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.
Question
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.
Solution
The problem is asking to return the minimum number of operations needed so that all elements of the array are greater than or equal to k. In one operation, you can remove one occurrence of the smallest element of nums.
Here is the C++ solution:
class Solution {
public:
int minOperations(vector<int>& nums, int k) {
int operations = 0;
sort(nums.begin(), nums.end());
for(int i = 0; i < nums.size(); i++) {
if(nums[i] < k) {
operations++;
} else {
break;
}
}
return operations;
}
};
In this solution, we first sort the array in ascending order. Then we iterate through the array. If the current element is less than k, we increment the operations counter and continue to the next element. If the current element is greater than or equal to k, we break the loop because all the following elements will also be greater than or equal to k due to the sorting. Finally, we return the operations counter which represents the minimum number of operations needed so that all elements of the array are greater than or equal to k.
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.
One or more method can be declared in a class.*
Write a function:class Solution { public int solution(int[] A); }that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.Given A = [1, 2, 3], the function should return 4.Given A = [−1, −3], the function should return 1.Write an efficient algorithm for the following assumptions:N is an integer within the range [1..100,000];each element of array A is an integer within the range [−1,000,000..1,000,000].
public class IntPair { // x int private final int x; // y int private final int y; /** * Constructor to create an instance of IntPair * @param x * @param y */ public IntPair(int x, int y) { this.x = x; this.y = y; } /** * Creates a new IntPair object that is the element-wise sum * * E.g., IntPair(2, 3).add(IntPair(-4, 5)) is equal to IntPair(-2, 8) * * @param other IntPair to add to this IntPair * @return a new IntPair which is the element-wise addition of this and other */
A basic feasible solution is called.................. if the value of at least one basic variable is zeroans.OptimumNo solutionDegenerateNon degenerate Previous Marked for Review Next
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