Given an array **A[]** consisting of **N **integers, the task is to construct an array **B[] **such that for every i^{th} index, **B[i] = X – Y**, where **X** and **Y **are the count of occurrences of** A[i]** after and before the i^{th} index.

**Examples:**

Input:A[] = {3, 2, 1, 2, 3}Output:1 1 0 -1 -1Explanation:

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

**Naive Approach:**

The **simplest approach **to solve the problem is to travere the array and consider every array element and compare it with all the elements on its left and right. For every array element, print the difference in its cunt of occurrences in its left and right.

**Time Complexity: **O(N^{2})**Auxiliary Space:** O(1)

**Efficient Approach:** Follow the steps below to optimize the above approach:

- Initialize two arrays
**left[]**and**right[]**to store frequencies of array elements present on the left and right indices of every array element. - Compute the left and right cumulative frequency tables.
- Print the difference of same indexed elements from the two frequency arrays.

Below is the implementation of the above approach:

## C++

`// C++ program of the above approach` `#include <bits/stdc++.h>` `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` |

**Output:**

1 1 0 -1 -1

**Time Complexity: **O(N)**Auxiliary Space: **O(N)

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