100232. Minimum Operations to Exceed Threshold Value II User Accepted:4338 User Tried:7626 Total Accepted:4413 Total Submissions:13969 Difficulty:Medium You are given a 0-indexed integer array nums, and an integer k. In one operation, you will: Take the two smallest integers x and y in nums. Remove x and y from nums. Add min(x, y) * 2 + max(x, y) anywhere in the array. Note that you can only apply the described operation if nums contains at least two elements. Return the minimum number of operations needed so that all elements of the array are greater than or equal to k. Example 1: Input: nums = [2,11,10,1,3], k = 10 Output: 2 Explanation: In the first operation, we remove elements 1 and 2, then add 1 * 2 + 2 to nums. nums becomes equal to [4, 11, 10, 3]. In the second operation, we remove elements 3 and 4, then add 3 * 2 + 4 to nums. nums becomes equal to [10, 11, 10]. At this stage, all the elements of nums are greater than or equal to 10 so we can stop. It can be shown that 2 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10. Example 2: Input: nums = [1,1,2,4,9], k = 20 Output: 4 Explanation: After one operation, nums becomes equal to [2, 4, 9, 3]. After two operations, nums becomes equal to [7, 4, 9]. After three operations, nums becomes equal to [15, 9]. After four operations, nums becomes equal to [33]. At this stage, all the elements of nums are greater than 20 so we can stop. It can be shown that 4 is the minimum number of operations needed so that all elements of the array are greater than or equal to 20. Constraints: 2 <= nums.length <= 2 * 105 1 <= nums[i] <= 109 1 <= k <= 109 The input is generated such that an answer always exists. That is, there exists some sequence of operations after which all elements of the array are greater than or equal to k. complete the function class Solution { public: int minOperations(vector<int>& nums, int k) { } };
Question
- Minimum Operations to Exceed Threshold Value II User Accepted:4338 User Tried:7626 Total Accepted:4413 Total Submissions:13969 Difficulty:Medium You are given a 0-indexed integer array nums, and an integer k.
In one operation, you will:
Take the two smallest integers x and y in nums. Remove x and y from nums. Add min(x, y) * 2 + max(x, y) anywhere in the array. Note that you can only apply the described operation if nums contains at least two elements.
Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.
Example 1:
Input: nums = [2,11,10,1,3], k = 10 Output: 2 Explanation: In the first operation, we remove elements 1 and 2, then add 1 * 2 + 2 to nums. nums becomes equal to [4, 11, 10, 3]. In the second operation, we remove elements 3 and 4, then add 3 * 2 + 4 to nums. nums becomes equal to [10, 11, 10]. At this stage, all the elements of nums are greater than or equal to 10 so we can stop. It can be shown that 2 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10. Example 2:
Input: nums = [1,1,2,4,9], k = 20 Output: 4 Explanation: After one operation, nums becomes equal to [2, 4, 9, 3]. After two operations, nums becomes equal to [7, 4, 9]. After three operations, nums becomes equal to [15, 9]. After four operations, nums becomes equal to [33]. At this stage, all the elements of nums are greater than 20 so we can stop. It can be shown that 4 is the minimum number of operations needed so that all elements of the array are greater than or equal to 20.
Constraints:
2 <= nums.length <= 2 * 105 1 <= nums[i] <= 109 1 <= k <= 109 The input is generated such that an answer always exists. That is, there exists some sequence of operations after which all elements of the array are greater than or equal to k. complete the function class Solution { public: int minOperations(vector<int>& nums, int k) {
}
};
Solution
This problem can be solved using a priority queue (min heap) in C++. The priority queue will always give us the smallest two elements in the array. We will keep performing the operation until the smallest element in the array is greater than or equal to k. Here is the code:
#include <queue>
#include <vector>
class Solution {
public:
int minOperations(vector<int>& nums, int k) {
priority_queue<int, vector<int>, greater<int>> pq(nums.begin(), nums.end());
int operations = 0;
while(pq.top() < k) {
int x = pq.top(); pq.pop();
int y = pq.top(); pq.pop();
int new_element = min(x, y) * 2 + max(x, y);
pq.push(new_element);
operations++;
}
return operations;
}
};
In this code, we first push all the elements of the array into the priority queue. Then, we keep popping the smallest two elements from the queue, perform the operation, and push the result back into the queue. We increment the operations counter each time we do this. We stop when the smallest element in the queue is greater than or equal to k. Finally, we return the operations counter.
Similar Questions
class Solution { public: int minOperations(vector<int>& nums, int k) { } }; complete the function using folloiwing statement User 100231. Minimum Operations to Exceed Threshold Value I User Accepted:1 User Tried:3 Total Accepted:1 Total Submissions:3 Difficulty:Easy You are given a 0-indexed integer array nums, and an integer k. In one operation, you can remove one occurrence of the smallest element of nums. Return the minimum number of operations needed so that all elements of the array are greater than or equal to k. Example 1: Input: nums = [2,11,10,1,3], k = 10 Output: 3 Explanation: After one operation, nums becomes equal to [2, 11, 10, 3]. After two operations, nums becomes equal to [11, 10, 3]. After three operations, nums becomes equal to [11, 10]. At this stage, all the elements of nums are greater than or equal to 10 so we can stop. It can be shown that 3 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10. Example 2: Input: nums = [1,1,2,4,9], k = 1 Output: 0 Explanation: All elements of the array are greater than or equal to 1 so we do not need to apply any operations on nums. Example 3: Input: nums = [1,1,2,4,9], k = 9 Output: 4 Explanation: only a single element of nums is greater than or equal to 9 so we need to apply the operations 4 times on nums. Constraints: 1 <= nums.length <= 50 1 <= nums[i] <= 109 1 <= k <= 109 The input is generated such that there is at least one index i such that nums[i] >= k.
You are given an array A of size NLet M be the minimum value present in the array initiallyIn one operation, you can choose an element Ai ( 1 ≤ i ≤ N ) and an integer X ( 1 ≤ X ≤ 100 ), and set Ai = X.Determine the minimum number of operations required to make M the maximum value in the array A.
Given an array of integers called nums, you can perform the following operation while nums contains at least 2 elements:Choose the first two elements of nums and delete them.The score of the operation is the sum of the deleted elements.Your task is to find the maximum number of operations that can be performed, such that all operations have the same score.Return the maximum number of operations possible that satisfy the condition mentioned above.
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You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.In one operation:choose an index i such that 0 <= i < nums.length,increase your score by nums[i], andreplace nums[i] with ceil(nums[i] / 3).Return the maximum possible score you can attain after applying exactly k operations.
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