2576

2576. Find the Maximum Number of Marked Indices (Medium)

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

 

Related Topics:
Array, Two Pointers, Binary Search, Greedy, Sorting

Solution 1. Binary Answer

// OJ: https://leetcode.com/problems/find-the-maximum-number-of-marked-indices
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int maxNumOfMarkedIndices(vector<int>& A) {
        sort(begin(A), end(A));
        int N = A.size(), L = 0, R = N / 2;
        auto valid = [&](int k) {
            int i = 0, j = N - k;
            for (; i < k; ++i, ++j) {
                if (2 * A[i] > A[j]) return false;
            }
            return true;
        };
        while (L <= R) {
            int M = (L + R) / 2;
            if (valid(M)) L = M + 1;
            else R = M - 1;
        }
        return R * 2;
    }
};