生成一个数组，该数组在其左侧和右侧的每个数组元素的出现次数之间存在差异

给定一个由N个整数组成的数组A [] ，任务是构造一个数组B [] ，使得对于每个^第i^个索引， B [i] = X – Y ，其中X和Y是A []的出现次数i]在第i^个索引之前和之后。

例子：

Input: A[] = {3, 2, 1, 2, 3}
Output: 1 1 0 -1 -1
Explanation: 
arr[0] = 3, X = 1, Y = 0. Therefore, print 1. 
arr[1] = 2, X = 1, Y = 0. Therefore, print 1. 
arr[2] = 1, X = 0, Y = 0. Therefore, print 0. 
arr[3] = 2, X = 0, Y = 1. Therefore, print -1. 
arr[4] = 3, X = 0, Y = 1. Therefore, print -1.

Input: A[] = {1, 2, 3, 4, 5}
Output: 0 0 0 0 0

天真的方法：
解决问题的最简单方法是遍历数组并考虑每个数组元素，并将其与左右两侧的所有元素进行比较。对于每个数组元素，在其左侧和右侧打印其出现的差异。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：请按照以下步骤优化上述方法：

1. 初始化两个数组left []和right []以存储每个数组元素的左和右索引上存在的数组元素的频率。
2. 计算左右累积频率表。
3. 打印两个频率阵列中相同索引元素的差。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to construct array of
// differences of counts on the left
// and right of the given array
void constructArray(int A[], int N)
{
    // Initialize left and right
    // frequency arrays
    int left[N + 1] = { 0 };
    int right[N + 1] = { 0 };
    int X[N + 1] = { 0 }, Y[N + 1] = { 0 };
 
    // Construct left cumulative
    // frequency table
    for (int i = 0; i < N; i++) {
        X[i] = left[A[i]];
        left[A[i]]++;
    }
 
    // Construct right cumulative
    // frequency table
    for (int i = N - 1; i >= 0; i--) {
        Y[i] = right[A[i]];
        right[A[i]]++;
    }
 
    // Print the result
    for (int i = 0; i < N; i++) {
        cout << Y[i] - X[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int A[] = { 3, 2, 1, 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    constructArray(A, N);
 
    return 0;
}

Java

// Java program of the above approach
import java.io.*;
 
class GFG{
 
// Function to construct array of
// differences of counts on the left
// and right of the given array
static void constructArray(int A[], int N)
{
     
    // Initialize left and right
    // frequency arrays
    int[] left = new int[N + 1];
    int[] right = new int[N + 1];
    int[] X = new int[N + 1];
    int[] Y = new int[N + 1];
 
    // Construct left cumulative
    // frequency table
    for(int i = 0; i < N; i++)
    {
        X[i] = left[A[i]];
        left[A[i]]++;
    }
 
    // Construct right cumulative
    // frequency table
    for(int i = N - 1; i >= 0; i--)
    {
        Y[i] = right[A[i]];
        right[A[i]]++;
    }
 
    // Print the result
    for (int i = 0; i < N; i++)
    {
        System.out.print(Y[i] - X[i] + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 3, 2, 1, 2, 3 };
    int N = A.length;
 
    // Function Call
    constructArray(A, N);
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program of the above approach
 
# Function to construct array of
# differences of counts on the left
# and right of the given array
def constructArray(A, N):
     
    # Initialize left and right
    # frequency arrays
    left = [0] * (N + 1)
    right = [0] * (N + 1)
    X = [0] * (N + 1)
    Y = [0] * (N + 1)
 
    # Construct left cumulative
    # frequency table
    for i in range(0, N):
        X[i] = left[A[i]]
        left[A[i]] += 1
 
    # Construct right cumulative
    # frequency table
    for i in range(N - 1, -1, -1):
        Y[i] = right[A[i]]
        right[A[i]] += 1
 
    # Print the result
    for i in range(0, N):
        print(Y[i] - X[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
 
    A = [ 3, 2, 1, 2, 3 ]
    N = len(A)
 
    # Function Call
    constructArray(A, N)
 
# This code is contributed by akhilsaini

C#

// C# program of the above approach
using System;
 
class GFG{
 
// Function to construct array of
// differences of counts on the left
// and right of the given array
static void constructArray(int[] A, int N)
{
     
    // Initialize left and right
    // frequency arrays
    int[] left = new int[N + 1];
    int[] right = new int[N + 1];
    int[] X = new int[N + 1];
    int[] Y = new int[N + 1];
 
    // Construct left cumulative
    // frequency table
    for(int i = 0; i < N; i++)
    {
        X[i] = left[A[i]];
        left[A[i]]++;
    }
 
    // Construct right cumulative
    // frequency table
    for(int i = N - 1; i >= 0; i--)
    {
        Y[i] = right[A[i]];
        right[A[i]]++;
    }
 
    // Print the result
    for(int i = 0; i < N; i++)
    {
        Console.Write(Y[i] - X[i] + " ");
    }
}
 
// Driver Code
public static void Main()
{
    int[] A = { 3, 2, 1, 2, 3 };
    int N = A.Length;
 
    // Function Call
    constructArray(A, N);
}
}
 
// This code is contributed by akhilsaini

Javascript

 
// JavaScript program of the above approach
 
// Function to construct array of
// differences of counts on the left
// and right of the given array
function constructArray(A, N)
{
      
    // Initialize left and right
    // frequency arrays
    let left = [];
    let right = [];
    let X = [];
    let Y = [];
     
    for(let i = 0; i < N; i++)
    {
        X[i] = 0;
        left[i] = 0;
        right[i] = 0;
        Y[i]= 0;
    }
  
    // Construct left cumulative
    // frequency table
    for(let i = 0; i < N; i++)
    {
        X[i] = left[A[i]];
        left[A[i]]++;
    }
  
    // Construct right cumulative
    // frequency table
    for(let i = N - 1; i >= 0; i--)
    {
        Y[i] = right[A[i]];
        right[A[i]]++;
    }
  
    // Print the result
    for(let i = 0; i < N; i++)
    {
        document.write(Y[i] - X[i] + " ");
    }
}
 
// Driver Code
let A = [ 3, 2, 1, 2, 3 ];
let N = A.length;
 
// Function Call
constructArray(A, N);
 
// This code is contribute by target_2
 

输出：

1 1 0 -1 -1

时间复杂度： O(N)
辅助空间： O(N)