equalizeArray.ts

/*
Given an array of integers, determine the minimum number of elements to delete to leave only elements of equal value.

Example
arr = [1, 2, 2, 3]

Delete the 2 elements 3 and 1 leaving 2. If both twos plus either the 1 or the 3 are deleted, it takes 2 deletions to leave either [3] or [1]. The minimum number of deletions is 2.

Function Description

Complete the equalizeArray function in the editor below.

equalizeArray has the following parameter(s):

int arr[n]: an array of integers
Returns

int: the minimum number of deletions required
Input Format

The first line contains an integer , the number of elements in .
The next line contains  space-separated integers .

Constraints

Sample Input

STDIN       Function
-----       --------
5           arr[] size n = 5
3 3 2 1 3   arr = [3, 3, 2, 1, 3]
Sample Output
2   
*/

// We create the map to save the number of repeat times of each element in the array
// Then we will find the largest repeating times
// The different between the length and the largest repeating times is the minimum number need to be deleted.
function equalizeArray(arr: number[]) {
  const map = new Map<number, number>();
  let maxRepeatingTime = 0;
  const length = arr.length;

  for (let i = 0; i < arr.length; i++) {
    if (map.has(arr[i])) {
      map.set(arr[i], map.get(arr[i]) + 1);
    } else {
      map.set(arr[i], 1);
    }

    if (maxRepeatingTime < map.get(arr[i])) {
      maxRepeatingTime = map.get(arr[i]);
		}
  }

  return length - maxRepeatingTime;
}

console.log(equalizeArray([3, 3, 2, 1, 3]));