给定一个由大小为N的正整数和一个正整数K 组成的数组arr[] ，任务是找到一个子数组的最大可能长度，通过向子数组的每个元素添加一些整数值，使得该长度相等添加元素的总和不超过K 。
例子：
Input: N = 5, arr[] = {1, 4, 9, 3, 6}, K = 9
Output: 3
Explanation:
{1, 4} : {1+3, 4} = {4, 4}
{4, 9} : {4+5, 9} = {9, 9}
{3, 6} : {3+3, 6} = {6, 6}
{9, 3, 6} : {9, 3+6, 6+3} = {9, 9, 9}
Hence, the maximum length of such a subarray is 3.
Input: N = 6, arr[] = {2, 4, 7, 3, 8, 5}, K = 10
Output: 4
方法：这个问题可以用动态规划解决。
- 初始化：
- dp[] ：存储添加到子数组的元素的总和。
- deque ：存储每个子数组的最大元素的索引。
- pos：子数组当前位置的索引。
- ans：最大子数组的长度。
- mx：子数组的最大元素
- pre：当前子数组的上一个索引。
- 遍历数组并检查双端队列是否为空。如果是，则更新最大元素和最大元素的索引以及pre和pos的索引。
- 检查当前添加的元素是否大于K 。如果是，则将其从dp[]数组中删除并更新pos和pre的索引。
- 最后，更新有效子数组的最大长度。
下面是上述方法的实现：
C++
// C++ code for the above approach
#include
using namespace std;
// Function to find maximum
// possible length of subarray
int validSubArrLength(int arr[],
int N, int K)
{
// Stores the sum of elements
// that needs to be added to
// the sub array
int dp[N + 1];
// Stores the index of the
// current position of subarray
int pos = 0;
// Stores the maximum
// length of subarray.
int ans = 0;
// Maximum element from
// each subarray length
int mx = 0;
// Previous index of the
// current subarray of
// maximum length
int pre = 0;
// Deque to store the indices
// of maximum element of
// each sub array
deque q;
// For each array element,
// find the maximum length of
// required subarray
for (int i = 0; i < N; i++) {
// Traverse the deque and
// update the index of
// maximum element.
while (!q.empty()
&& arr[q.back()] < arr[i])
q.pop_back();
q.push_back(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0) {
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i]) {
// Update max and dp[]
dp[i] = dp[i - 1] + arr[i];
mx = arr[i];
}
else {
dp[i] = dp[i - 1] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0)
pos = 0;
else
pos = pre - 1;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1) * mx
- (dp[i] - dp[pos])
> K
&& pre < i) {
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements
// from deque and
// update the
// maximum element
while (!q.empty()
&& q.front() < pre
&& pre < i) {
q.pop_front();
mx = arr[q.front()];
}
}
// Update the maximum length
// of the required subarray.
ans = max(ans, i - pre + 1);
}
return ans;
}
// Driver Program
int main()
{
int N = 6;
int K = 8;
int arr[] = { 2, 7, 1, 3, 4, 5 };
cout << validSubArrLength(arr, N, K);
return 0;
} Java
// Java code for the above approach
import java.util.*;
class GFG{
// Function to find maximum
// possible length of subarray
static int validSubArrLength(int arr[],
int N, int K)
{
// Stores the sum of elements
// that needs to be added to
// the sub array
int []dp = new int[N + 1];
// Stores the index of the
// current position of subarray
int pos = 0;
// Stores the maximum
// length of subarray.
int ans = 0;
// Maximum element from
// each subarray length
int mx = 0;
// Previous index of the
// current subarray of
// maximum length
int pre = 0;
// Deque to store the indices
// of maximum element of
// each sub array
Deque q = new LinkedList<>();
// For each array element,
// find the maximum length of
// required subarray
for(int i = 0; i < N; i++)
{
// Traverse the deque and
// update the index of
// maximum element.
while (!q.isEmpty() &&
arr[q.getLast()] < arr[i])
q.removeLast();
q.add(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0)
{
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i])
{
// Update max and dp[]
dp[i] = dp[i - 1] + arr[i];
mx = arr[i];
}
else
{
dp[i] = dp[i - 1] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0)
pos = 0;
else
pos = pre - 1;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1) * mx -
(dp[i] - dp[pos]) > K && pre < i)
{
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements from
// deque and update the
// maximum element
while (!q.isEmpty() &&
q.peek() < pre && pre < i)
{
q.removeFirst();
mx = arr[q.peek()];
}
}
// Update the maximum length
// of the required subarray.
ans = Math.max(ans, i - pre + 1);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 6;
int K = 8;
int arr[] = { 2, 7, 1, 3, 4, 5 };
System.out.print(validSubArrLength(arr, N, K));
}
}
// This code is contributed by amal kumar choubey Python3
# Python3 code for the above approach
# Function to find maximum
# possible length of subarray
def validSubArrLength(arr, N, K):
# Stores the sum of elements
# that needs to be added to
# the sub array
dp = [0 for i in range(N + 1)]
# Stores the index of the
# current position of subarray
pos = 0
# Stores the maximum
# length of subarray.
ans = 0
# Maximum element from
# each subarray length
mx = 0
# Previous index of the
# current subarray of
# maximum length
pre = 0
# Deque to store the indices
# of maximum element of
# each sub array
q = []
# For each array element,
# find the maximum length of
# required subarray
for i in range(N):
# Traverse the deque and
# update the index of
# maximum element.
while (len(q) and arr[len(q) - 1] < arr[i]):
q.remove(q[len(q) - 1])
q.append(i)
# If it is first element
# then update maximum
# and dp[]
if (i == 0):
mx = arr[i]
dp[i] = arr[i]
# Else check if current
# element exceeds max
elif (mx <= arr[i]):
# Update max and dp[]
dp[i] = dp[i - 1] + arr[i]
mx = arr[i]
else:
dp[i] = dp[i - 1] + arr[i]
# Update the index of the
# current maximum length
# subarray
if (pre == 0):
pos = 0
else:
pos = pre - 1
# While current element
# being added to dp[] array
# exceeds K
while ((i - pre + 1) *
mx - (dp[i] - dp[pos]) > K and
pre < i):
# Update index of
# current position and
# the previous position
pos = pre
pre += 1
# Remove elements
# from deque and
# update the
# maximum element
while (len(q) and
q[0] < pre and
pre < i):
q.remove(q[0])
mx = arr[q[0]]
# Update the maximum length
# of the required subarray.
ans = max(ans, i - pre + 1)
return ans
# Driver code
if __name__ == '__main__':
N = 6
K = 8
arr = [ 2, 7, 1, 3, 4, 5 ]
print(validSubArrLength(arr, N, K))
# This code is contributed by ipg2016107Javascript
输出：
4时间复杂度： O(N 2 )
辅助空间复杂度： O(N)
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