给定一个由N个正整数和一个正整数K组成的数组arr [] ，任务是通过将最多K个数组元素拆分为两个等于其值的数字，以最小化数组中存在的最大元素。
例子：
Input: arr[] = {2, 4, 8, 2}, K = 4
Output: 2
Explanation:
Following sequence of operations are required to be performed:
Operation 1: Splitting arr[1] (= 4) to {2, 2} modifies the array to {2, 2, 2, 8, 2}.
Operation 2: Splitting arr[3] (= 8) to {2, 6} modifies the array to {2, 2, 2, 2, 6, 2}.
Operation 3: Splitting arr[4] (= 6) to {2, 4} modifies the array to {2, 2, 2, 2, 2, 4, 2}.
Operation 4: Splitting arr[5] (= 4) to {2, 2} modifies the array to {2, 2, 2, 2, 2, 2, 2, 2}.
After completing the above operations, the maximum element present in the array is 2.
Input: arr[] = {7, 17}, K = 2
Output: 7
方法：可以根据以下观察结果解决给定问题：
- 如果通过执行最多K次操作， X可以成为数组arr []中的最大元素，则存在某个值K(K> X) ，也可以是数组arr []中存在的最大元素 通过最多对数组元素进行K个拆分。
- 如果通过执行最多K次运算， X不能成为数组A []中的最大元素，则存在某个值K(K
- 因此，我们的想法是使用二进制搜索来查找[1， INT_MAX ]范围内的值，该值可以是最多K个分割后的可能最大值。
请按照以下步骤解决问题：
- 初始化两个变量，例如low和high分别为1和数组arr []中的最大元素 分别。
- 迭代直到low小于high并执行以下步骤：
- 找到范围[low，high]的中间值，作为mid =(low + high)/ 2 。
- 初始化一个变量，例如count ，以存储使最大元素等于mid所需的数组元素拆分的最大数目。
- 遍历给定的数组arr []并将count的值更新为(arr [i] – 1)/ mid以计算所需的分割数。
- 如果count的值最多为K ，则将high的值更新为mid 。
- 否则，将low的值更新为(mid +1) 。
- 完成上述步骤后，将high的值打印为所获得的数组中存在的结果最大元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
int possible(int A[], int N,
int mid, int K)
{
// Stores the number
// of splits required
int count = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
// Update count
count += (A[i] - 1) / mid;
}
// If possible, return true.
// Otherwise return false
return count <= K;
}
// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
int minimumMaximum(int A[], int N, int K)
{
// Set lower and upper limits
int lo = 1;
int hi = *max_element(A, A + N);
int mid;
// Perform Binary Search
while (lo < hi) {
// Calculate mid
mid = (lo + hi) / 2;
// Check if all array elements
// can be reduced to at most
// mid value by at most K splits
if (possible(A, N, mid, K)) {
// Update the value of hi
hi = mid;
}
// Otherwise
else {
// Update the value of lo
lo = mid + 1;
}
}
// Return the minimized maximum
// element in the array
return hi;
}
// Driver Code
int main()
{
int arr[] = { 2, 4, 8, 2 };
int K = 4;
int N = sizeof(arr) / sizeof(arr[0]);
cout << minimumMaximum(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
static boolean possible(int A[], int N,
int mid, int K)
{
// Stores the number
// of splits required
int count = 0;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Update count
count += (A[i] - 1) / mid;
}
// If possible, return true.
// Otherwise return false
return count <= K;
}
// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
static int minimumMaximum(int A[], int N, int K)
{
// Set lower and upper limits
int lo = 1;
Arrays.sort(A);
int hi = A[N - 1];
int mid;
// Perform Binary Search
while (lo < hi)
{
// Calculate mid
mid = (lo + hi) / 2;
// Check if all array elements
// can be reduced to at most
// mid value by at most K splits
if (possible(A, N, mid, K))
{
// Update the value of hi
hi = mid;
}
// Otherwise
else
{
// Update the value of lo
lo = mid + 1;
}
}
// Return the minimized maximum
// element in the array
return hi;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 2, 4, 8, 2 };
int K = 4;
int N = arr.length;
System.out.println(minimumMaximum(arr, N, K));
}
}
// This code is contributed by AnkThonPython3
# Python3 program for the above approach
# Function to check if all array
# elements can be reduced to at
# most mid by at most K splits
def possible(A, N, mid, K):
# Stores the number
# of splits required
count = 0
# Traverse the array arr[]
for i in range(N):
# Update count
count += (A[i] - 1) // mid
# If possible, return true.
# Otherwise return false
return count <= K
# Function to find the minimum possible
# value of maximum array element that
# can be obtained by at most K splits
def minimumMaximum(A, N, K):
# Set lower and upper limits
lo = 1
hi = max(A)
# Perform Binary Search
while (lo < hi):
# Calculate mid
mid = (lo + hi) // 2
# Check if all array elements
# can be reduced to at most
# mid value by at most K splits
if (possible(A, N, mid, K)):
# Update the value of hi
hi = mid
# Otherwise
else:
# Update the value of lo
lo = mid + 1
# Return the minimized maximum
# element in the array
return hi
# Driver Code
if __name__ == '__main__':
arr = [ 2, 4, 8, 2 ]
K = 4
N = len(arr)
print(minimumMaximum(arr, N, K))
# This code is contributed by ipg2016107C#
// C# program for the above approach
using System;
class GFG{
// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
static bool possible(int[] A, int N,
int mid, int K)
{
// Stores the number
// of splits required
int count = 0;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Update count
count += (A[i] - 1) / mid;
}
// If possible, return true.
// Otherwise return false
return count <= K;
}
// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
static int minimumMaximum(int[] A, int N, int K)
{
// Set lower and upper limits
int lo = 1;
Array.Sort(A);
int hi = A[N - 1];
int mid;
// Perform Binary Search
while (lo < hi)
{
// Calculate mid
mid = (lo + hi) / 2;
// Check if all array elements
// can be reduced to at most
// mid value by at most K splits
if (possible(A, N, mid, K))
{
// Update the value of hi
hi = mid;
}
// Otherwise
else
{
// Update the value of lo
lo = mid + 1;
}
}
// Return the minimized maximum
// element in the array
return hi;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 2, 4, 8, 2 };
int K = 4;
int N = arr.Length;
Console.WriteLine(minimumMaximum(arr, N, K));
}
}
// This code is contributed by ukasp输出：
2时间复杂度： O(N * log M)，其中M是数组的最大元素。
辅助空间： O(1)