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 program for the above approach import 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);
}
} |
from typing import List
class Item:
def __init__( self , val: int , index: int ):
self .val = val
self .index = index
def 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 res
def 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_items
def print_array(count_list: List [ int ]) - > None :
print ( ' ' .join( str (i) for i in count_list))
# Driver Code if __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 |
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 equivalent //Define an object Item class Item { constructor(val, index) {
this .val = val
this .index = index
}
} //Function to count large elements function 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 function function 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 function function 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 array function 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)
|
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.
#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 vector void PrintArray(vector< int >& list) {
for ( int i = 0; i < list.size(); i++) {
cout << list[i] << " " ;
}
cout << endl;
} // Main function int 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;
} |
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);
}
} |
def CountLarge( list ):
sortedList = sorted ( list )
for i in range ( len ( list )):
DeleteItemFromSortedList(sortedList, list [i])
list [i] = CountLargeNumbers( list [i], sortedList)
return list
def 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 ) - l
def 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 - 1
def PrintArray( list ):
for item in list :
print (item, end = " " )
arr = [ 3 , 7 , 1 , 5 , 9 , 2 ]
res = CountLarge(arr)
PrintArray(res) |
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 + " " );
}
} |
<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> |
3 1 3 1 0 0
Time Complexity: O(N^2)
Auxiliary Space: O(N)