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2576. Find the Maximum Number of Marked Indices

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 {
    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;
    }
};

  • 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


  • 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;
}


  • impl Solution {
    pub fn max_num_of_marked_indices(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let mut i = 0;
        let n = nums.len();
        for j in (n + 1) / 2..n {
            if nums[i] * 2 <= nums[j] {
                i += 1;
            }
        }
        (i * 2) as i32
    }
}

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