给定大小为N且整数Y的数组arr [] ，任务是在大小为Y的所有子数组中的最大元素和最小元素之间找到所有差值中的最小值。
例子：
Input: arr[] = { 3, 2, 4, 5, 6, 1, 9 } Y = 3
Output: 2
Explanation:
All subarrays of length = 3 are:
{3, 2, 4} where maximum element = 4 and minimum element = 2 difference = 2
{2, 4, 5} where maximum element = 5 and minimum element = 2 difference = 3
{4, 5, 6} where maximum element = 6 and minimum element = 4 difference = 2
{5, 6, 1} where maximum element = 6 and minimum element = 1 difference = 5
{6, 1, 9} where maximum element = 9 and minimum element = 6 difference = 3
Out of these minimum is 2.
Input: arr[] = { 1, 2, 3, 3, 2, 2 } Y = 4
Output: 1
Explanation:
All subarrays of length = 4 are:
{1, 2, 3, 3} maximum element = 3 and minimum element = 1 difference = 2
{2, 3, 3, 2} maximum element = 3 and minimum element = 2 difference = 1
{3, 3, 2, 2} maximum element = 3 and minimum element = 2 difference = 1
Out of these minimum is 1.
天真的方法：天真的想法是遍历范围[0，N – Y]中的每个索引i，使用另一个循环从第i个索引遍历到(i + Y – 1)个索引，然后计算最小和最大元素在大小为Y的子数组中，因此计算第i次迭代的最大元素和最小元素之差。最后，通过检查差异，评估最小差异。
时间复杂度： O(N * Y)
辅助空间： O(1)
高效的方法：其想法是使用本手册中讨论的方法的概念 下一个更大元素文章。步骤如下：
- 构建两个数组maxarr []和minarr [] ，其中maxarr []将存储元素的索引，该索引比第i个索引的元素大一个，而minarr []将存储下一个元素的索引，该索引小于该元素的索引在第i个索引处。
- 在上述两种情况下，都使用0初始化堆栈以存储索引。
- 对于每个索引， i在[0，N – Y]范围内，使用嵌套循环和滑动窗口方法形成两个数组submax和submin。这些数组将在第i次迭代中将最大和最小元素存储在子数组中。
- 最后，计算最小差异。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to get the maximum of all
// the subarrays of size Y
vector get_submaxarr(int* arr,
int n, int y)
{
int j = 0;
stack stk;
// ith index of maxarr array
// will be the index upto which
// Arr[i] is maximum
vector maxarr(n);
stk.push(0);
for (int i = 1; i < n; i++) {
// Stack is used to find the
// next larger element and
// keeps track of index of
// current iteration
while (stk.empty() == false
and arr[i] > arr[stk.top()]) {
maxarr[stk.top()] = i - 1;
stk.pop();
}
stk.push(i);
}
// Loop for remaining indexes
while (!stk.empty()) {
maxarr[stk.top()] = n - 1;
stk.pop();
}
vector submax;
for (int i = 0; i <= n - y; i++) {
// j < i used to keep track
// whether jth element is
// inside or outside the window
while (maxarr[j] < i + y - 1
or j < i) {
j++;
}
submax.push_back(arr[j]);
}
// Return submax
return submax;
}
// Function to get the minimum for
// all subarrays of size Y
vector get_subminarr(int* arr,
int n, int y)
{
int j = 0;
stack stk;
// ith index of minarr array
// will be the index upto which
// Arr[i] is minimum
vector minarr(n);
stk.push(0);
for (int i = 1; i < n; i++) {
// Stack is used to find the
// next smaller element and
// keeping track of index of
// current iteration
while (stk.empty() == false
and arr[i] < arr[stk.top()]) {
minarr[stk.top()] = i;
stk.pop();
}
stk.push(i);
}
// Loop for remaining indexes
while (!stk.empty()) {
minarr[stk.top()] = n;
stk.pop();
}
vector submin;
for (int i = 0; i <= n - y; i++) {
// j < i used to keep track
// whether jth element is inside
// or outside the window
while (minarr[j] <= i + y - 1
or j < i) {
j++;
}
submin.push_back(arr[j]);
}
// Return submin
return submin;
}
// Function to get minimum difference
void getMinDifference(int Arr[], int N,
int Y)
{
// Create submin and submax arrays
vector submin
= get_subminarr(Arr, N, Y);
vector submax
= get_submaxarr(Arr, N, Y);
// Store initial difference
int minn = submax[0] - submin[0];
int b = submax.size();
for (int i = 1; i < b; i++) {
// Calculate temporary difference
int dif = submax[i] - submin[i];
minn = min(minn, dif);
}
// Final minimum difference
cout << minn << "\n";
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, 3, 3, 2, 2 };
int N = sizeof arr / sizeof arr[0];
// Given subarray size
int Y = 4;
// Function Call
getMinDifference(arr, N, Y);
return 0;
} Python3
# Python3 program for the above approach
# Function to get the maximum of all
# the subarrays of size Y
def get_submaxarr(arr, n, y):
j = 0
stk = []
# ith index of maxarr array
# will be the index upto which
# Arr[i] is maximum
maxarr = [0] * n
stk.append(0)
for i in range(1, n):
# Stack is used to find the
# next larger element and
# keeps track of index of
# current iteration
while (len(stk) > 0 and
arr[i] > arr[stk[-1]]):
maxarr[stk[-1]] = i - 1
stk.pop()
stk.append(i)
# Loop for remaining indexes
while (stk):
maxarr[stk[-1]] = n - 1
stk.pop()
submax = []
for i in range(n - y + 1):
# j < i used to keep track
# whether jth element is
# inside or outside the window
while (maxarr[j] < i + y - 1 or
j < i):
j += 1
submax.append(arr[j])
# Return submax
return submax
# Function to get the minimum for
# all subarrays of size Y
def get_subminarr(arr, n, y):
j = 0
stk = []
# ith index of minarr array
# will be the index upto which
# Arr[i] is minimum
minarr = [0] * n
stk.append(0)
for i in range(1 , n):
# Stack is used to find the
# next smaller element and
# keeping track of index of
# current iteration
while (stk and arr[i] < arr[stk[-1]]):
minarr[stk[-1]] = i
stk.pop()
stk.append(i)
# Loop for remaining indexes
while (stk):
minarr[stk[-1]] = n
stk.pop()
submin = []
for i in range(n - y + 1):
# j < i used to keep track
# whether jth element is inside
# or outside the window
while (minarr[j] <= i + y - 1 or
j < i):
j += 1
submin.append(arr[j])
# Return submin
return submin
# Function to get minimum difference
def getMinDifference(Arr, N, Y):
# Create submin and submax arrays
submin = get_subminarr(Arr, N, Y)
submax = get_submaxarr(Arr, N, Y)
# Store initial difference
minn = submax[0] - submin[0]
b = len(submax)
for i in range(1, b):
# Calculate temporary difference
diff = submax[i] - submin[i]
minn = min(minn, diff)
# Final minimum difference
print(minn)
# Driver code
# Given array arr[]
arr = [ 1, 2, 3, 3, 2, 2 ]
N = len(arr)
# Given subarray size
Y = 4
# Function call
getMinDifference(arr, N, Y)
# This code is contributed by Stuti Pathak1
时间复杂度： O(N)
辅助空间： O(N)
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