Fill an array based on frequency where elements are in range from 0 to n-1
Given an array of positive integers with duplicate allowed. The array contains elements from range 0 to n-1. The task is to fill the array such that arr[i] contains the frequency of i.
Examples :
Input : arr[] = {1, 4, 3, 4, 1, 1, 4, 4, 4, 7}
Output : arr[] = {0, 3, 0, 1, 5, 0, 0, 1, 0, 0}
Here 0 appears 0 times, so arr[0] is 0
1 appears 3 times
2 appears 0 times
3 appears 0 times
4 appears 5 times
..........
Input : arr[] = {1, 2, 3, 4, 1, 1, 4, 5, 6, 7}
Output : arr[] = {0, 3, 1, 1, 2, 1, 1, 1, 0, 0}
A simple solution of this problem is to run two loops. The outer loop picks elements one by one. The inner loop counts frequency of the picked element and stores frequency in final array.
An efficient solution is to use an auxiliary array of size n.
1) Create an array temp[0..n-1] and
initialize it as 0.
2) Traverse given array and do following
for every element arr[i].
temp[arr[i]]++
3) Copy temp[] to arr[].
Below is the implementation of above steps.
C++
#include<bits/stdc++.h>
using namespace std;
void fillWithFreq( int arr[], int n)
{
int temp[n];
memset (temp, 0, sizeof (temp));
for ( int i=0; i<n; i++)
temp[arr[i]] += 1;
for ( int i=0; i<n; i++)
arr[i] = temp[i];
}
int main()
{
int arr[] = {5, 2, 3, 4, 5, 5, 4, 5, 6, 7};
int n = sizeof (arr)/ sizeof (arr[0]);
fillWithFreq(arr, n);
for ( int i=0; i<n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java.util.Arrays;
class GFG {
static void fillWithFreq( int arr[], int n)
{
int temp[]= new int [n];
Arrays.fill(temp, 0 );
for ( int i = 0 ; i < n; i++)
temp[arr[i]] += 1 ;
for ( int i = 0 ; i < n; i++)
arr[i] = temp[i];
}
public static void main(String[] args)
{
int arr[] = { 5 , 2 , 3 , 4 , 5 , 5 , 4 , 5 , 6 , 7 };
int n = arr.length;
fillWithFreq(arr, n);
for ( int i= 0 ; i<n; i++)
System.out.print(arr[i] + " " );
}
}
|
Python3
def fillWithFreq(arr, n):
temp = [ 0 for i in range (n)]
for i in range (n):
temp[arr[i]] + = 1
for i in range (n):
arr[i] = temp[i]
arr = [ 5 , 2 , 3 , 4 , 5 , 5 , 4 , 5 , 6 , 7 ]
n = len (arr)
fillWithFreq(arr, n)
for i in range (n):
print (arr[i], end = " " )
|
C#
using System;
class GFG {
static void fillWithFreq( int []arr, int n)
{
int []temp = new int [n];
for ( int i = 0; i < n; i++)
temp[i] = 0;
for ( int i = 0; i < n; i++)
temp[arr[i]] += 1;
for ( int i = 0; i < n; i++)
arr[i] = temp[i];
}
public static void Main()
{
int []arr = {5, 2, 3, 4, 5,
5, 4, 5, 6, 7};
int n = arr.Length;
fillWithFreq(arr, n);
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
}
|
PHP
<?php
function fillWithFreq( $arr , $n )
{
$temp = array ();
for ( $i = 0; $i < $n ; $i ++)
$temp [ $i ] = 0;
for ( $i = 0; $i < $n ; $i ++)
$temp [ $arr [ $i ]] += 1;
for ( $i = 0; $i < $n ; $i ++)
$arr [ $i ] = $temp [ $i ];
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ] . " " ;
}
$arr = array (5, 2, 3, 4, 5,
5, 4, 5, 6, 7);
$n = count ( $arr );
fillWithFreq( $arr , $n );
?>
|
Javascript
<script>
function fillWithFreq(arr, n)
{
let temp = new Array(n);
for (let i = 0; i < n; i++)
temp[i] = 0;
for (let i = 0; i < n; i++)
temp[arr[i]] += 1;
for (let i = 0; i < n; i++)
arr[i] = temp[i];
}
let arr = [5, 2, 3, 4, 5, 5, 4, 5, 6, 7];
let n = arr.length;
fillWithFreq(arr, n);
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
</script>
|
Output
0 0 1 1 2 4 1 1 0 0
Time Complexity : O(n)
Auxiliary Space : O(n)
Last Updated :
13 Sep, 2023
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