Given an array arr[] consisting of N integers, the task is to count the number of greater elements on the right side of each array element.
Examples:
Input: arr[] = {3, 7, 1, 5, 9, 2}
Output: {3, 1, 3, 1, 0, 0}
Explanation: For arr[0], the elements greater than it on the right are {7, 5, 9}. For arr[1], the only element greater than it on the right is {9}. For arr[2], the elements greater than it on the right are {5, 9, 2}. For arr[3], the only element greater than it on the right is {9}. For arr[4] and arr[5], no greater elements exist on the right.Input: arr[] = {5, 4, 3, 2}
Output: {0, 0, 0, 0}
Naive Approach: The simplest approach is to iterate all array elements using two loops and for each array element, count the number of elements greater than it on its right side and then print it.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The problem can be solved using the concept of Merge Sort in descending order. Follow the steps given below to solve the problem:
- Initialize an array count[] where count[i] store the respective count of greater elements on the right for every arr[i]
- Take the indexes i and j, and compare the elements in an array.
- If higher index element is greater than the lower index element then, all the higher index element will be greater than all the elements after that lower index.
- Since the left part is already sorted, add the count of elements after the lower index element to the count[] array for the lower index.
- Repeat the above steps until the entire array is sorted.
- Finally print the values of count[] array.
Below is the implementation of the above approach:
Java
// Java program for the above approachimport java.util.*;public class GFG { // Stores the index & value pairs static class Item { int val; int index; public Item(int val, int index) { this.val = val; this.index = index; } } // Function to count the number of // greater elements on the right // of every array element public static ArrayList<Integer> countLarge(int[] a) { // Length of the array int len = a.length; // Stores the index-value pairs Item[] items = new Item[len]; for (int i = 0; i < len; i++) { items[i] = new Item(a[i], i); } // Stores the count of greater // elements on right int[] count = new int[len]; // Perform MergeSort operation mergeSort(items, 0, len - 1, count); ArrayList<Integer> res = new ArrayList<>(); for (int i : count) { res.add(i); } return res; } // Function to sort the array // using Merge Sort public static void mergeSort( Item[] items, int low ,int high, int[] count) { // Base Case if (low >= high) { return; } // Find Mid int mid = low + (high - low) / 2; mergeSort(items, low, mid, count); mergeSort(items, mid + 1, high, count); // Merging step merge(items, low, mid, mid + 1, high, count); } // Utility function to merge sorted // subarrays and find the count of // greater elements on the right public static void merge( Item[] items, int low, int lowEnd, int high, int highEnd, int[] count) { int m = highEnd - low + 1; // mid Item[] sorted = new Item[m]; int rightCounter = 0; int lowInd = low, highInd = high; int index = 0; // Loop to store the count of // larger elements on right side // when both array have elements while (lowInd <= lowEnd && highInd <= highEnd) { if (items[lowInd].val < items[highInd].val) { rightCounter++; sorted[index++] = items[highInd++]; } else { count[items[lowInd].index] += rightCounter; sorted[index++] = items[lowInd++]; } } // Loop to store the count of // larger elements in right side // when only left array have // some element while (lowInd <= lowEnd) { count[items[lowInd].index] += rightCounter; sorted[index++] = items[lowInd++]; } // Loop to store the count of // larger elements in right side // when only right array have // some element while (highInd <= highEnd) { sorted[index++] = items[highInd++]; } System.arraycopy(sorted, 0, items, low, m); } // Utility function that prints // the count of greater elements // on the right public static void printArray(ArrayList<Integer> countList) { for (Integer i : countList) System.out.print(i + " "); System.out.println(); } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 3, 7, 1, 5, 9, 2 }; int n = arr.length; // Function Call ArrayList<Integer> countList = countLarge(arr); printArray(countList); }} |
Python3
