为每个数组元素计算数组中存在的较小元素

给定一个由N个整数组成的数组arr [] ，对于每个数组元素(例如arr [i]) ，其任务是查找小于arr [i]的数组元素的数量。

例子：

Input: arr[] = {3, 4, 1, 1, 2}
Output: 3 4 0 0 2
Explanation:
The elements which are smaller than arr[0](= 3) are {1, 1, 2}. Hence, the count is 3.
The elements which are smaller than arr[1](= 4) are {1, 1, 2, 3}. Hence, the count is 4.
The elements arr[2](= 1) and arr[3](= 1) are the smallest possible. Hence, the count is 0.
The elements which are smaller than arr[4](= 2) are {1, 1}. Hence, the count is 2.

Input: arr[] = {1, 2, 3, 4}
Output: 0 1 2 3

天真的方法：最简单的方法是遍历数组，并对每个数组元素计数小于它们的数组元素的数量，然后打印获得的计数。

下面是上述方法的实现：

C++

// C++ program for the above approch
 
#include 
using namespace std;
 
// Function to count for each array
// element, the number of elements
// that are smaller than that element
void smallerNumbers(int arr[], int N)
{
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Stores the count
        int count = 0;
 
        // Traverse the array
        for (int j = 0; j < N; j++) {
 
            // Increment count
            if (arr[j] < arr[i]) {
                count++;
            }
        }
 
        // Print the count of smaller
        // elements for the current element
        cout << count << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 4, 1, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    smallerNumbers(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count for each array
// element, the number of elements
// that are smaller than that element
static void smallerNumbers(int arr[], int N)
{
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count
        int count = 0;
 
        // Traverse the array
        for(int j = 0; j < N; j++)
        {
             
            // Increment count
            if (arr[j] < arr[i])
            {
                count++;
            }
        }
 
        // Print the count of smaller
        // elements for the current element
        System.out.print(count + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 1, 1, 2 };
    int N = arr.length;
 
    smallerNumbers(arr, N);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approch
 
# Function to count for each array
# element, the number of elements
# that are smaller than that element
def smallerNumbers(arr, N):
 
    # Traverse the array
    for i in range(N):
 
        # Stores the count
        count = 0
 
        # Traverse the array
        for j in range(N):
 
            # Increment count
            if (arr[j] < arr[i]):
                count += 1
 
        # Print the count of smaller
        # elements for the current element
        print(count, end=" ")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 4, 1, 1, 2]
    N = len(arr)
 
    smallerNumbers(arr, N)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
public class GFG
{
 
  // Function to count for each array
  // element, the number of elements
  // that are smaller than that element
  static void smallerNumbers(int[] arr, int N)
  {
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
 
      // Stores the count
      int count = 0;
 
      // Traverse the array
      for(int j = 0; j < N; j++)
      {
 
        // Increment count
        if (arr[j] < arr[i])
        {
          count++;
        }
      }
 
      // Print the count of smaller
      // elements for the current element
      Console.Write(count + " ");
    }
  }
 
  // Driver Code
  static public void Main()
  {
    int[] arr = { 3, 4, 1, 1, 2 };
    int N = arr.Length;
 
    smallerNumbers(arr, N);
  }
}
 
// This code is contributed by sanjoy_62,

输出：

3 4 0 0 2

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：可以通过使用哈希优化上述方法。请按照以下步骤解决问题：

• 初始化大小为10 ⁵的辅助数组hash [] ，并使用0初始化所有数组元素，以存储每个数组元素的频率。
• 遍历给定的数组arr []并以hash [arr [i]] ++的形式将数组hash []中的arr [i]的频率增加1 。
• 找到数组hash []的前缀和。
• 再次遍历给定的数组，并对每个数组元素arr [] ，打印hash [arr [i] – 1]的值作为较小元素的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count for each array
// element, the number of elements
// that are smaller than that element
void smallerNumbers(int arr[], int N)
{
    // Stores the frequencies
    // of array elements
    int hash[100000] = { 0 };
 
    // Traverse the array
    for (int i = 0; i < N; i++)
 
        // Update frequency of arr[i]
        hash[arr[i]]++;
 
    // Initialize sum with 0
    int sum = 0;
 
    // Compute prefix sum of the array hash[]
    for (int i = 1; i < 100000; i++) {
        hash[i] += hash[i - 1];
    }
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If current element is 0
        if (arr[i] == 0) {
            cout << "0";
            continue;
        }
 
        // Print the resultant count
        cout << hash[arr[i] - 1]
             << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 4, 1, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    smallerNumbers(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count for each array
// element, the number of elements
// that are smaller than that element
static void smallerNumbers(int arr[], int N)
{
     
    // Stores the frequencies
    // of array elements
    int hash[] = new int[100000];
 
    // Traverse the array
    for(int i = 0; i < N; i++)
 
        // Update frequency of arr[i]
        hash[arr[i]]++;
 
    // Initialize sum with 0
    int sum = 0;
 
    // Compute prefix sum of the array hash[]
    for(int i = 1; i < 100000; i++)
    {
        hash[i] += hash[i - 1];
    }
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 0
        if (arr[i] == 0)
        {
            System.out.println("0");
            continue;
        }
 
        // Print the resultant count
        System.out.print(hash[arr[i] - 1] + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 1, 1, 2 };
    int N = arr.length;
 
    smallerNumbers(arr, N);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to count for each array
# element, the number of elements
# that are smaller than that element
def smallerNumbers(arr, N):
 
    # Stores the frequencies
    # of array elements
    hash = [0] * 100000
 
    # Traverse the array
    for i in range(N):
 
        # Update frequency of arr[i]
        hash[arr[i]] += 1
 
    # Initialize sum with 0
    sum = 0
 
    # Compute prefix sum of the array hash[]
    for i in range(1, 100000):
        hash[i] += hash[i - 1]
 
    # Traverse the array arr[]
    for i in range(N):
 
        # If current element is 0
        if (arr[i] == 0):
            print("0")
            continue
 
        # Print the resultant count
        print(hash[arr[i] - 1], end = " ")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 3, 4, 1, 1, 2 ]
    N = len(arr)
 
    smallerNumbers(arr, N)
 
# This code is contributed by AnkThon

C＃

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count for each array
// element, the number of elements
// that are smaller than that element
static void smallerNumbers(int []arr, int N)
{
     
    // Stores the frequencies
    // of array elements
    int []hash = new int[100000];
 
    // Traverse the array
    for(int i = 0; i < N; i++)
     
        // Update frequency of arr[i]
        hash[arr[i]]++;
 
    // Initialize sum with 0
    //int sum = 0;
 
    // Compute prefix sum of the array hash[]
    for(int i = 1; i < 100000; i++)
    {
        hash[i] += hash[i - 1];
    }
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 0
        if (arr[i] == 0)
        {
            Console.WriteLine("0");
            continue;
        }
 
        // Print the resultant count
        Console.Write(hash[arr[i] - 1] + " ");
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int []arr = { 3, 4, 1, 1, 2 };
    int N = arr.Length;
 
    smallerNumbers(arr, N);
}
}
 
// This code is contributed by AnkThon

输出：

3 4 0 0 2

时间复杂度： O(N)
辅助空间： O(10 ⁵ )