给定一个大小为N的数组arr[] ，任务是找到最大值小于剩余后缀数组中最大元素的前缀数组的数量。
例子：
Input: arr[] = {2, 3, 4, 8, 1, 4}
Output: 3
Explanation:
Prefix array = {2}, {2, 3}, {2, 3, 4}, {2, 3, 4, 8}, {2, 3, 4, 8, 1}
Respective Suffix = {3, 4, 8, 1, 4}, {4, 8, 1, 4}, {8, 1, 4}, {1, 4}, {4}
Only the first 3 the prefix arrays have maximum valued element less than the maximum value element in the respective suffix array.
Input: arr[] = {4, 4, 4, 4, 5}
Output: 4
朴素的方法：最简单的方法是找到给定数组的所有可能的前缀和后缀，并计算前缀数组中最大元素小于后缀数组中最大元素的索引数量。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效方法：对上述方法进行优化，其思想是存储数组中每个前缀和后缀的最大值，然后计算最大值小于其对应后缀数组的前缀数组的数量。请按照以下步骤解决问题：
- 初始化一个变量，比如ans和两个大小为N 的数组prefix[]和suffix[] 。
- 在[0, N – 1]范围内遍历数组arr[]并且对于prefix[] 中的每个第i个索引，将最大元素存储为prefix[i] = max(prefix[i – 1], arr[i] ) 。
- 在[N – 1, 0]范围内以相反的顺序遍历给定数组，对于suffix[] 中的每个第i个索引，将最大元素存储为suffix[i] = max(suffix[i + 1], arr[我]） 。
- 现在，在[0, N – 2]范围内遍历数组arr[] ，如果prefix[i]小于suffix[i] ，则将ans增加1 。
- 完成以上步骤后，打印ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
void count(int a[], int n)
{
// If size of array is 1
if (n == 1) {
cout << 0;
return;
}
// pre[]: Prefix array
// suf[]: Suffix array
int pre[n - 1], suf[n - 1];
int max = a[0];
// Stores the required count
int ans = 0, i;
pre[0] = a[0];
// Find the maximum in prefix array
for (i = 1; i < n - 1; i++) {
if (a[i] > max)
max = a[i];
pre[i] = max;
}
max = a[n - 1];
suf[n - 2] = a[n - 1];
// Find the maximum in suffix array
for (i = n - 2; i >= 1; i--) {
if (a[i] > max)
max = a[i];
suf[i - 1] = max;
}
// Traverse the array
for (i = 0; i < n - 1; i++) {
// If maximum in prefix array
// is less than maximum in
// the suffix array
if (pre[i] < suf[i])
ans++;
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 4, 8, 1, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
count(arr, N);
return 0;
} Java
// Java program for the above approach
public class gfg
{
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
static void count(int a[], int n)
{
// If size of array is 1
if (n == 1)
{
System.out.print(0);
return;
}
// pre[]: Prefix array
// suf[]: Suffix array
int[] pre = new int[n - 1];
int[] suf = new int[n - 1];
int max = a[0];
// Stores the required count
int ans = 0, i;
pre[0] = a[0];
// Find the maximum in prefix array
for(i = 1; i < n - 1; i++)
{
if (a[i] > max)
max = a[i];
pre[i] = max;
}
max = a[n - 1];
suf[n - 2] = a[n - 1];
// Find the maximum in suffix array
for(i = n - 2; i >= 1; i--)
{
if (a[i] > max)
max = a[i];
suf[i - 1] = max;
}
// Traverse the array
for(i = 0; i < n - 1; i++)
{
// If maximum in prefix array
// is less than maximum in
// the suffix array
if (pre[i] < suf[i])
ans++;
}
// Print the answer
System.out.print(ans);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 8, 1, 4 };
int N = arr.length;
// Function Call
count(arr, N);
}
}
// This code is contributed by divyesh072019.Python3
# Python program for the above approach
# Function to print the count of indices
# in which the maximum in prefix arrays
# is less than that in the suffix array
def count(a, n) :
# If size of array is 1
if (n == 1) :
print(0)
return
# pre[]: Prefix array
# suf[]: Suffix array
pre = [0] * (n - 1)
suf = [0] * (n - 1)
max = a[0]
# Stores the required count
ans = 0
pre[0] = a[0]
# Find the maximum in prefix array
for i in range(n-1):
if (a[i] > max):
max = a[i]
pre[i] = max
max = a[n - 1]
suf[n - 2] = a[n - 1]
# Find the maximum in suffix array
for i in range(n-2, 0, -1):
if (a[i] > max):
max = a[i]
suf[i - 1] = max
# Traverse the array
for i in range(n - 1):
# If maximum in prefix array
# is less than maximum in
# the suffix array
if (pre[i] < suf[i]) :
ans += 1
# Print the answer
print(ans)
# Driver Code
arr = [ 2, 3, 4, 8, 1, 4 ]
N = len(arr)
# Function Call
count(arr, N)
# This code is contributed by code_hunt.C#
// C# program for the above approach
using System;
class GFG{
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
static void count(int[] a, int n)
{
// If size of array is 1
if (n == 1)
{
Console.Write(0);
return;
}
// pre[]: Prefix array
// suf[]: Suffix array
int[] pre = new int[n - 1];
int[] suf = new int[n - 1];
int max = a[0];
// Stores the required count
int ans = 0, i;
pre[0] = a[0];
// Find the maximum in prefix array
for(i = 1; i < n - 1; i++)
{
if (a[i] > max)
max = a[i];
pre[i] = max;
}
max = a[n - 1];
suf[n - 2] = a[n - 1];
// Find the maximum in suffix array
for(i = n - 2; i >= 1; i--)
{
if (a[i] > max)
max = a[i];
suf[i - 1] = max;
}
// Traverse the array
for(i = 0; i < n - 1; i++)
{
// If maximum in prefix array
// is less than maximum in
// the suffix array
if (pre[i] < suf[i])
ans++;
}
// Print the answer
Console.Write(ans);
}
// Driver code
static void Main()
{
int[] arr = { 2, 3, 4, 8, 1, 4 };
int N = arr.Length;
// Function Call
count(arr, N);
}
}
// This code is contributed by divyeshrabadiya07Javascript
输出：
3时间复杂度： O(N)
辅助空间： O(N)
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