from typing import Listclass Item: def __init__(self, val: int, index: int): self.val = val self.index = indexdef count_large(a: List[int]) -> List[int]: # Length of the array length = len(a) # Stores the index-value pairs items = [Item(a[i], i) for i in range(length)] # Stores the count of greater elements on right count = [0] * length # Perform MergeSort operation merge_sort(items, 0, length - 1, count) res = count.copy() return resdef merge_sort(items: List[Item], low: int, high: int, count: List[int]) -> None: # Base Case if low >= high: return # Find Mid mid = low + (high - low) // 2 merge_sort(items, low, mid, count) merge_sort(items, mid + 1, high, count) # Merging step merge(items, low, mid, mid + 1, high, count)def merge(items: List[Item], low: int, low_end: int, high: int, high_end: int, count: List[int]) -> None: m = high_end - low + 1 # mid sorted_items = [None] * m right_counter = 0 low_ind = low high_ind = high index = 0 # Loop to store the count of larger elements on right side # when both array have elements while low_ind <= low_end and high_ind <= high_end: if items[low_ind].val < items[high_ind].val: right_counter += 1 sorted_items[index] = items[high_ind] index += 1 high_ind += 1 else: count[items[low_ind].index] += right_counter sorted_items[index] = items[low_ind] index += 1 low_ind += 1 # Loop to store the count of larger elements in right side # when only left array have elements while low_ind <= low_end: count[items[low_ind].index] += right_counter sorted_items[index] = items[low_ind] index += 1 low_ind += 1 # Loop to store the count of larger elements in right side # when only right array have elements while high_ind <= high_end: sorted_items[index] = items[high_ind] index += 1 high_ind += 1 items[low:low + m] = sorted_itemsdef print_array(count_list: List[int]) -> None: print(' '.join(str(i) for i in count_list))# Driver Codeif __name__ == '__main__': # Given array arr = [3, 7, 1, 5, 9, 2] # Function Call count_list = count_large(arr) print_array(count_list)# This code is contributed by Aditya Sharma |
C#
using System;using System.Collections.Generic;public class GFG { // Stores the index & 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 // greater elements on the right // of every array element public static List<int> CountLarge(int[] a) { // Length of the array int len = a.Length; // Stores the index-value pairs Item[] items = new Item[len]; for (int i = 0; i < len; i++) { items[i] = new Item(a[i], i); } // Stores the count of greater // elements on right int[] count = new int[len]; // Perform MergeSort operation MergeSort(items, 0, len - 1, count); List<int> res = new List<int>(); foreach(int i in count) { res.Add(i); } return res; } // Function to sort the array // using Merge Sort public static void MergeSort(Item[] items, int low, int high, int[] count) { // Base Case if (low >= high) { return; } // Find Mid int mid = low + (high - low) / 2; MergeSort(items, low, mid, count); MergeSort(items, mid + 1, high, count); // Merging step Merge(items, low, mid, mid + 1, high, count); } // Utility function to merge sorted // subarrays and find the count of // greater elements on the right public static void Merge(Item[] items, int low, int lowEnd, int high, int highEnd, int[] count) { int m = highEnd - low + 1; // mid Item[] sorted = new Item[m]; int rightCounter = 0; int lowInd = low, highInd = high; int index = 0; // Loop to store the count of // larger elements on right side // when both array have elements while (lowInd <= lowEnd && highInd <= highEnd) { if (items[lowInd].val < items[highInd].val) { rightCounter++; sorted[index++] = items[highInd++]; } else { count[items[lowInd].index] += rightCounter; sorted[index++] = items[lowInd++]; } } // Loop to store the count of // larger elements in right side // when only left array have // some element while (lowInd <= lowEnd) { count[items[lowInd].index] += rightCounter; sorted[index++] = items[lowInd++]; } // Loop to store the count of // larger elements in right side // when only right array have // some element while (highInd <= highEnd) { sorted[index++] = items[highInd++]; } Array.Copy(sorted, 0, items, low, m); } // Utility function that prints // the count of greater elements // on the right public static void PrintArray(List<int> countList) { foreach(int i in countList) { Console.Write(i + " "); } Console.WriteLine(); } // Driver Code public static void Main(string[] args) { // Given array int[] arr = { 3, 7, 1, 5, 9, 2 }; int n = arr.Length; // Function Call List<int> countList = CountLarge(arr); PrintArray(countList); }}// This code is contributed by akashish__ |
Javascript
