Question

2576. Find the Maximum Number of Marked Indices

中文文档

Description

You are given a 0-indexed integer array nums.

Initially, all of the indices are unmarked. You are allowed to make this operation any number of times:

• Pick two different unmarked indices i and j such that 2 * nums[i] <= nums[j], then mark i and j.

Return _the maximum possible number of marked indices in nums using the above operation any number of times_.

 

Example 1:

Input: nums = [3,5,2,4]
Output: 2
Explanation: In the first operation: pick i = 2 and j = 1, the operation is allowed because 2 * nums[2] <= nums[1]. Then mark index 2 and 1.
It can be shown that there's no other valid operation so the answer is 2.

Example 2:

Input: nums = [9,2,5,4]
Output: 4
Explanation: In the first operation: pick i = 3 and j = 0, the operation is allowed because 2 * nums[3] <= nums[0]. Then mark index 3 and 0.
In the second operation: pick i = 1 and j = 2, the operation is allowed because 2 * nums[1] <= nums[2]. Then mark index 1 and 2.
Since there is no other operation, the answer is 4.

Example 3:

Input: nums = [7,6,8]
Output: 0
Explanation: There is no valid operation to do, so the answer is 0.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

 

Solutions

Solution 1: Greedy + Two Pointers

In order to mark as many indices as possible, we can sort the array nums, and then traverse the array from left to right. For each index $i$, we find the first index $j$ in the right half of the array that satisfies $2 \times nums[i] \leq nums[j]$, and then mark indices $i$ and $j$. Continue to traverse the next index $i$. When we have traversed the right half of the array, it means that the marking is complete, and the number of marked indices is the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array nums.

class Solution:
  def maxNumOfMarkedIndices(self, nums: List[int]) -> int:
    nums.sort()
    n = len(nums)
    i, j = 0, (n + 1) // 2
    ans = 0
    while j < n:
      while j < n and nums[i] * 2 > nums[j]:
        j += 1
      if j < n:
        ans += 2
      i, j = i + 1, j + 1
    return ans

class Solution {
  public int maxNumOfMarkedIndices(int[] nums) {
    Arrays.sort(nums);
    int n = nums.length;
    int ans = 0;
    for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
      while (j < n && nums[i] * 2 > nums[j]) {
        ++j;
      }
      if (j < n) {
        ans += 2;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int maxNumOfMarkedIndices(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int n = nums.size();
    int ans = 0;
    for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
      while (j < n && nums[i] * 2 > nums[j]) {
        ++j;
      }
      if (j < n) {
        ans += 2;
      }
    }
    return ans;
  }
};

func maxNumOfMarkedIndices(nums []int) (ans int) {
  sort.Ints(nums)
  n := len(nums)
  for i, j := 0, (n+1)/2; j < n; i, j = i+1, j+1 {
    for j < n && nums[i]*2 > nums[j] {
      j++
    }
    if j < n {
      ans += 2
    }
  }
  return
}

function maxNumOfMarkedIndices(nums: number[]): number {
  nums.sort((a, b) => a - b);
  const n = nums.length;
  let ans = 0;
  for (let i = 0, j = Math.floor((n + 1) / 2); j < n; ++i, ++j) {
    while (j < n && nums[i] * 2 > nums[j]) {
      ++j;
    }
    if (j < n) {
      ans += 2;
    }
  }
  return ans;
}