2024-08-17: In go, given an array of integers nums starting at 0 and an integer k, the
Each operation removes the smallest element of the array.
Your goal is to make all the elements in the array greater than or equal to k with these operations.
Calculate the minimum number of operations required to achieve this.
Input: nums = [2,11,10,1,3], k = 10.
Output: 3.
Explanation: After the first operation, nums becomes [2, 11, 10, 3].
After the second operation, nums becomes [11, 10, 3].
After the third operation, nums becomes [11, 10].
At this point, all the elements in the array are greater than or equal to 10, so we stop the operation.
The minimum number of operations required to make all elements of an array greater than or equal to 10 is 3.
Answer 2024-08-17:
chatgpt
Title from leetcode3065.
The broad steps are as follows:
1. Iterate through the arraynums, for elements less than k, the number of operations will beansPlus 1.
2. In the given example, at the initial timenumsis [2, 11, 10, 1, 3] and k is 10. After the first operation, the smallest element 1 is deleted to get [2, 11, 10, 3] with operation number 1.
3. After the second operation, the smallest element 2 is deleted, giving [11, 10, 3], and the number of operations is 2.
4. After the third operation, the smallest element 3 is deleted to get [11, 10], and the number of operations is 3.
5. At this point, all elements in the array are greater than or equal to 10, the operation stops, so that all elements in the array is greater than or equal to 10 the minimum number of operations required is 3.
The total time complexity is O(n), where n is the arraynumslength, each element will be traversed at most once.
The total additional space complexity is O(1), no additional data structures are used to store the intermediate results, and only a constant level of additional space is consumed.
The full Go code is below:
package main
import (
"fmt"
)
func minOperations(nums []int, k int) (ans int) {
for _, x := range nums {
if x < k {
ans++
}
}
return
}
func main() {
nums := []int{2,11,10,1,3}
k:=10
(minOperations(nums, k))
}
The full Python code is below:
# -*-coding:utf-8-*-
def min_operations(nums, k):
ans = sum(1 for x in nums if x < k)
return ans
nums = [2, 11, 10, 1, 3]
k = 10
print(min_operations(nums, k))