//Javascript equivalent//Define an object Itemclass Item { constructor(val, index) { this.val = val this.index = index }}//Function to count large elementsfunction countLarge(arr) { // Length of the array const length = arr.length // Stores the index-value pairs const items = [] for (let i = 0; i < length; i++) { items.push(new Item(arr[i], i)) } // Stores the count of greater elements on right const count = [] for (let i = 0; i < length; i++) { count.push(0) } // Perform MergeSort operation mergeSort(items, 0, length - 1, count) const res = count.slice(0) return res}//Merge Sort functionfunction mergeSort(items, low, high, count) { // Base Case if (low >= high) return // Find Mid const mid = low + Math.floor((high - low) / 2) mergeSort(items, low, mid, count) mergeSort(items, mid + 1, high, count) // Merging step merge(items, low, mid, mid + 1, high, count)}//Merge functionfunction merge(items, low, lowEnd, high, highEnd, count) { const mid = highEnd - low + 1 // mid const sortedItems = [] for (let i = 0; i < mid; i++) { sortedItems.push(null) } let rightCounter = 0 let lowInd = low let highInd = high let index = 0 // Loop to store the count of larger elements on right side // when both array have elements while (lowInd <= lowEnd && highInd <= highEnd) { if (items[lowInd].val < items[highInd].val) { rightCounter++ sortedItems[index] = items[highInd] index++ highInd++ } else { count[items[lowInd].index] += rightCounter sortedItems[index] = items[lowInd] index++ lowInd++ } } // Loop to store the count of larger elements in right side // when only left array have elements while (lowInd <= lowEnd) { count[items[lowInd].index] += rightCounter sortedItems[index] = items[lowInd] index++ lowInd++ } // Loop to store the count of larger elements in right side // when only right array have elements while (highInd <= highEnd) { sortedItems[index] = items[highInd] index++ highInd++ } for (let i = 0; i < mid; i++) { items[low + i] = sortedItems[i] }}//Function to print arrayfunction printArray(countList) { let str = '' for (let i = 0; i < countList.length; i++) { str += countList[i] + ' ' } console.log(str)}// Driver Code // Given array const arr = [3, 7, 1, 5, 9, 2] // Function Call const countList = countLarge(arr) printArray(countList) |
Output
3 1 3 1 0 0
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
Another approach: We can use binary search to solve this. The idea is to create a sorted list of input and then for each element of input we first remove that element from the sorted list and then apply the modified binary search to find the element just greater than the current element and then the number of large elements will be the difference between the found index & the length of sorted list.
C++
#include <iostream>#include <algorithm>#include <vector>using namespace std;// Helper function to count the number of elements greater than 'item' in the sorted 'list'int CountLargeNumbers(int item, vector<int>& list) { int l=0; int r=list.size()-1; int mid = 0; while(l<r){ mid = l + (r-l)/2; if(list[mid] > item){ r = mid; } else{ l = mid + 1; } } if(l==r && item > list[l]){ return 0; } return list.size()-l;}// Helper function to delete the first occurrence of 'item' from the sorted 'list'void DeleteItemFromSortedList(vector<int>& list, int item) { int index = lower_bound(list.begin(), list.end(), item) - list.begin(); list.erase(list.begin() + index);}// Function to count the number of elements greater than each element in the input list 'list'vector<int> CountLarge(vector<int>& list) { // Create a sorted copy of the input list vector<int> sortedList = list; sort(sortedList.begin(), sortedList.end()); // For each element in the input list, count the number of elements greater than it for (int i = 0; i < list.size(); i++) { DeleteItemFromSortedList(sortedList, list[i]); list[i] = CountLargeNumbers(list[i], sortedList); } return list;}// Helper function to print the contents of a vectorvoid PrintArray(vector<int>& list) { for (int i = 0; i < list.size(); i++) { cout << list[i] << " "; } cout << endl;}// Main functionint main() { // Create an input vector vector<int> arr = {3, 7, 1, 5, 9, 2}; // Call the 'CountLarge' function to count the number of elements greater than each element in 'arr' vector<int> res = CountLarge(arr); // Print the result vector PrintArray(res); return 0;} |
Java
