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 = 0Input: 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
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
C++
// C++ program to find the count of // smaller elements on right side of // each element in an Array // using Merge sort #include <bits/stdc++.h> using namespace std; const int N = 100001; int ans[N]; // Utility function that merge the array // and count smaller element on right side void merge(pair< int , int > a[], int start, int mid, int end) { pair< int , int > f[mid - start + 1], s[end - mid]; int n = mid - start + 1; int m = end - mid; for ( int i = start; i <= mid; i++) f[i - start] = a[i]; for ( int i = mid + 1; i <= end; i++) s[i - mid - 1] = a[i]; int i = 0, j = 0, k = start; int cnt = 0; // Loop to store the count of smaller // Elements on right side when both // Array have some elements while (i < n && j < m) { if (f[i].second <= s[j].second) { ans[f[i].first] += cnt; a[k++] = f[i++]; } else { cnt++; a[k++] = s[j++]; } } // Loop to store the count of smaller // elements in right side when only // left array have some element while (i < n) { ans[f[i].first] += cnt; a[k++] = f[i++]; } // Loop to store the count of smaller // elements in right side when only // right array have some element while (j < m) { a[k++] = s[j++]; } } // Function to invoke merge sort. void mergesort(pair< int , int > item[], int low, int high) { if (low >= high) return ; int mid = (low + high) / 2; mergesort(item, low, mid); mergesort(item, mid + 1, high); merge(item, low, mid, high); } // Utility function that prints // out an array on a line void print( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver code. int main() { int arr[] = { 10, 9, 5, 2, 7, 6, 11, 0, 2 }; int n = sizeof (arr) / sizeof ( int ); pair< int , int > a[n]; memset (ans, 0, sizeof (ans)); for ( int i = 0; i < n; i++) { a[i].second = arr[i]; a[i].first = i; } mergesort(a, 0, n - 1); print(ans, n); return 0; } // This code is contributed by rishabhtyagi2306 |
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); } } |
Python3
# Python3 program to find the count of # smaller elements on right side of # each element in an Array # using Merge sort N = 100001 ans = [ 0 ] * N # Utility function that merge the array # and count smaller element on right side def merge(a, start, mid, end): f = [ 0 ] * (mid - start + 1 ) s = [ 0 ] * (end - mid) n = mid - start + 1 m = end - mid for i in range (start, mid + 1 ): f[i - start] = a[i] for i in range ( mid + 1 , end + 1 ): s[i - mid - 1 ] = a[i] i = 0 j = 0 k = start cnt = 0 # Loop to store the count of smaller # Elements on right side when both # Array have some elements while (i < n and j < m): if (f[i][ 1 ] < = s[j][ 1 ]): ans[f[i][ 0 ]] + = cnt a[k] = f[i] k + = 1 i + = 1 else : cnt + = 1 a[k] = s[j] k + = 1 j + = 1 # Loop to store the count of smaller # elements in right side when only # left array have some element while (i < n): ans[f[i][ 0 ]] + = cnt a[k] = f[i]; k + = 1 i + = 1 # Loop to store the count of smaller # elements in right side when only # right array have some element while (j < m): a[k] = s[j] k + = 1 j + = 1 # Function to invoke merge sort. def mergesort(item, low, high): if (low > = high): return mid = (low + high) / / 2 mergesort(item, low, mid) mergesort(item, mid + 1 , high) merge(item, low, mid, high) # Utility function that prints # out an array on a line def print_(arr, n): for i in range (n): print (arr[i], end = " " ) # Driver code. if __name__ = = "__main__" : arr = [ 10 , 9 , 5 , 2 , 7 , 6 , 11 , 0 , 2 ] n = len (arr) a = [[ 0 for x in range ( 2 )] for y in range (n)] for i in range (n): a[i][ 1 ] = arr[i] a[i][ 0 ] = i mergesort(a, 0 , n - 1 ) print_(ans, n) # This code is contributed by chitranayal |
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 |
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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