Count of smaller elements on right side of each element in an Array using Merge sort

Given an array arr[] of N integers, the task is to count the number of smaller elements on the right side for each of the element in the array

Examples:

Input: arr[] = {6, 3, 7, 2}
Output: 2, 1, 1, 0
Explanation:
Smaller elements after 6 = 2 [3, 2]
Smaller elements after 3 = 1 [2]
Smaller elements after 7 = 1 [2]
Smaller elements after 2 = 0

Input: arr[] = {6, 19, 111, 13}
Output: 0, 1, 1, 0
Explanation:
Smaller elements after 6 = 0
Smaller elements after 19 = 1 [13]
Smaller elements after 111 = 1 [13]
Smaller elements after 13 = 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
Use the idea of the merge sort at the time of merging two arrays. When higher index element is less than the lower index element, it represents that the higher index element is smaller than all the elements after that lower index because the left part is already sorted. Hence add up to all the elements after the lower index element for the required count.

Below is the implementation of the above approach

Java

// Java program to find the count of smaller elements 
// on right side of each element in an Array 
// using Merge sort 
  
import java.util.*; 
  
public class GFG { 
  
    // Class for storing the index 
    // and Value pairs 
    class Item { 
  
        int val; 
        int index; 
  
        public Item(int val, int index) 
        { 
            this.val = val; 
            this.index = index; 
        } 
    } 
  
    // Function to count the number of 
    // smaller elements on right side 
    public ArrayList<Integer> countSmall(int[] A) 
    { 
  
        int len = A.length; 
        Item[] items = new Item[len]; 
  
        for (int i = 0; i < len; i++) { 
            items[i] = new Item(A[i], i); 
        } 
  
        int[] count = new int[len]; 
        mergeSort(items, 0, len - 1, count); 
        ArrayList<Integer> res = new ArrayList<>(); 
  
        for (int i : count) { 
            res.add(i); 
        } 
        return res; 
    } 
  
    // Function for Merge Sort 
    private void mergeSort(Item[] items, 
                           int low, int high, 
                           int[] count) 
    { 
  
        if (low >= high) { 
            return; 
        } 
  
        int mid = low + (high - low) / 2; 
        mergeSort(items, low, mid, count); 
        mergeSort(items, mid + 1, high, count); 
        merge(items, low, mid, mid + 1, high, count); 
    } 
  
    // Utility function that merge the array 
    // and count smaller element on right side 
    private void merge(Item[] items, int low, 
                       int lowEnd, int high, 
                       int highEnd, int[] count) 
    { 
        int m = highEnd - low + 1; 
        Item[] sorted = new Item[m]; 
        int rightCounter = 0; 
        int lowPtr = low, highPtr = high; 
        int index = 0; 
  
        // Loop to store the count of smaller 
        // Elements on right side when both 
        // Array have some elements 
        while (lowPtr <= lowEnd && highPtr <= highEnd) { 
            if (items[lowPtr].val > items[highPtr].val) { 
                rightCounter++; 
                sorted[index++] = items[highPtr++]; 
            } 
            else { 
                count[items[lowPtr].index] += rightCounter; 
                sorted[index++] = items[lowPtr++]; 
            } 
        } 
  
        // Loop to store the count of smaller 
        // elements in right side when only 
        // left array have some element 
        while (lowPtr <= lowEnd) { 
            count[items[lowPtr].index] += rightCounter; 
            sorted[index++] = items[lowPtr++]; 
        } 
  
        // Loop to store the count of smaller 
        // elements in right side when only 
        // right array have some element 
        while (highPtr <= highEnd) { 
            sorted[index++] = items[highPtr++]; 
        } 
  
        System.arraycopy(sorted, 0, items, low, m); 
    } 
  
    // Utility function that prints 
    // out an array on a line 
    void printArray(ArrayList<Integer> countList) 
    { 
  
        for (Integer i : countList) 
            System.out.print(i + " "); 
  
        System.out.println(""); 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        GFG cntSmall = new GFG(); 
        int arr[] = { 10, 9, 5, 2, 7, 6, 11, 0, 2 }; 
        int n = arr.length; 
        ArrayList<Integer> countList 
            = cntSmall.countSmall(arr); 
        cntSmall.printArray(countList); 
    } 
} 

C#

// C# program to find the count of smaller elements 
// on right side of each element in an Array 
// using Merge sort 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
  
    // Class for storing the index 
    // and Value pairs 
    public class Item  
    { 
  
        public int val; 
        public int index; 
  
        public Item(int val, int index) 
        { 
            this.val = val; 
            this.index = index; 
        } 
    } 
  
    // Function to count the number of 
    // smaller elements on right side 
    public List<int> countSmall(int[] A) 
    { 
  
        int len = A.Length; 
        Item[] items = new Item[len]; 
  
        for (int i = 0; i < len; i++)  
        { 
            items[i] = new Item(A[i], i); 
        } 
  
        int[] count = new int[len]; 
        mergeSort(items, 0, len - 1, count); 
        List<int> res = new List<int>(); 
  
        foreach (int i in count)  
        { 
            res.Add(i); 
        } 
        return res; 
    } 
  
    // Function for Merge Sort 
    private void mergeSort(Item[] items, 
                        int low, int high, 
                        int[] count) 
    { 
  
        if (low >= high)  
        { 
            return; 
        } 
  
        int mid = low + (high - low) / 2; 
        mergeSort(items, low, mid, count); 
        mergeSort(items, mid + 1, high, count); 
        merge(items, low, mid, mid + 1, high, count); 
    } 
  
    // Utility function that merge the array 
    // and count smaller element on right side 
    private void merge(Item[] items, int low, 
                    int lowEnd, int high, 
                    int highEnd, int[] count) 
    { 
        int m = highEnd - low + 1; 
        Item[] sorted = new Item[m]; 
        int rightCounter = 0; 
        int lowPtr = low, highPtr = high; 
        int index = 0; 
  
        // Loop to store the count of smaller 
        // Elements on right side when both 
        // Array have some elements 
        while (lowPtr <= lowEnd && highPtr <= highEnd) 
        { 
            if (items[lowPtr].val > items[highPtr].val)  
            { 
                rightCounter++; 
                sorted[index++] = items[highPtr++]; 
            } 
            else
            { 
                count[items[lowPtr].index] += rightCounter; 
                sorted[index++] = items[lowPtr++]; 
            } 
        } 
  
        // Loop to store the count of smaller 
        // elements in right side when only 
        // left array have some element 
        while (lowPtr <= lowEnd)  
        { 
            count[items[lowPtr].index] += rightCounter; 
            sorted[index++] = items[lowPtr++]; 
        } 
  
        // Loop to store the count of smaller 
        // elements in right side when only 
        // right array have some element 
        while (highPtr <= highEnd) 
        { 
            sorted[index++] = items[highPtr++]; 
        } 
  
        Array.Copy(sorted, 0, items, low, m); 
    } 
  
    // Utility function that prints 
    // out an array on a line 
    void printArray(List<int> countList) 
    { 
  
        foreach (int i in countList) 
            Console.Write(i + " "); 
  
        Console.WriteLine(""); 
    } 
  
    // Driver Code 
    public static void Main(String[] args) 
    { 
        GFG cntSmall = new GFG(); 
        int []arr = { 10, 9, 5, 2, 7, 6, 11, 0, 2 }; 
        int n = arr.Length; 
        List<int> countList 
            = cntSmall.countSmall(arr); 
        cntSmall.printArray(countList); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Output:

7 6 3 1 3 2 2 0 0

Time Complexity: O(N log N)

Related Article: Count smaller elements on right side

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Java

C#

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