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 = -1
Input: 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++) { // 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> |
1 1 0 -1 -1
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.
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