通过将任何元素替换为其他元素最多 K 次来最小化 Array 的最大值
给定一个大小为N的数组arr[]和一个整数K ，任务是在将数组的任何元素替换为该数组的任何其他元素最多K次后，最小化数组arr[]的最大元素的值。
例子：
Input: arr[] = {5, 3, 3, 2, 1}, K = 3
Output: 2
Explanation: Replace the elements at index 0, 1 and 2 with the value 1.
The array becomes {1, 1, 1, 2, 1}. The maximum is 2.
This is the minimum possible maximum.
Input: arr[] = {1, -2, 3}, K = 2
Output: -2
解决方法：这个问题可以使用Hash map的概念来解决，思路如下：
If number of elements with value greater than any array elements arr[i] is K and arr[i] is the largest element fulfilling this criteria, then with at most K replacements all those values can be made at least equal to arr[i].
请按照下图进行更好的理解。
插图：
Consider arr[] = {5, 3, 3, 2, 1}, K = 3
For element 5:
=> Number of elements greater than or equal to 5 is 1.
=> K > 1
For element 3:
=> Number of elements greater than or equal to 3 is 3.
=> K ≥ 3. So not this. Because this can be converted to some other element
For element 2:
=> Number of elements greater than or equal to 2 is 4.
=> K < 4. This is the highest such element which satisfies the criteria.
So 2 is the answer.
请按照以下步骤解决问题：
- 如果K = 0 ，则无法替换任何元素，则数组的最大元素保持不变。
- 如果K ≥ N – 1 ，则将所有元素替换为最小元素，因此可以使最大值与数组的最小值相同。
- 否则，使用映射来存储数组元素的出现次数。
- 从数组的最大值开始遍历：
- 如果元素的计数小于K ，则将K递减计数。
- 否则，该元素是最小可能的最大值。
- 返回最小可能最大值的值。
下面是上述方法的实现。
C++
// C++ code to implement the above approach
#include
using namespace std;
// Function to minimize the maximum
// element of the array
int minimizeMax(int* arr, int N, int K)
{
int Ans;
// If K >= (N - 1), then maximum
// element changes to minimum
// element of array
if (K >= (N - 1)) {
Ans = INT_MAX;
for (int i = 0; i < N; i++)
Ans = min(Ans, arr[i]);
}
// If K==0, then maximum element
// remains same
else if (K == 0) {
Ans = INT_MIN;
for (int i = 0; i < N; i++)
Ans = max(Ans, arr[i]);
}
else {
map mp;
for (int i = 0; i < N; i++) {
mp[arr[i]]++;
}
// Create a map reverse iterator
map::reverse_iterator it;
// Traverse map from reverse and
// if K >= Count of current
// element then substract it from
// k and move to next element
// else return the element
for (it = mp.rbegin();
it != mp.rend(); it++) {
if (K >= it->second)
K -= it->second;
else
return it->first;
}
}
// If any of first two conditions
// satisfied then return Ans
return Ans;
}
// Driver Code
int main()
{
int arr[] = { 5, 3, 3, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
// Function call
cout << minimizeMax(arr, N, K) << endl;
return 0;
} Java
// Java code to implement the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to minimize the maximum
// element of the array
public static int minimizeMax(int arr[], int N, int K)
{
int ans = 0;
// If K >= (N - 1), then maximum
// element changes to minimum
// element of array
if (K >= (N - 1)) {
ans = Integer.MAX_VALUE;
for (int i = 0; i < N; i++)
ans = Math.min(ans, arr[i]);
}
// If K==0, then maximum element
// remains same
else if (K == 0) {
ans = Integer.MIN_VALUE;
for (int i = 0; i < N; i++)
ans = Math.max(ans, arr[i]);
}
else
{
// Creating a map in descending order of keys
TreeMap mp
= new TreeMap<>(Collections.reverseOrder());
for (int i = 0; i < N; i++) {
if (mp.get(arr[i]) != null)
mp.put(arr[i], mp.get(arr[i]) + 1);
else
mp.put(arr[i], 1);
}
// Traverse map and
// if K >= Count of current
// element then substract it from
// k and move to next element
// else return the element
for (Map.Entry ele :
mp.entrySet()) {
if (K >= ele.getValue())
K -= ele.getValue();
else
return ele.getKey();
}
}
// If any of first two conditions
// satisfied then return Ans
return ans;
}
public static void main(String[] args)
{
int arr[] = { 5, 3, 3, 2, 1 };
int N = 5;
int K = 3;
// Function call
System.out.println(minimizeMax(arr, N, K));
}
}
// This code is contributed by Rohit Pradhan. Python3
# Python code to implement the above approach
# Function to minimize the maximum
# element of the array
def minimizeMax(arr, N, K):
Ans = 0
# If K >= (N - 1), then maximum
# element changes to minimum
# element of array
if (K >= (N - 1)):
Ans = sys.maxsize
for i in range(0, N):
Ans = min(Ans, arr[i])
# If K==0, then maximum element
# remains same
elif (K == 0):
Ans = -1*sys.maxsize
for i in range(0, N):
Ans = max(Ans, arr[i])
else:
mp = dict()
for i in range(N):
if arr[i] in mp.keys():
mp[arr[i]] += 1
else:
mp[arr[i]] = 1
# Traverse map from reverse and
# if K >= Count of current
# element then substract it from
# k and move to next element
# else return the element
for x in mp:
if (K >= mp[x]):
K -= mp[x]
else:
return x
# If any of first two conditions
# satisfied then return Ans
return Ans
# Driver Code
arr = [5, 3, 3, 2, 1]
N = len(arr)
K = 3
# Function call
print(minimizeMax(arr, N, K))
# This code is contributed by Taranpreet2时间复杂度：O(N)
辅助空间：O(N)