# Find the difference of count of equal elements on the right and the left for each element

Given an array **arr[]** of size **N**. The task is to find **X – Y** for each of the element where **X** is the count of **j** such that **arr[i] = arr[j]** and **j > i**. **Y** is the count of **j** such that **arr[i] = arr[j]** and **j < i**.**Examples:**

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

For index 0, X – Y = 1 – 0 = 1

For index 1, X – Y = 1 – 0 = 1

For index 2, X – Y = 0 – 0 = 0

For index 3, X – Y = 0 – 1 = -1

For index 4, X – Y = 0 – 1 = -1Input:arr[] = {1, 1, 1, 1, 1}Output:4 2 0 -2 -4

**Approach:** An efficient approach is to use a map. One map is to store the count of each element in the array and another map to count the number of same elements left to each element.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the count of equal` `// elements to the right - count of equal` `// elements to the left for each of the element` `void` `right_left(` `int` `a[], ` `int` `n)` `{` ` ` `// Maps to store the frequency and same` ` ` `// elements to the left of an element` ` ` `unordered_map<` `int` `, ` `int` `> total, left;` ` ` `// Count the frequency of each element` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `total[a[i]]++;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Print the answer for each element` ` ` `cout << (total[a[i]] - 1 - (2 * left[a[i]])) << ` `" "` `;` ` ` `// Increment it's left frequency` ` ` `left[a[i]]++;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `a[] = { 1, 2, 3, 2, 1 };` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]);` ` ` `right_left(a, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` `// Function to find the count of equal` `// elements to the right - count of equal` `// elements to the left for each of the element` `static` `void` `right_left(` `int` `a[], ` `int` `n)` `{` ` ` `// Maps to store the frequency and same` ` ` `// elements to the left of an element` ` ` `Map<Integer, Integer> total = ` `new` `HashMap<>();` ` ` `Map<Integer, Integer> left = ` `new` `HashMap<>();` ` ` `// Count the frequency of each element` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `total.put(a[i],` ` ` `total.get(a[i]) == ` `null` `? ` `1` `:` ` ` `total.get(a[i]) + ` `1` `);` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `// Print the answer for each element` ` ` `System.out.print((total.get(a[i]) - ` `1` `-` ` ` `(` `2` `* (left.containsKey(a[i]) == ` `true` `?` ` ` `left.get(a[i]) : ` `0` `))) + ` `" "` `);` ` ` `// Increment it's left frequency` ` ` `left.put(a[i],` ` ` `left.get(a[i]) == ` `null` `? ` `1` `:` ` ` `left.get(a[i]) + ` `1` `);` ` ` `}` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `a[] = { ` `1` `, ` `2` `, ` `3` `, ` `2` `, ` `1` `};` ` ` `int` `n = a.length;` ` ` `right_left(a, n);` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python3 implementation of the approach` `# Function to find the count of equal` `# elements to the right - count of equal` `# elements to the left for each of the element` `def` `right_left(a, n) :` ` ` `# Maps to store the frequency and same` ` ` `# elements to the left of an element` ` ` `total ` `=` `dict` `.fromkeys(a, ` `0` `);` ` ` `left ` `=` `dict` `.fromkeys(a, ` `0` `);` ` ` `# Count the frequency of each element` ` ` `for` `i ` `in` `range` `(n) :` ` ` `if` `a[i] ` `not` `in` `total :` ` ` `total[a[i]] ` `=` `1` ` ` `total[a[i]] ` `+` `=` `1` `;` ` ` `for` `i ` `in` `range` `(n) :` ` ` `# Print the answer for each element` ` ` `print` `(total[a[i]] ` `-` `1` `-` `(` `2` `*` `left[a[i]]),` ` ` `end ` `=` `" "` `);` ` ` `# Increment it's left frequency` ` ` `left[a[i]] ` `+` `=` `1` `;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `a ` `=` `[ ` `1` `, ` `2` `, ` `3` `, ` `2` `, ` `1` `];` ` ` `n ` `=` `len` `(a);` ` ` `right_left(a, n);` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;` ` ` `class` `GFG` `{` `// Function to find the count of equal` `// elements to the right - count of equal` `// elements to the left for each of the element` `static` `void` `right_left(` `int` `[]a, ` `int` `n)` `{` ` ` `// Maps to store the frequency and same` ` ` `// elements to the left of an element` ` ` `Dictionary<` `int` `, ` `int` `> total = ` `new` `Dictionary<` `int` `, ` `int` `>();` ` ` `Dictionary<` `int` `, ` `int` `> left = ` `new` `Dictionary<` `int` `, ` `int` `>();` ` ` `// Count the frequency of each element` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `if` `(total.ContainsKey(a[i]))` ` ` `{` ` ` `total[a[i]] = total[a[i]] + 1;` ` ` `}` ` ` `else` `{` ` ` `total.Add(a[i], 1);` ` ` `}` ` ` `}` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `// Print the answer for each element` ` ` `Console.Write((total[a[i]] - 1 -` ` ` `(2 * (left.ContainsKey(a[i]) == ` `true` `?` ` ` `left[a[i]] : 0))) + ` `" "` `);` ` ` `// Increment it's left frequency` ` ` `if` `(left.ContainsKey(a[i]))` ` ` `{` ` ` `left[a[i]] = left[a[i]] + 1;` ` ` `}` ` ` `else` ` ` `{` ` ` `left.Add(a[i], 1);` ` ` `}` ` ` `}` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]a = { 1, 2, 3, 2, 1 };` ` ` `int` `n = a.Length;` ` ` `right_left(a, n);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` `// Function to find the count of equal` `// elements to the right - count of equal` `// elements to the left for each of the element` `function` `right_left(a, n)` `{` ` ` ` ` `// Maps to store the frequency and same` ` ` `// elements to the left of an element` ` ` `let total = ` `new` `Map();` ` ` `let left = ` `new` `Map();` ` ` ` ` `// Count the frequency of each element` ` ` `for` `(let i = 0; i < n; i++)` ` ` `total.set(a[i],` ` ` `total.get(a[i]) == ` `null` `? 1 :` ` ` `total.get(a[i]) + 1);` ` ` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// Prlet the answer for each element` ` ` `document.write((total.get(a[i]) - 1 -` ` ` `(2 * (left.has(a[i]) == ` `true` `?` ` ` `left.get(a[i]) : 0))) + ` `" "` `);` ` ` ` ` `// Increment it's left frequency` ` ` `left.set(a[i],` ` ` `left.get(a[i]) == ` `null` `? 1 :` ` ` `left.get(a[i]) + 1);` ` ` `}` `}` ` ` ` ` `// Driver code` ` ` ` ` `let a = [ 1, 2, 3, 2, 1 ];` ` ` `let n = a.length;` ` ` ` ` `right_left(a, n);` ` ` ` ` `// This code is contributed by susmitakundugoaldanga.` `</script>` |

**Output:**

1 1 0 -1 -1

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