为每个数组元素的最小子数组计数 | 2套

给定一个由N 个整数组成的数组arr[] ，任务是创建一个大小为N的数组brr[] ，其中brr[i]表示其中arr[i]是最小元素的子数组的数量。

例子：

Input: arr[] = {3, 2, 4}
Output: {1, 3, 1}
Explanation: 
For arr[0], there is only one subarray in which 3 is the smallest({3}). 
For arr[1], there are three such subarrays where 2 is the smallest({2}, {3, 2}, {2, 4}). 
For arr[2], there is only one subarray in which 4 is the smallest({4}).

Input: arr[] = {1, 2, 3, 4, 5}
Output: {5, 4, 3, 2, 1}

哈希方法：遍历数组以找到所有可能子数组的最小元素并将它们的计数存储在 Map 中。请按照以下步骤解决问题：

1. 创建一个 Map 来存储每个元素的子数组计数。
2. 遍历所有可能的子数组，找到子数组中的最小元素。
3. 在Map 中获得的最小元素的增量计数。
4. 最后，打印每个数组元素在Map 中获得的频率。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate total number
// of sub-arrays for each element
// where that element is occurring
// as the minimum element
void minSubarray(int* arr, int N)
{
 
    // Map for storing the number of
    // sub-arrays for each element
    unordered_map m;
 
    // Traverse over all possible subarrays
    for (int i = 0; i < N; i++) {
 
        int mini = INT_MAX;
 
        for (int j = i; j < N; j++) {
 
            // Minimum in each subarray
            mini = min(mini, arr[j]);
            m[mini]++;
        }
    }
 
    // Print the result
    for (int i = 0; i < N; i++) {
 
        cout << m[arr[i]] << " ";
    }
}
 
// Driver Code
int main()
{
 
    int arr[] = { 3, 2, 1, 4 };
    int N = sizeof(arr) / sizeof(int);
 
    // Function Call
    minSubarray(arr, N);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to calculate total
// number of sub-arrays for each
// element where that element is
// occurring as the minimum element
static void minSubarray(int []arr,
                        int N)
{
  // Map for storing the number of
  // sub-arrays for each element
  HashMap mp = new HashMap();
 
  // Traverse over all possible
  // subarrays
  for (int i = 0; i < N; i++)
  {
    int mini = Integer.MAX_VALUE;
 
    for (int j = i; j < N; j++)
    {
      // Minimum in each subarray
      mini = Math.min(mini, arr[j]);
      if(mp.containsKey(mini))
      {
        mp.put(mini, mp.get(mini) + 1);
      }
      else
      {
        mp.put(mini, 1);
      }
    }
  }
 
  // Print the result
  for (int i = 0; i < N; i++)
  {
    System.out.print(mp.get(arr[i]) + " ");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int arr[] = {3, 2, 1, 4};
  int N = arr.length;
 
  // Function Call
  minSubarray(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for
# the above approach
 
# Function to calculate total
# number of sub-arrays for
# each element where that
# element is occurring as the
# minimum element
def minSubarray(arr, N):
 
    # Map for storing the
    # number of sub-arrays
    # for each element
    m = {}
 
    # Traverse over all
    # possible subarrays
    for i in range(N):
 
        mini = 10 ** 9
 
        for j in range(i, N):
 
            # Minimum in each subarray
            mini = min(mini, arr[j])
            m[mini] = m.get(mini, 0) + 1
 
    # Print result
    for i in arr:
        print(m[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
 
    arr = [3, 2, 1, 4]
    N = len(arr)
 
    # Function Call
    minSubarray(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to calculate total
// number of sub-arrays for each
// element where that element is
// occurring as the minimum element
static void minSubarray(int []arr,
                        int N)
{
   
  // Map for storing the number of
  // sub-arrays for each element
  Dictionary mp = new Dictionary();
   
  // Traverse over all possible
  // subarrays
  for(int i = 0; i < N; i++)
  {
    int mini = int.MaxValue;
 
    for(int j = i; j < N; j++)
    {
       
      // Minimum in each subarray
      mini = Math.Min(mini, arr[j]);
       
      if (mp.ContainsKey(mini))
      {
        mp[mini] = mp[mini] + 1;
      }
      else
      {
        mp.Add(mini, 1);
      }
    }
  }
 
  // Print the result
  for(int i = 0; i < N; i++)
  {
    Console.Write(mp[arr[i]] + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = { 3, 2, 1, 4 };
  int N = arr.Length;
 
  // Function Call
  minSubarray(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

输出

1 2 6 1 

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：这种方法基于下一个更大的元素和前一个更大的元素的概念。请按照以下步骤解决问题：

1. 为了找到该元件的发生为最小，首先找到x和y，其中x是严格更大数目的上长度左改编的[i]和y是更大的数量的上ARR的右侧的长度[我] 。
2. 因此， x * y是arr[i]最小的子数组的总数。
3. 要查找x和y，请使用带有下一个更大元素和前一个更大元素概念的堆栈。
4. 对于下一个更大的元素，创建一个对的堆栈，并将第一个数组元素和一个计数器压入堆栈，以将子数组作为一对计数。
5. 遍历数组并一一选取数组元素：
   • 如果当前的数组元素比堆栈的顶部元件更大，则对于在堆栈顶部元件，当前元素是下一个元素更大。所以，弹出堆栈的顶部元件和由递增计数器值的计数器的堆栈的顶部，然后下一个栈顶元件与当前的数组元素等进行比较。
   • 否则，将带有计数器的当前数组元素压入堆栈并将计数器插入结果数组。
6. 对前一个更大的元素重复步骤 4、5。
7. 最后，打印结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count subarrays
// for each element where it is
// the minimum
void minSubarray(int* arr, int N)
{
    int result[N];
    stack > l, r;
 
