给定一个整数数组和 k，找出大小为 k 的窗口中严格增加的子数组(大小大于 1)的数量与严格减少的子数组的数量之间的差。
例子：
Input : nums = {10, 20, 30, 15, 15};
k = 3;
Output : 3, 0, -1
Explanation
For the first window of [10, 20, 30], there are
3 increasing subranges ([10, 20, 30], [10, 20],
and [20, 30]) and 0 decreasing, so the answer
is 3. For the second window of [20, 30, 15],
there is 1 increasing subrange and 1 decreasing,
so the answer is 0. For the third window of
[20, 15, 15], there is 1 decreasing subrange
and 0 increasing, so the answer is -1. 我们需要计算每个大小为 K 的窗口的递增和递减子数组之间的差异。 我们遍历数组，对于每个大小为 k 的窗口，我们计算递增和递减子数组的数量并插入它们的差异在输出数组中。
- 创建两个大小为 N 的数组并将它们初始化为零并创建输出数组。
- 遍历数组
- 计算递增子数组的个数
- 计算递减的子数组
- 在输出数组中插入递增和递减子数组之间的差异。
解决方案的时间复杂度 – O(n*k)
n = 数组的大小
k = 窗口大小
C++
// CPP program to count the differences
#include
#include
using namespace std;
// function to calculate the difference
vector Diffs(vector const& a, int k)
{
vector out;
vector inc, dec;
// initializing inc and dec with 0 and resizing
// equal to the size of main array
inc.resize(a.size(), 0);
dec.resize(a.size(), 0);
int inc_sum = 0;
int dec_sum = 0;
// iterate through the array
for (int i = 0; i < a.size(); ++i) {
// finding number of increasing
// subarrays in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] > a[j]; --j) {
++inc[j];
++inc_sum;
}
// Finding number of decreasing subarrays
// in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] < a[j]; --j) {
++dec[j];
++dec_sum;
}
// calculate the difference
if (i >= k - 1) {
// if this is not the first window then
// calculate inc_sum and dec_sum
if (i >= k) {
inc_sum -= inc[i - k];
dec_sum -= dec[i - k];
}
// insert the difference in k size window
// in the output vector
out.push_back(inc_sum - dec_sum);
}
}
return out;
}
// driver program
int main()
{
vector out = Diffs({ 10, 20, 30, 15, 15}, 3);
for (int n : out)
cout << n << ", ";
return 0;
} Java
// JAVA program to count the differences
import java.util.*;
class GFG
{
// function to calculate the difference
static Vector Diffs(int []a, int k)
{
Vector out = new Vector();
int [] inc, dec;
// initializing inc and dec with 0 and resizing
// equal to the size of main array
inc = new int[a.length];
dec = new int[a.length];
int inc_sum = 0;
int dec_sum = 0;
// iterate through the array
for (int i = 0; i < a.length; ++i)
{
// finding number of increasing
// subarrays in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] > a[j]; --j)
{
++inc[j];
++inc_sum;
}
// Finding number of decreasing subarrays
// in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] < a[j]; --j)
{
++dec[j];
++dec_sum;
}
// calculate the difference
if (i >= k - 1)
{
// if this is not the first window then
// calculate inc_sum and dec_sum
if (i >= k)
{
inc_sum -= inc[i - k];
dec_sum -= dec[i - k];
}
// insert the difference in k size window
// in the output vector
out.add(inc_sum - dec_sum);
}
}
return out;
}
// Driver code
public static void main(String[] args)
{
int []arr = { 10, 20, 30, 15, 15};
Vectorout = Diffs(arr, 3);
for (int n : out)
System.out.print(n + ", ");
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program to count the differences
# Function to calculate the difference
def Diffs(a, k):
out, inc, dec = [], [0] * len(a), [0] * len(a)
# Initializing inc and dec with 0 and
# resizing equal to the size of main array
inc_sum, dec_sum = 0, 0
# Iterate through the array
for i in range(0, len(a)):
# Finding number of increasing
# subarrays in a window size k
j = i - 1
while (j >= 0 and j > i - k and
a[j + 1] > a[j]):
inc[j] += 1
inc_sum += 1
j -= 1
# Finding number of decreasing
# subarrays in a window size k
j = i - 1
while (j >= 0 and j > i - k and
a[j + 1] < a[j]):
dec[j] += 1
dec_sum += 1
j -= 1
# calculate the difference
if i >= k - 1:
# if this is not the first window then
# calculate inc_sum and dec_sum
if i >= k:
inc_sum -= inc[i - k]
dec_sum -= dec[i - k]
# insert the difference in k size window
# in the output vector
out.append(inc_sum - dec_sum)
return out
# Driver Code
if __name__ == "__main__":
out = Diffs([10, 20, 30, 15, 15], 3)
for n in out:
print(n, end = ", ")
# This code is contributed by Rituraj JainC#
// C# program to count the differences
using System;
using System.Collections.Generic;
class GFG
{
// function to calculate the difference
static List Diffs(int []a, int k)
{
List Out = new List();
int [] inc, dec;
// initializing inc and dec with 0 and resizing
// equal to the size of main array
inc = new int[a.Length];
dec = new int[a.Length];
int inc_sum = 0;
int dec_sum = 0;
// iterate through the array
for (int i = 0; i < a.Length; ++i)
{
// finding number of increasing
// subarrays in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] > a[j]; --j)
{
++inc[j];
++inc_sum;
}
// Finding number of decreasing subarrays
// in a window size k
for (int j = i - 1; j >= 0 && j > i - k &&
a[j + 1] < a[j]; --j)
{
++dec[j];
++dec_sum;
}
// calculate the difference
if (i >= k - 1)
{
// if this is not the first window then
// calculate inc_sum and dec_sum
if (i >= k)
{
inc_sum -= inc[i - k];
dec_sum -= dec[i - k];
}
// insert the difference in k size window
// in the output vector
Out.Add(inc_sum - dec_sum);
}
}
return Out;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 10, 20, 30, 15, 15};
ListOut = Diffs(arr, 3);
foreach (int n in Out)
Console.Write(n + ", ");
}
}
// This code is contributed by Rajput-Ji Javascript
输出：
3, 0, -1, 如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。