You are given an array nums of positive integers. In one operation, you can choose any number from nums and reduce it to exactly half the number. (Note that you may choose this reduced number in future operations.)

Return the minimum number of operations to reduce the sum of nums by at least half.

Example 1:

Input: nums = [5,19,8,1]
Output: 3
Explanation: The initial sum of nums is equal to 5 + 19 + 8 + 1 = 33.
The following is one of the ways to reduce the sum by at least half:
Pick the number 19 and reduce it to 9.5.
Pick the number 9.5 and reduce it to 4.75.
Pick the number 8 and reduce it to 4.
The final array is [5, 4.75, 4, 1] with a total sum of 5 + 4.75 + 4 + 1 = 14.75.
The sum of nums has been reduced by 33 - 14.75 = 18.25, which is at least half of the initial sum, 18.25 >= 33/2 = 16.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Example 2:

Input: nums = [3,8,20]
Output: 3
Explanation: The initial sum of nums is equal to 3 + 8 + 20 = 31.
The following is one of the ways to reduce the sum by at least half:
Pick the number 20 and reduce it to 10.
Pick the number 10 and reduce it to 5.
Pick the number 3 and reduce it to 1.5.
The final array is [1.5, 8, 5] with a total sum of 1.5 + 8 + 5 = 14.5.
The sum of nums has been reduced by 31 - 14.5 = 16.5, which is at least half of the initial sum, 16.5 >= 31/2 = 16.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^7

Approach 1: Max Heap

We can keep track of the current largest number, and perform the division on the largest number while keeping track of the sum

Written by@heiheihang

class Solution:
    def halveArray(self, nums: List[int]) -> int:
        s = sum(nums)

        target = s / 2
        h = []

        for n in nums:
            heappush(h, -n)

        val = s
        res = 0
        while(val > target):
            res += 1
            top = -(h[0] / 2)
            val -= top
            heappop(h)
            heappush(h, - top)

        return res

Approach 2: Priority Queue

First we calculate $sum$ and $half$ . We use priority queue to store the elements. We pop out the maximum element $top$ , add the half of it (i.e. $top / 2$ ) to $reduce$ , and push it to priority queue. We keep doing that until $reduce$ is greater than / equal to $half$ . The answer is how many rounds we have performed.

Written by@wkw

class Solution {
public:
    int halveArray(vector<int>& nums) {
        int n = nums.size(), ans = 0;
        // calculate the sum
        long double sum = accumulate(nums.begin(), nums.end(), 0LL);
        // calculate the half
        long double half = sum / 2;
        // init priority queue
        priority_queue<double> pq;
        for (auto x : nums) pq.push(x);
        // used to compare with half
        double reduce = 0;
        // final goal: reduce >= half
        while (reduce < half) {
            // take the max one and pop it out
            double top = pq.top(); pq.pop();
            // add half of it
            reduce += top / 2;
            // push the half to the queue
            pq.push(top / 2);
            // 1 operation is done here
            ans++;
        }
        return ans;
    }
};

2208 - Minimum Operations to Halve Array Sum (Medium)

Problem Link

Problem Statement

Approach 1: Max Heap

Approach 2: Priority Queue

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