2024-08-21: In go, given an array of integers nums indexed from 0 and an integer k, design an algorithm to make all elements in the array greater than or equal to k, returning the minimum number of operations required.
The following steps can be performed for each operation:
1. Select the two smallest integers x and y in the array.
2. Remove x and y from the array.
3. Calculate the value of min(x, y) * 2 + max(x, y) and add it back to any position in the array.
Repeat the above steps until all the elements in the array are greater than or equal to k.
Make sure there are at least two elements in the array to perform the operation.
Redesign an algorithm based on the above requirements such that all array elements are greater than or equal to k in the least number of operations.
Input: nums = [2,11,10,1,3], k = 10.
Output: 2.
Explanation: In the first operation, we delete the elements 1 and 2 and add 1 * 2 + 2 to nums, which becomes [4, 11, 10, 3].
In the second operation, we delete the elements 3 and 4 and add 3 * 2 + 4 to nums, which becomes [10, 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 2.
Answer 2024-08-21:
chatgpt
Title from leetcode3066.
The broad steps are as follows:
1. Create a structure hp containing an array to store the incoming integer array nums.
2. Initialize the hp structure, store the nums in it, and convert it to a minimal heap structure.
3. Enter the loop, determine whether the smallest value in the smallest heap is less than or equal to k, if so, perform the following steps, otherwise end the loop:
. Pop the minimum value x from the minimum heap.
. Double the value of x and put it back on top of the smallest heap pair and correct the heap structure.
. The counter ans increments itself by 1 to indicate that an operation was performed.
4. Return the minimum number of operations ans.
Total time complexity:
-
Initializing the heap structure has a time complexity of O(n).
-
The time complexity of popping elements from the heap and modifying the heap structure in each loop is O(log(n)) and is executed at most n times.
Therefore, the total time complexity is O(n log n).
Total extra space complexity:
- In addition to storing the input array, an additional heap structure is used to maintain the minimum value, resulting in an additional space complexity of O(n).
The full Go code is below:
package main
import (
"fmt"
"container/heap"
"sort"
)
func minOperations(nums []int, k int) (ans int) {
h := &hp{nums}
(h)
for [0] < k {
x := (h).(int)
[0] += x * 2
(h, 0)
ans++
}
return
}
type hp struct{ }
func (hp) Push(any) {}
func (h *hp) Pop() any { a := ; v := a[len(a)-1]; = a[:len(a)-1]; return v }
func main() {
nums := []int{2, 11, 10, 1, 3}
k := 10
(minOperations(nums, k))
}
The full Python code is below:
# -*-coding:utf-8-*-
import heapq
class Hp(list):
def push(self, item):
(self, item)
def pop(self):
return (self)
def minOperations(nums, k):
h = Hp(nums)
(h)
ans = 0
while h[0] < k:
x = ()
h[0] += x * 2
(h, h[0])
ans += 1
return ans
nums = [2, 11, 10, 1, 3]
k = 10
print(minOperations(nums, k))