import java.util.*;public class Main { // Helper function to count the number of elements greater than 'item' in the sorted 'list' public static int countLargeNumbers(int item, List<Integer> list) { int l = 0; int r = list.size() - 1; int mid = 0; while (l < r) { mid = l + (r - l) / 2; if (list.get(mid) > item) { r = mid; } else { l = mid + 1; } } if (l == r && item > list.get(l)) { return 0; } return list.size() - l; } // Helper function to delete the first occurrence of 'item' from the sorted 'list' public static void deleteItemFromSortedList(List<Integer> list, int item) { int index = Collections.binarySearch(list, item); if (index >= 0) { list.remove(index); } } // Function to count the number of elements greater than each element in the input list 'list' public static List<Integer> countLarge(List<Integer> list) { // Create a sorted copy of the input list List<Integer> sortedList = new ArrayList<>(list); Collections.sort(sortedList); // For each element in the input list, count the number of elements greater than it for (int i = 0; i < list.size(); i++) { deleteItemFromSortedList(sortedList, list.get(i)); list.set(i, countLargeNumbers(list.get(i), sortedList)); } return list; } // Helper function to print the contents of a list public static void printList(List<Integer> list) { for (int i = 0; i < list.size(); i++) { System.out.print(list.get(i) + " "); } System.out.println(); } // Main function public static void main(String[] args) { // Create an input list List<Integer> arr = new ArrayList<>(Arrays.asList(3, 7, 1, 5, 9, 2)); // Call the 'countLarge' function to count the number of elements greater than each element in 'arr' List<Integer> res = countLarge(arr); // Print the result list printList(res); }} |
Python3
def CountLarge(list): sortedList = sorted(list) for i in range(len(list)): DeleteItemFromSortedList(sortedList, list[i]) list[i] = CountLargeNumbers(list[i], sortedList) return listdef CountLargeNumbers(item, list): l=0 r=len(list)-1 mid = 0 while(l<r): mid = l + (r-l)//2 if(list[mid] > item): r = mid else: l = mid + 1 if(l==r and item > list[l]): return 0 return len(list)-ldef DeleteItemFromSortedList(list, item): index = BinarySearch(list, item) list.pop(index)def BinarySearch(list, item): l=0 r=len(list)-1 mid = 0 while(l<=r): mid = l + (r-l)//2 if(list[mid] == item): return mid elif(list[mid] < item): l = mid + 1 else: r = mid - 1 return -1def PrintArray(list): for item in list: print(item, end=" ")arr = [3, 7, 1, 5, 9, 2]res = CountLarge(arr)PrintArray(res) |
C#
using System;using System.Collections.Generic;public class GFG{ static public void Main (){ //Code var arr = new List<int>(){3, 7, 1, 5, 9, 2}; var res = CountLarge(arr); PrintArray(res); } public static List<int> CountLarge(List<int> list) { var sortedList = new List<int>(list); sortedList.Sort(); for(int i=0;i<list.Count;i++){ DeleteItemFromSortedList(sortedList, list[i]); list[i] = CountLargeNumbers(list[i], sortedList); } return list; } public static int CountLargeNumbers(int item, List<int> list){ int l=0,r=list.Count-1,mid; while(l<r){ mid = l + (r-l)/2; if(list[mid] > item) r = mid; else l = mid + 1; } if(l==r && item > list[l]) return 0; return list.Count-l; } public static void DeleteItemFromSortedList(List<int> list, int item){ var index = BinarySearch(list, item); list.RemoveAt(index); } public static int BinarySearch(List<int> list, int item){ int l=0,r=list.Count-1,mid; while(l<=r){ mid = l + (r-l)/2; if(list[mid] == item) return mid; else if(list[mid] < item) l = mid + 1; else r = mid - 1; } return -1; } public static void PrintArray(List<int> list) { foreach(var item in list) Console.Write(item + " "); }} |
Javascript
<script>function main() { //Code const arr = [3, 7, 1, 5, 9, 2]; const res = countLarge(arr); printArray(res);}function countLarge(list) { const sortedList = [...list]; sortedList.sort(); for (let i = 0; i < list.length; i++) { deleteItemFromSortedList(sortedList, list[i]); list[i] = countLargeNumbers(list[i], sortedList); } return list;}function countLargeNumbers(item, list) { let l = 0; let r = list.length - 1; let mid; while (l < r) { mid = l + Math.floor((r - l) / 2); if (list[mid] > item) r = mid; else l = mid + 1; } if (l === r && item > list[l]) return 0; return list.length - l;}function deleteItemFromSortedList(list, item) { const index = binarySearch(list, item); list.splice(index, 1);}function binarySearch(list, item) { let l = 0; let r = list.length - 1; let mid; while (l <= r) { mid = l + Math.floor((r - l) / 2); if (list[mid] === item) return mid; else if (list[mid] < item) l = mid + 1; else r = mid - 1; } return -1;}function printArray(list) { for (const item of list) { document.write(item + " "); }}main();</script> |
Output
3 1 3 1 0 0
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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