArr = []
  for (let i = 0; i < slicedArr.length; i++) {
    let m = max
    let n = min
    if (slicedArr[i][0] === max || slicedArr[i][1] === max) {
      if (slicedArr[i][0] === secondMax || slicedArr[i][1] === secondMax) {
        m = thirdMax
      } else {
        m = secondMax
      }
    }
    if (slicedArr[i][0] === min || slicedArr[i][1] === min) {
      if (slicedArr[i][0] === secondMin || slicedArr[i][1] === secondMin) {
        n = thirdMin
      } else {
        n = secondMin
      }
    }
    amplitudeArr.push(m - n)

  }
  return Math.min(...amplitudeArr)
}

console.log(minAmplitude(arr))
console.log(minAmplitude(arr2))
console.log(minAmplitude(arr3))

Determine the lowest amplitude possible by eliminating K consecutive elements from a given array

Answer №1

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