Given an array arr, the task is to print the elements of the array in descending order along with their frequencies.
Examples:
Input: arr[] = {1, 3, 3, 3, 4, 4, 5}
Output: 5 occurs 1 times
4 occurs 2 times
3 occurs 3 times
1 occurs 1 times
Input: arr[] = {1, 1, 1, 2, 3, 4, 9, 9, 10}
Output: 10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times
Naive approach: Use some Data-Structure (e.g. multiset) that stores elements in decreasing order and then print the elements one by one with its count and then erase it from the Data-structure. The time complexity will be O(N log N) and the auxiliary space will be O(N) for the Data-structure used.
Below is the implementation of the above approach:
// C++ program to print the elements in // descending along with their frequencies #include <bits/stdc++.h> using namespace std;
// Function to print the elements in descending // along with their frequencies void printElements( int a[], int n)
{ // A multiset to store elements in decreasing order
multiset< int , greater< int > > ms;
// Insert elements in the multiset
for ( int i = 0; i < n; i++) {
ms.insert(a[i]);
}
// Print the elements along with their frequencies
while (!ms.empty()) {
// Find the maximum element
int maxel = *ms.begin();
// Number of times it occurs
int times = ms.count(maxel);
cout << maxel << " occurs " << times << " times\n" ;
// Erase the maxel
ms.erase(maxel);
}
} // Driver Code int main()
{ int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = sizeof (a) / sizeof (a[0]);
printElements(a, n);
return 0;
} |
# Function to print the elements in descending # along with their frequencies def printElements(a):
# A multiset to store elements in decreasing order
ms = sorted (a, reverse = True )
# Print the elements along with their frequencies
while ms:
# Find the maximum element
maxel = ms[ 0 ]
# Number of times it occurs
times = ms.count(maxel)
print (f "{maxel} occurs {times} times" )
# Remove all occurrences of maxel
ms = [x for x in ms if x ! = maxel]
# Driver code if __name__ = = '__main__' :
a = [ 1 , 1 , 1 , 2 , 3 , 4 , 9 , 9 , 10 ]
printElements(a)
|
// JavaScript program to print the elements in // descending along with their frequencies in reverse order // Function to print the elements in descending // along with their frequencies in reverse order function printElements(a) {
// A Map to store elements in decreasing order
const ms = new Map();
// Insert elements in the Map
for (let i = 0; i < a.length; i++) {
const key = a[i];
ms.set(key, (ms.get(key) || 0) + 1);
}
// Array to store the elements along with their frequencies
const elements = [];
// Store the elements and their frequencies in the array
for (const [key, value] of ms.entries()) {
elements.push(`${key} occurs ${value} times`);
}
// Print the elements along with their frequencies in reverse order
for (let i = elements.length - 1; i >= 0; i--) {
console.log(elements[i]);
}
} // Driver Code const a = [1, 1, 1, 2, 3, 4, 9, 9, 10]; printElements(a); |
import java.util.*;
public class Main {
// Function to print the elements in descending
// along with their frequencies
static void printElements( int [] a, int n) {
// A TreeMap to store elements in decreasing order
TreeMap<Integer, Integer> tm = new TreeMap<>(Collections.reverseOrder());
// Insert elements in the TreeMap
for ( int i = 0 ; i < n; i++) {
if (tm.containsKey(a[i])) {
tm.put(a[i], tm.get(a[i]) + 1 );
} else {
tm.put(a[i], 1 );
}
}
// Print the elements along with their frequencies
for (Map.Entry<Integer, Integer> entry : tm.entrySet()) {
int element = entry.getKey();
int frequency = entry.getValue();
System.out.println(element + " occurs " + frequency + " times" );
}
}
// Driver Code
public static void main(String[] args) {
int [] a = { 1 , 1 , 1 , 2 , 3 , 4 , 9 , 9 , 10 };
int n = a.length;
printElements(a, n);
}
} |
using System;
using System.Collections.Generic;
public class Program {
// Function to print the elements in descending
// along with their frequencies
static void PrintElements( int [] a, int n) {
// A SortedDictionary to store elements in decreasing order
SortedDictionary< int , int > sd = new SortedDictionary< int , int >( new DescendingComparer());
// Insert elements in the SortedDictionary
for ( int i = 0; i < n; i++) {
if (sd.ContainsKey(a[i])) {
sd[a[i]]++;
} else {
sd[a[i]] = 1;
}
}
// Print the elements along with their frequencies
foreach (KeyValuePair< int , int > entry in sd) {
int element = entry.Key;
int frequency = entry.Value;
Console.WriteLine(element + " occurs " + frequency + " times" );
}
}
// Custom comparer to sort the keys in descending order
class DescendingComparer : IComparer< int > {
public int Compare( int x, int y) {
return y.CompareTo(x);
}
}
// Driver Code
public static void Main() {
int [] a = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = a.Length;
PrintElements(a, n);
}
} |
10 occurs 1 times 9 occurs 2 times 4 occurs 1 times 3 occurs 1 times 2 occurs 1 times 1 occurs 3 times
Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal, we are doing a multiset operation which will cost us logN time.
Auxiliary Space: O(N), as we are using extra space for the multiset.
Efficient Approach: Sort the array in descending order and then start printing the elements from the beginning along with their frequencies.
