1338. Reduce Array Size to The Half

Problem

You are given an integer array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return **the minimum size of the set so that *at least* half of the integers of the array are removed**.

  Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

  Constraints:

• 2 <= arr.length <= 10^5

• arr.length is even.

• 1 <= arr[i] <= 10^5

Solution

/**
 * @param {number[]} arr
 * @return {number}
 */
var minSetSize = function(arr) {
  const n = arr.length

  const occurrences = arr.reduce(
    (acc, el) => acc.set(el, (acc.get(el) || 0) + 1),
    new Map()
  )

  const pairsSortedByOccurencesDesc = [...occurrences.entries()].sort(
    ([_, occurrenceA], [__, occurrenceB]) => occurrenceA - occurrenceB
  )

  let i = n
  let setRemovedLength = 0

  while (i > Math.floor(n / 2)) {
    i -= pairsSortedByOccurencesDesc.pop()[1]
    setRemovedLength++
  }

  return setRemovedLength
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).