Find the difference of count of equal elements on the right and the left for each element
Last Updated :
12 Jul, 2025
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 = -1
Input: arr[] = {1, 1, 1, 1, 1}
Output: 4 2 0 -2 -4
Brute Force Approach:
Brute force approach to solve this problem would be to use nested loops to count the number of elements to the right and left of each element that are equal to it. For each element, we can initialize two counters, one for counting the number of equal elements to the right and the other for counting the number of equal elements to the left. Then we can use nested loops to traverse the array and count the number of equal elements to the right and left of each element. Finally, we can subtract the two counts to get the required result.
- Iterate over the array using a for loop starting from index 0 to index n-1.
- For each element at index i, initialize the variables 'right_count' and 'left_count' to 0.
- Use another for loop to iterate over the elements from i+1 to n-1 to count the number of elements to the right of i that are equal to a[i]. Increment 'right_count' for each such element.
- Use another for loop to iterate over the elements from i-1 to 0 to count the number of elements to the left of i that are equal to a[i]. Increment 'left_count' for each such element.
- Calculate the difference between 'right_count' and 'left_count' for each i.
Below is the implementation of the above approach:
C++
// C++ implementation of the brute force 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)
{
for(int i=0; i<n; i++) {
int right_count = 0, left_count = 0;
for(int j=i+1; j<n; j++) {
if(a[i] == a[j]) {
right_count++;
}
}
for(int j=i-1; j>=0; j--) {
if(a[i] == a[j]) {
left_count++;
}
}
cout << right_count - left_count << " ";
}
}
// Driver code
int main()
{
int a[] = { 1, 2, 3, 2, 1 };
int n = sizeof(a) / sizeof(a[0]);
right_left(a, n);
return 0;
}
import java.util.*;
public class GFG {
// Function to find the count of equal elements to the right
// minus the count of equal elements to the left for each element
static void rightLeft(int[] a, int n) {
for (int i = 0; i < n; i++) {
int rightCount = 0, leftCount = 0;
// Count equal elements to the right
for (int j = i + 1; j < n; j++) {
if (a[i] == a[j]) {
rightCount++;
}
}
// Count equal elements to the left
for (int j = i - 1; j >= 0; j--) {
if (a[i] == a[j]) {
leftCount++;
}
}
System.out.print(rightCount - leftCount + " ");
}
}
public static void main(String[] args) {
int[] a = { 1, 2, 3, 2, 1 };
int n = a.length;
rightLeft(a, n);
}
}
# Function to find the count of equal elements to the right
# minus the count of equal elements to the left for each element
def right_left_counts(arr):
n = len(arr)
result = []
for i in range(n):
right_count = 0
left_count = 0
# Count equal elements to the right
for j in range(i + 1, n):
if arr[i] == arr[j]:
right_count += 1
# Count equal elements to the left
for j in range(i - 1, -1, -1):
if arr[i] == arr[j]:
left_count += 1
result.append(right_count - left_count)
return result
# Driver code
if __name__ == "__main__":
arr = [1, 2, 3, 2, 1]
result = right_left_counts(arr)
print(result)
using System;
public class GFG {
// Function to find the count of equal elements to the
// right minus the count of equal elements to the left
// for each element
static void RightLeft(int[] a, int n)
{
for (int i = 0; i < n; i++) {
int rightCount = 0, leftCount = 0;
// Count equal elements to the right
for (int j = i + 1; j < n; j++) {
if (a[i] == a[j]) {
rightCount++;
}
}
// Count equal elements to the left
for (int j = i - 1; j >= 0; j--) {
if (a[i] == a[j]) {
leftCount++;
}
}
Console.Write(rightCount - leftCount + " ");
}
}
public static void Main(string[] args)
{
int[] a = { 1, 2, 3, 2, 1 };
int n = a.Length;
RightLeft(a, n);
}
}
// Function to find the count of equal
// elements to the right - count of equal
// elements to the left for each of the element
function rightLeft(arr) {
const n = arr.length;
for (let i = 0; i < n; i++) {
let rightCount = 0, leftCount = 0;
// Count equal elements to the right
for (let j = i + 1; j < n; j++) {
if (arr[i] === arr[j]) {
rightCount++;
}
}
// Count equal elements to the left
for (let j = i - 1; j >= 0; j--) {
if (arr[i] === arr[j]) {
leftCount++;
}
}
// Print the result
console.log(rightCount - leftCount);
}
}
// Driver code
const arr = [1, 2, 3, 2, 1];
rightLeft(arr);
Time Complexity: O(n^2) because it uses two nested loops to count the equal elements to the right and left of each element.
Space Complexity: O(1), as we are not using extra space.
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 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 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# 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
<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++)
{
// Print 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>
Time complexity: O(N), where N is the size of the given array.
Auxiliary space: O(N), as two hashmaps are required to store the frequency of the elements.
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem