Check if frequency of each digit is less than the digit
Last Updated :
13 Sep, 2023
Given an integer n, the task is to check if frequency of each digit of the number is less than or equal to digit itself.
Examples:
Input : 51241
Output : False
Input : 1425243
Output : True
Naive Approach: Start from 0 and count the frequency for every digit upto 9, if at any place frequency is more than the digit value then return false, else return true.
C++
#include<bits/stdc++.h>
using namespace std;
bool validate( long long int n)
{
for ( int i=0; i<10; i++)
{
long long int temp = n;
int count = 0;
while (temp)
{
if (temp % 10 == i)
count++;
if (count > i)
return false ;
temp /= 10;
}
}
return true ;
}
int main()
{
long long int n = 1552793;
if (validate(n))
cout << "True" ;
else
cout << "False" ;
return 0;
}
|
Java
import java .io.*;
public class GFG {
static boolean validate( long n)
{
for ( int i = 0 ; i < 10 ; i++)
{
long temp = n;
int count = 0 ;
while (temp > 0 )
{
if (temp % 10 == i)
count++;
if (count > i)
return false ;
temp /= 10 ;
}
}
return true ;
}
static public void main (String[] args)
{
long n = 1552793 ;
if (validate(n))
System.out.println( "True" );
else
System.out.println( "False" );
}
}
|
Python3
def validate(n):
for i in range ( 10 ):
temp = n;
count = 0 ;
while (temp):
if (temp % 10 = = i):
count + = 1 ;
if (count > i):
return - 1 ;
temp / / = 10 ;
return 1 ;
n = 1552793 ;
geek = "True" if validate(n) else "False" ;
print (geek);
|
C#
using System;
public class GFG {
static bool validate( long n)
{
for ( int i = 0; i < 10; i++)
{
long temp = n;
int count = 0;
while (temp > 0)
{
if (temp % 10 == i)
count++;
if (count > i)
return false ;
temp /= 10;
}
}
return true ;
}
static public void Main(String[] args)
{
long n = 1552793;
if (validate(n))
Console.WriteLine( "True" );
else
Console.WriteLine( "False" );
}
}
|
PHP
<?php
function validate( $n )
{
for ( $i = 0; $i < 10; $i ++)
{
$temp = $n ;
$count = 0;
while ( $temp )
{
if ( $temp % 10 == $i )
$count ++;
if ( $count > $i )
return -1;
$temp /= 10;
}
}
return 1;
}
$n = 1552793;
$geek = validate( $n ) ? "True" : "False" ;
echo ( $geek );
?>
|
Javascript
<script>
function validate(n)
{
for (let i = 0; i < 10; i++)
{
let temp = n;
let count = 0;
while (temp > 0)
{
if (temp % 10 == i)
count++;
if (count > i)
return false ;
temp /= 10;
}
}
return true ;
}
let n = 1552793;
if (validate(n))
document.write( "True" );
else
document.write( "False" );
</script>
|
Output:
True
Time Complexity: O(10 * log10n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach: is to store the frequency of each digit and if at any place frequency is more than the digit value then return false, else return true.
C++
#include<bits/stdc++.h>
using namespace std;
bool validate( long long int n)
{
int count[10] = {0};
while (n)
{
int r = n % 10;
if (count[r] == r)
return false ;
count[r]++;
n /= 10;
}
return true ;
}
int main()
{
long long int n = 1552793;
if (validate(n))
cout << "True" ;
else
cout << "False" ;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static boolean validate( long n)
{
int count[] = new int [ 10 ] ;
while (n > 0 )
{
int r = ( int )n % 10 ;
if (count[r] == r)
return false ;
count[r]++;
n /= 10 ;
}
return true ;
}
public static void main (String[] args)
{
long n = 1552793 ;
if (validate(n))
System.out.println( "True" );
else
System.out.println( "False" );
}
}
|
Python3
import math as mt
def validate(n):
count = [ 0 for i in range ( 10 )]
while (n > 0 ):
r = n % 10
if (count[r] = = r):
return False
count[r] + = 1
n = n / / 10
return True
n = 1552793
if (validate(n)):
print ( "True" )
else :
print ( "False" )
|
C#
using System;
class GFG
{
static bool validate( long n)
{
int []count = new int [10] ;
while (n > 0)
{
int r = ( int )n % 10;
if (count[r] == r)
return false ;
count[r]++;
n /= 10;
}
return true ;
}
static public void Main ()
{
long n = 1552793;
if (validate(n))
Console.WriteLine( "True" );
else
Console.WriteLine( "False" );
}
}
|
PHP
<?php
function validate( $n )
{
$count = array (10);
while ( $n )
{
$r = $n % 10;
if (( $count [ $r ] == $r ))
{
return false;
}
$count [ $r ] = $count [ $r ] + 1;
$n = $n / 10;
}
return true;
}
$n = 1552793;
$geek = validate( $n ) ? "True" : "False" ;
echo ( $geek );
?>
|
Javascript
<script>
function validate(n)
{
let count = new Uint8Array(10);
while (n)
{
let r = n % 10;
if (count[r] == r)
return false ;
count[r]++;
n = Math.floor(n / 10);
}
return true ;
}
let n = 1552793;
if (validate(n))
document.write( "True" );
else
document.write( "False" );
</script>
|
Output:
True
Time Complexity: O(log10n), where n represents the given integer.
Auxiliary Space: O(10), no extra space is required, so it is a constant.
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