3072. Distribute Elements Into Two Arrays II

Description

You are given a 1-indexed array of integers nums of length n.

We define a function greaterCount such that greaterCount(arr, val) returns the number of elements in arr that are strictly greater than val.

You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the ith operation:

• If greaterCount(arr1, nums[i]) > greaterCount(arr2, nums[i]), append nums[i] to arr1.
• If greaterCount(arr1, nums[i]) < greaterCount(arr2, nums[i]), append nums[i] to arr2.
• If greaterCount(arr1, nums[i]) == greaterCount(arr2, nums[i]), append nums[i] to the array with a lesser number of elements.
• If there is still a tie, append nums[i] to arr1.

The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].

Return the integer array result.

 

Example 1:

Input: nums = [2,1,3,3]
Output: [2,3,1,3]
Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3rd operation, the number of elements greater than 3 is zero in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1.
In the 4th operation, the number of elements greater than 3 is zero in both arrays. As the length of arr2 is lesser, hence, append nums[4] to arr2.
After 4 operations, arr1 = [2,3] and arr2 = [1,3].
Hence, the array result formed by concatenation is [2,3,1,3].

Example 2:

Input: nums = [5,14,3,1,2]
Output: [5,3,1,2,14]
Explanation: After the first 2 operations, arr1 = [5] and arr2 = [14].
In the 3rd operation, the number of elements greater than 3 is one in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1.
In the 4th operation, the number of elements greater than 1 is greater in arr1 than arr2 (2 > 1). Hence, append nums[4] to arr1.
In the 5th operation, the number of elements greater than 2 is greater in arr1 than arr2 (2 > 1). Hence, append nums[5] to arr1.
After 5 operations, arr1 = [5,3,1,2] and arr2 = [14].
Hence, the array result formed by concatenation is [5,3,1,2,14].

Example 3:

Input: nums = [3,3,3,3]
Output: [3,3,3,3]
Explanation: At the end of 4 operations, arr1 = [3,3] and arr2 = [3,3].
Hence, the array result formed by concatenation is [3,3,3,3].

 

Constraints:

• 3 <= n <= 105
• 1 <= nums[i] <= 109

 

Solutions

Solution: Binary Indexed Tree

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

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
const resultArray = function (nums) {
  const n = nums.length;
  const arr1 = [];
  const arr2 = [];
  const rankMap = createRankMap(nums);
  const bit1 = new BIT(rankMap.size);
  const bit2 = new BIT(rankMap.size);

  const addElement = (num, arr, bit) => {
    arr.push(num);
    bit.update(rankMap.get(num), 1);
  };

  addElement(nums[0], arr1, bit1);
  addElement(nums[1], arr2, bit2);

  for (let index = 2; index < n; index++) {
    const num = nums[index];
    const rank = rankMap.get(num);
    const greaterCount1 = arr1.length - bit1.query(rank);
    const greaterCount2 = arr2.length - bit2.query(rank);

    if (greaterCount1 > greaterCount2) {
      addElement(num, arr1, bit1);
    } else if (greaterCount1 < greaterCount2) {
      addElement(num, arr2, bit2);
    } else if (arr1.length <= arr2.length) {
      addElement(num, arr1, bit1);
    } else {
      addElement(num, arr2, bit2);
    }
  }

  return [...arr1, ...arr2];
};

class BIT {
  constructor(n) {
    this.bit = Array.from({ length: n + 2 }, () => 0);
  }

  update(num, delta) {
    while (num < this.bit.length) {
      this.bit[num] += delta;
      num += num & -num;
    }
  }

  query(num) {
    let result = 0;

    while (num) {
      result += this.bit[num];
      num -= num & -num;
    }

    return result;
  }
}

function createRankMap(nums) {
  const sorted = [...new Set(nums)].toSorted((a, b) => a - b);
  const rankMap = new Map();
  let rank = 1;

  for (const num of sorted) {
    rankMap.set(num, rank);
    rank += 1;
  }

  return rankMap;
}