Given an array arr of integers of size N, the task is to find, for every element, the number of elements that are greater than it.
Examples:
Input: arr[] = {4, 6, 2, 1, 8, 7}
Output: {3, 2, 4, 5, 0, 1}
Input: arr[] = {2, 3, 4, 5, 6, 7, 8}
Output: {6, 5, 4, 3, 2, 1, 0}
Approach: Store the frequencies of every array element using a Map. Iterate the Map in reverse and store the sum of the frequency of all previously traversed elements (i.e. elements greater than it) for every element.
Below code is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
void countOfGreaterElements( int arr[], int n)
{ // Store the frequency of the
// array elements
map< int , int > mp;
for ( int i = 0; i < n; i++) {
mp[arr[i]]++;
}
int x = 0;
// Store the sum of frequency of elements
// greater than the current element
for ( auto it = mp.rbegin(); it != mp.rend(); it++) {
int temp = it->second;
mp[it->first] = x;
x += temp;
}
for ( int i = 0; i < n; i++)
cout << mp[arr[i]] << " " ;
} // Driver code int main()
{ int arr[] = { 7, 9, 5, 2, 1, 3, 4, 8, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
countOfGreaterElements(arr, n);
return 0;
} |
// Java implementation of the above approach import java.util.*;
class GfG {
public static void countOfGreaterElements( int arr[])
{
int n = arr.length;
TreeMap<Integer, Integer> mp = new TreeMap<Integer, Integer>(Collections.reverseOrder());
// Store the frequency of the
// array elements
for ( int i = 0 ; i < n; i++) {
mp.put(arr[i], mp.getOrDefault(arr[i], 0 ) + 1 );
}
// Store the sum of frequency of elements
// greater than the current element
int x = 0 ;
for (Map.Entry<Integer, Integer> e : mp.entrySet()) {
Integer temp = e.getValue();
mp.put(e.getKey(), x);
x += temp;
}
for ( int i = 0 ; i < n; i++)
System.out.print(mp.get(arr[i]) + " " );
}
public static void main(String args[])
{
int arr[] = { 7 , 9 , 5 , 2 , 1 , 3 , 4 , 8 , 6 };
countOfGreaterElements(arr);
}
} |
# Python 3 implementation of the above approach def countOfGreaterElements(arr, n):
# Store the frequency of the
# array elements
mp = {i: 0 for i in range ( 1000 )}
for i in range (n):
mp[arr[i]] + = 1
x = 0
# Store the sum of frequency of elements
# greater than the current element
p = []
q = []
m = []
for key, value in mp.items():
m.append([key, value])
m = m[:: - 1 ]
for p in m:
temp = p[ 1 ]
mp[p[ 0 ]] = x
x + = temp
for i in range (n):
print (mp[arr[i]], end = " " )
# Driver code if __name__ = = '__main__' :
arr = [ 7 , 9 , 5 , 2 , 1 , 3 , 4 , 8 , 6 ]
n = len (arr)
countOfGreaterElements(arr, n)
# This code is contributed by Surendra_Gangwar |
// C# implementation of the above approach using System;
using System.Collections.Generic;
class GfG {
public static void countOfGreaterElements( int [] arr)
{
int n = arr.Length;
SortedDictionary< int , int > mp =
new SortedDictionary< int , int >();
// Store the frequency of the
// array elements
for ( int i = 0; i < n; i++) {
int temp = 0;
if (mp.ContainsKey(arr[i]))
{
temp = mp[arr[i]];
mp.Remove(arr[i]);
}
mp[arr[i]] = temp + 1;
}
// Store the sum of frequency of elements
// greater than the current element
int x = 0;
var k1 = mp.Keys;
int [] keys = new int [mp.Count];
k1.CopyTo(keys, 0);
Array.Reverse(keys);
foreach ( var e in keys) {
int temp = mp[e];
mp.Remove(e);
mp[e] = x;
x += temp;
}
for ( int i = 0; i < n; i++)
Console.Write(mp[arr[i]] + " " );
}
public static void Main( string [] args)
{
int [] arr = { 7, 9, 5, 2, 1, 3, 4, 8, 6 };
countOfGreaterElements(arr);
}
} // This code is contributed by phasing17 |
// JavaScript implementation of the above approach function countOfGreaterElements(arr, n)
{ // Store the frequency of the
// array elements
let mp = {}
for ( var i = 0; i < n; i++)
{
if (!mp.hasOwnProperty(arr[i]))
mp[arr[i]] = 0
mp[arr[i]] = mp[arr[i]] + 1
}
let x = 0
// Store the sum of frequency of elements
// greater than the current element
let m = []
for ( var [key, value] of Object.entries(mp))
{
key = parseInt(key)
value = parseInt(value)
m.push([key, value])
}
for ( var i = m.length - 1; i >= 0; i--)
{
var p = m[i]
var temp = p[1]
mp[p[0]] = x
x += temp
}
for ( var i = 0; i < n; i++)
process.stdout.write(mp[arr[i]] + " " )
} // Driver code let arr = [7, 9, 5, 2, 1, 3, 4, 8, 6] let n = arr.length countOfGreaterElements(arr, n) // This code is contributed by phasing17 |
2 0 4 7 8 6 5 1 3
Time Complexity: O(N log(N))
Auxiliary Space: O(N)