给定由N个整数和整数K组成的数组A [] ，任务是在对数组元素执行最多K个增量1之后，最大化具有相等元素的子数组的长度。
注意：同一数组元素可以增加一次以上。
例子：
Input: A[] = {2, 4, 8, 5, 9, 6}, K = 6
Output: 3
Explanation:
Subarray [8, 5, 9] can be modified to [9, 9, 9].
Total number of increments required is 5 which is less than K(= 6).
Input: A[] = {2, 2, 4}, K = 10
Output: 3
天真的方法：解决该问题的最简单方法是生成所有可能的子数组，并针对每个子数组检查所有元素是否最多可以K个增量相等。打印获得的此类子数组的最大长度。
时间复杂度： O(N 3 )
辅助空间： O(1)
方法：可以使用“滑动窗口”技术优化上述方法。请按照以下步骤解决问题：
- 跟踪窗口的最大元素。
- 特定窗口所需的全部操作可通过以下公式获得：
Count of operations = (Length of the window calculated so far + 1) * (Maximum element from the window) – Sum of the window
- 现在，检查以上计算的值是否超过K。如果是这样，则将窗口的起始指针向右滑动，否则增加到目前为止计算的窗口的长度。
- 重复上述步骤，以获得满足所需条件的最长窗口。
下面是上述方法的实现：
C++14
// C++14 program for above approach
#include
using namespace std;
#define newl "\n"
// Function to find the maximum length
// of subarray of equal elements after
// performing at most K increments
int maxSubarray(int a[], int k, int n)
{
// Stores the size
// of required subarray
int answer = 0;
// Starting point of a window
int start = 0;
// Stores the sum of window
long int s = 0;
deque dq;
// Iterate over arrray
for(int i = 0; i < n; i++)
{
// Current element
int x = a[i];
// Remove index of minimun elements
// from deque which are less than
// the current element
while (!dq.empty() &&
a[dq.front()] <= x)
dq.pop_front();
// Insert current index in deque
dq.push_back(i);
// Update current window sum
s += x;
// Calculate required operation to
// make current window elements equal
long int cost = (long int)a[dq.front()] *
(answer + 1) - s;
// If cost is less than k
if (cost <= (long int)k)
answer++;
// Shift window start pointer towards
// right and update current window sum
else
{
if (dq.front() == start)
dq.pop_front();
s -= a[start++];
}
}
// Return answer
return answer;
}
// Driver Code
int main()
{
int a[] = { 2, 2, 4 };
int k = 10;
// Length of array
int n = sizeof(a) / sizeof(a[0]);
cout << (maxSubarray(a, k, n));
return 0;
}
// This code is contributed by jojo9911 Java
// Java Program for above approach
import java.util.*;
class GFG {
// Function to find the maximum length
// of subarray of equal elements after
// performing at most K increments
static int maxSubarray(int[] a, int k)
{
// Length of array
int n = a.length;
// Stores the size
// of required subarray
int answer = 0;
// Starting point of a window
int start = 0;
// Stores the sum of window
long s = 0;
Deque dq = new LinkedList<>();
// Iterate over arrray
for (int i = 0; i < n; i++) {
// Current element
int x = a[i];
// Remove index of minimun elements
// from deque which are less than
// the current element
while (!dq.isEmpty() && a[dq.peek()] <= x)
dq.poll();
// Insert current index in deque
dq.add(i);
// Update current window sum
s += x;
// Calculate required operation to
// make current window elements equal
long cost
= (long)a[dq.peekFirst()] * (answer + 1)
- s;
// If cost is less than k
if (cost <= (long)k)
answer++;
// Shift window start pointer towards
// right and update current window sum
else {
if (dq.peekFirst() == start)
dq.pollFirst();
s -= a[start++];
}
}
// Return answer
return answer;
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 2, 2, 4 };
int k = 10;
// Funtion call
System.out.println(maxSubarray(a, k));
}
} Python3
# Python3 program for above approach
from collections import deque
# Function to find the maximum length
# of subarray of equal elements after
# performing at most K increments
def maxSubarray(a, k):
# Length of array
n = len(a)
# Stores the size
# of required subarray
answer = 0
# Starting po of a window
start = 0
# Stores the sum of window
s = 0
dq = deque()
# Iterate over arrray
for i in range(n):
# Current element
x = a[i]
# Remove index of minimun elements
# from deque which are less than
# the current element
while (len(dq) > 0 and a[dq[-1]] <= x):
dq.popleft()
# Insert current index in deque
dq.append(i)
# Update current window sum
s += x
# Calculate required operation to
# make current window elements equal
cost = a[dq[0]] * (answer + 1) - s
# If cost is less than k
if (cost <= k):
answer += 1
# Shift window start poer towards
# right and update current window sum
else:
if (dq[0] == start):
dq.popleft()
s -= a[start]
start += 1
# Return answer
return answer
# Driver Code
if __name__ == '__main__':
a = [ 2, 2, 4 ]
k = 10
# Funtion call
print(maxSubarray(a, k))
# This code is contributed by mohit kumar 29C#
// C# Program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum length
// of subarray of equal elements after
// performing at most K increments
static int maxSubarray(int[] a, int k)
{
// Length of array
int n = a.Length;
// Stores the size
// of required subarray
int answer = 0;
// Starting point of a window
int start = 0;
// Stores the sum of window
long s = 0;
Queue dq = new Queue();
// Iterate over arrray
for (int i = 0; i < n; i++)
{
// Current element
int x = a[i];
// Remove index of minimun
// elements from deque
// which are less than
// the current element
while (dq.Count!=0 &&
a[dq.Peek()] <= x)
dq.Dequeue();
// Insert current
// index in deque
dq.Enqueue(i);
// Update current window sum
s += x;
// Calculate required operation to
// make current window elements equal
long cost = (long)a[dq.Peek()] *
(answer + 1) - s;
// If cost is less than k
if (cost <= (long)k)
answer++;
// Shift window start pointer towards
// right and update current window sum
else
{
if (dq.Peek() == start)
dq.Dequeue();
s -= a[start++];
}
}
// Return answer
return answer;
}
// Driver Code
public static void Main(String[] args)
{
int[] a = {2, 2, 4};
int k = 10;
// Funtion call
Console.WriteLine(maxSubarray(a, k));
}
}
// This code is contributed by gauravrajput1 输出：
3
时间复杂度： O(N)
辅助空间： O(N)