    // For the length of strictly larger
    // numbers on the left of A[i]
    for (int i = 0; i < N; i++) {
 
        int count = 1;
        while (!l.empty()
               && l.top().first > arr[i]) {
 
            count += l.top().second;
            l.pop();
        }
        l.push({ arr[i], count });
 
        // Storing x in result[i]
        result[i] = count;
    }
 
    // For the length of strictly larger
    // numbers on the right of A[i]
    for (int i = N - 1; i >= 0; i--) {
 
        int count = 1;
        while (!r.empty()
               && r.top().first >= arr[i]) {
 
            count += r.top().second;
            r.pop();
        }
        r.push({ arr[i], count });
 
        // Store x*y in result array
        result[i] *= count;
    }
 
    // Print the result
    for (int i = 0; i < N; i++) {
 
        cout << result[i] << " ";
    }
}
 
// Driver Code
int main()
{
 
    int arr[] = { 3, 2, 1, 4 };
    int N = sizeof(arr) / sizeof(int);
 
    // Function Call
    minSubarray(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
static class pair
{
    int first, second;
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
// Function to count subarrays
// for each element where it is
// the minimum
static void minSubarray(int []arr, int N)
{
    int []result = new int[N];
    Stack l = new Stack<>();
    Stack r = new Stack<>();
 
    // For the length of strictly larger
    // numbers on the left of A[i]
    for(int i = 0; i < N; i++)
    {
        int count = 1;
        while (!l.isEmpty() &&
                l.peek().first > arr[i])
        {
            count += l.peek().second;
            l.pop();
        }
        l.add(new pair(arr[i], count));
 
        // Storing x in result[i]
        result[i] = count;
    }
 
    // For the length of strictly larger
    // numbers on the right of A[i]
    for(int i = N - 1; i >= 0; i--)
    {
        int count = 1;
        while (!r.isEmpty() &&
                r.peek().first >= arr[i])
        {
            count += r.peek().second;
            r.pop();
        }
        r.add(new pair(arr[i], count));
         
        // Store x*y in result array
        result[i] *= count;
    }
 
    // Print the result
    for(int i = 0; i < N; i++)
    {
        System.out.print(result[i] + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 1, 4 };
    int N = arr.length;
 
    // Function Call
    minSubarray(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

蟒蛇3

# Python3 program for the
# above approach
 
# Function to count subarrays
# for each element where it is
# the minimum
def minSubarray(arr, N):
   
    result = [0] * N
    l = []
    r = []
 
    # For the length of strictly
    # larger numbers on the left
    # of A[i]
    for i in range(N):
        count = 1;
        while (len(l) != 0 and
               l[-1][0] > arr[i]):
            count += l[-1][1]
            l.pop()
 
        l.append([arr[i], count])
 
        # Storing x in result[i]
        result[i] = count;
 
    # For the length of
    # strictly larger
    # numbers on the
    # right of A[i]
    for i in range(N - 1,
                   -1, -1):
        count = 1;
        while (len(r) != 0 and
               r[-1][0] >= arr[i]):
            count += r[-1][1]
            r.pop();
 
        r.append([arr[i], count]);
 
        # Store x*y in result
        # array
        result[i] *= count
 
    # Print the result
    for i in range(N):
        print(result[i],
              end = " ")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 2, 1, 4]
    N = len(arr)
 
    # Function Call
    minSubarray(arr, N)
 
# This code is contributed by Chitranayal

C＃

// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
 
public class pair
{
  public int first,
  second;
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
 
// Function to count subarrays
// for each element where it is
// the minimum
static void minSubarray(int []arr,
                        int N)
{
  int []result = new int[N];
  Stack l = new Stack();
  Stack r = new Stack();
 
  // For the length of strictly
  // larger numbers on the left
  // of A[i]
  for(int i = 0; i < N; i++)
  {
    int count = 1;
    while (l.Count != 0 &&
           l.Peek().first > arr[i])
    {
      count += l.Peek().second;
      l.Pop();
    }
    l.Push(new pair(arr[i],
                    count));
 
    // Storing x in result[i]
    result[i] = count;
  }
 
  // For the length of strictly
  // larger numbers on the right
  // of A[i]
  for(int i = N - 1; i >= 0; i--)
  {
    int count = 1;
    while (r.Count != 0 &&
           r.Peek().first >= arr[i])
    {
      count += r.Peek().second;
      r.Pop();
    }
    r.Push(new pair(arr[i],
                    count));
 
    // Store x*y in result array
    result[i] *= count;
  }
 
  // Print the result
  for(int i = 0; i < N; i++)
  {
    Console.Write(result[i] + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = {3, 2, 1, 4};
  int N = arr.Length;
 
  // Function Call
  minSubarray(arr, N);
}
}
 
// This code is contributed by Princi Singh

输出

1 2 6 1 

时间复杂度： O(N)
辅助空间： O(N)

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