Below is the implementation of the above approach:
// C++ program to print the elements in // descending along with their frequencies #include <bits/stdc++.h> using namespace std;
// Function to print the elements in descending // along with their frequencies void printElements( int a[], int n)
{ // Sorts the element in decreasing order
sort(a, a + n, greater< int >());
int cnt = 1;
// traverse the array elements
for ( int i = 0; i < n - 1; i++) {
// Prints the number and count
if (a[i] != a[i + 1]) {
cout << a[i] << " occurs " << cnt << " times\n" ;
cnt = 1;
}
else
cnt += 1;
}
// Prints the last step
cout << a[n - 1] << " occurs " << cnt << " times\n" ;
} // Driver Code int main()
{ int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = sizeof (a) / sizeof (a[0]);
printElements(a, n);
return 0;
} |
// Java program to print the elements in // descending along with their frequencies import java.util.*;
class GFG
{ // Function to print the elements in descending // along with their frequencies static void printElements( int a[], int n)
{ // Sorts the element in decreasing order
Arrays.sort(a);
a = reverse(a);
int cnt = 1 ;
// traverse the array elements
for ( int i = 0 ; i < n - 1 ; i++)
{
// Prints the number and count
if (a[i] != a[i + 1 ])
{
System.out.print(a[i]+ " occurs " +
cnt + " times\n" );
cnt = 1 ;
}
else
cnt += 1 ;
}
// Prints the last step
System.out.print(a[n - 1 ]+ " occurs " +
cnt + " times\n" );
} static int [] reverse( int a[])
{ int i, n = a.length, t;
for (i = 0 ; i < n / 2 ; i++)
{
t = a[i];
a[i] = a[n - i - 1 ];
a[n - i - 1 ] = t;
}
return a;
} // Driver Code public static void main(String[] args)
{ int a[] = { 1 , 1 , 1 , 2 , 3 , 4 , 9 , 9 , 10 };
int n = a.length;
printElements(a, n);
} } // This code is contributed by PrinciRaj1992 |
# Python3 program to print the elements in # descending along with their frequencies # Function to print the elements in # descending along with their frequencies def printElements(a, n) :
# Sorts the element in decreasing order
a.sort(reverse = True )
cnt = 1
# traverse the array elements
for i in range (n - 1 ) :
# Prints the number and count
if (a[i] ! = a[i + 1 ]) :
print (a[i], " occurs " , cnt, "times" )
cnt = 1
else :
cnt + = 1
# Prints the last step
print (a[n - 1 ], "occurs" , cnt, "times" )
# Driver Code if __name__ = = "__main__" :
a = [ 1 , 1 , 1 , 2 ,
3 , 4 , 9 , 9 , 10 ]
n = len (a)
printElements(a, n)
# This code is contributed by Ryuga |
// C# program to print the elements in // descending along with their frequencies using System;
class GFG
{ // Function to print the elements in descending // along with their frequencies static void printElements( int []a, int n)
{ // Sorts the element in decreasing order
Array.Sort(a);
a = reverse(a);
int cnt = 1;
// traverse the array elements
for ( int i = 0; i < n - 1; i++)
{
// Prints the number and count
if (a[i] != a[i + 1])
{
Console.Write(a[i]+ " occurs " +
cnt + " times\n" );
cnt = 1;
}
else
cnt += 1;
}
// Prints the last step
Console.Write(a[n - 1]+ " occurs " +
cnt + " times\n" );
} static int [] reverse( int []a)
{ int i, n = a.Length, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
} // Driver Code public static void Main(String[] args)
{ int []a = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = a.Length;
printElements(a, n);
} } // This code is contributed by PrinciRaj1992 |
<?php // PHP program to print the elements in // descending along with their frequencies // Function to print the elements in // descending along with their frequencies function printElements(& $a , $n )
{ // Sorts the element in
// decreasing order
rsort( $a );
$cnt = 1;
// traverse the array elements
for ( $i = 0; $i < $n - 1; $i ++)
{
// Prints the number and count
if ( $a [ $i ] != $a [ $i + 1])
{
echo ( $a [ $i ]);
echo ( " occurs " );
echo $cnt ;
echo ( " times\n" );
$cnt = 1;
}
else
$cnt += 1;
}
// Prints the last step
echo ( $a [ $n - 1]);
echo ( " occurs " );
echo $cnt ;
echo ( " times\n" );
} // Driver Code $a = array (1, 1, 1, 2, 3,
4, 9, 9, 10 );
$n = sizeof( $a );
printElements( $a , $n );
// This code is contributed // by Shivi_Aggarwal ?> |
<script> // javascript program to print the elements in // descending along with their frequencies // Function to print the elements in descending // along with their frequencies function printElements(a, n)
{ // Sorts the element in decreasing order
a=a.sort(compare);
a = reverse(a);
var cnt = 1;
// traverse the array elements
for ( var i = 0; i < n - 1; i++)
{
// Prints the number and count
if (a[i] != a[i + 1])
{
document.write(a[i]+ " occurs " + cnt + " times" + "<br>" );
cnt = 1;
}
else
cnt += 1;
}
// Prints the last step
document.write(a[n - 1]+ " occurs " + cnt + " times" + "<br>" );
} function reverse(a){
var i, n = a.length, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
} function compare(a, b) {
if (a < b) {
return -1;
} else if (a > b) {
return 1;
} else {
return 0;
}
} // Driver Code var a = [ 1, 1, 1, 2, 3, 4, 9, 9, 10 ];
var n = a.length;
printElements(a, n);
// This code is contributed by bunnyram19.
</script> |
10 occurs 1 times 9 occurs 2 times 4 occurs 1 times 3 occurs 1 times 2 occurs 1 times 1 occurs 3 times
Time Complexity: O(N*logN), as we are using the sort function which will cost us O(N*logN) time.
Auxiliary Space: O(1), as we are not using any extra space.