Find the most frequent digit without using array/string

Given an integer, find the most occurring digit in it. If two or more digits occur same number of times, then return the highest of them. Input integer is given as an int variable, not as a string or array. Use of hash or array or string is not allowed. 

Example:

Input:  x = 12234
Output: The most frequent digit is 2
Input:  x = 1223377
Output: The most frequent digit is 7
Input:  x = 5
Output: The most frequent digit is 5
Input:  x = 1000
Output: The most frequent digit is 0

We strongly recommend you to minimize your browser and try this yourself first.

We could create a map of size 10 and store count of all digits, but use of any array/string is not allowed.
The idea is simple, we write a function that counts occurrences of a given digit in a given integer. Then we count all digits from 0 to 9 in given integer. We keep updating maximum count whenever count becomes more or same as previous count. Below is the implementation.

Implementation:

C++

// Finds maximum occurring digit without using any array/string
#include <bits/stdc++.h>
using namespace std;

// Simple function to count occurrences of digit d in x
int countOccurrences(long int x, int d)
{
    int count = 0;  // Initialize count of digit d
    while (x)
    {
        // Increment count if current digit is same as d
        if (x%10 == d)
           count++;
        x = x/10;
    }
    return count;
}

// Returns the max occurring digit in x
int maxOccurring(long int x)
{
   // Handle negative number
   if (x < 0)
      x = -x;

   int result = 0; // Initialize result which is a digit
   int max_count = 1; // Initialize count of result

   // Traverse through all digits
   for (int d=0; d<=9; d++)
   {
      // Count occurrences of current digit
      int count = countOccurrences(x, d);

      // Update max_count and result if needed
      if (count >= max_count)
      {
         max_count = count;
         result = d;
      }
   }
   return result;
}

// Driver program
int main()
{
    long int x = 1223355;
    cout << "Max occurring digit is " << maxOccurring(x);
    return 0;
}

Java

// Finds maximum occurring digit
// without using any array/string
import java.io.*;

class GFG
{

// Simple function to count
// occurrences of digit d in x
static int countOccurrences(int x,
                            int d)
{
    // Initialize count
    // of digit d
    int count = 0;
    while (x > 0)
    {
        // Increment count if
        // current digit is
        // same as d
        if (x % 10 == d)
        count++;
        x = x / 10;
    }
    return count;
}

// Returns the max
// occurring digit in x
static int maxOccurring( int x)
{

// Handle negative number
if (x < 0)
    x = -x;

// Initialize result
// which is a digit
int result = 0;

// Initialize count
// of result
int max_count = 1;

// Traverse through
// all digits
for (int d = 0; d <= 9; d++)
{
    // Count occurrences
    // of current digit
    int count = countOccurrences(x, d);

    // Update max_count
    // and result if needed
    if (count >= max_count)
    {
        max_count = count;
        result = d;
    }
}
return result;
}

// Driver Code
public static void main (String[] args)
{
    int x = 1223355;
    System.out.println("Max occurring digit is " +
                                 maxOccurring(x));

}
}

// This code is contributed
// by akt_mit

Python3

# Finds maximum occurring digit
# without using any array/string

# Simple function to count
# occurrences of digit d in x
def countOccurrences(x, d):
    count = 0; # Initialize count
               # of digit d
    while (x):

        # Increment count if current
        # digit is same as d
        if (x % 10 == d):
            count += 1;
        x = int(x / 10);

    return count;

# Returns the max occurring
# digit in x
def maxOccurring(x):

    # Handle negative number
    if (x < 0):
        x = -x;

    result = 0; # Initialize result
                # which is a digit
    max_count = 1; # Initialize count
                   # of result

    # Traverse through all digits
    for d in range(10):

        # Count occurrences of current digit
        count = countOccurrences(x, d);

        # Update max_count and
        # result if needed
        if (count >= max_count):
            max_count = count;
            result = d;

    return result;

# Driver Code
x = 1223355;
print("Max occurring digit is",
              maxOccurring(x));

# This code is contributed by mits.

C#

// Finds maximum occurring digit
// without using any array/string
class GFG
{

// Simple function to count
// occurrences of digit d in x
static int countOccurrences(int x, int d)
{
    // Initialize count
    // of digit d
    int count = 0;
    while (x > 0)
    {
        // Increment count if
        // current digit is
        // same as d
        if (x % 10 == d)
        count++;
        x = x / 10;
    }
    return count;
}

// Returns the max
// occurring digit in x
static int maxOccurring( int x)
{

// Handle negative number
if (x < 0)
    x = -x;

// Initialize result
// which is a digit
int result = 0;

// Initialize count
// of result
int max_count = 1;

// Traverse through
// all digits
for (int d = 0; d <= 9; d++)
{
    // Count occurrences
    // of current digit
    int count = countOccurrences(x, d);

    // Update max_count
    // and result if needed
    if (count >= max_count)
    {
        max_count = count;
        result = d;
    }
}
return result;
}

// Driver Code
static void Main()
{
    int x = 1223355;
    System.Console.WriteLine("Max occurring digit is " +
                                       maxOccurring(x));
}
}

// This code is contributed by mits

Javascript

<script>

// Finds maximum occurring digit
// without using any array/string


// Simple function to count
// occurrences of digit d in x
function countOccurrences(x, d)
{
    // Initialize count
    // of digit d
    var count = 0;
    while (x > 0)
    {
        // Increment count if
        // current digit is
        // same as d
        if (x % 10 == d)
        count++;
        x = parseInt(x / 10);
    }
    return count;
}

// Returns the max
// occurring digit in x
function maxOccurring(x)
{

    // Handle negative number
    if (x < 0)
        x = -x;

    // Initialize result
    // which is a digit
    var result = 0;

    // Initialize count
    // of result
    var max_count = 1;

    // Traverse through
    // all digits
    for (d = 0; d <= 9; d++)
    {
        // Count occurrences
        // of current digit
        var count = countOccurrences(x, d);

        // Update max_count
        // and result if needed
        if (count >= max_count)
        {
            max_count = count;
            result = d;
        }
    }
    return result;
}

// Driver Code
var x = 1223355;
document.write("Max occurring digit is " +
                             maxOccurring(x));

// This code contributed by shikhasingrajput

</script>

PHP

<?php
// Finds maximum occurring digit
// without using any array/string

// Simple function to count
// occurrences of digit d in x
function countOccurrences($x, $d)
{
    // Initialize count of digit d
    $count = 0;
    while ($x)
    {
        // Increment count if current
        // digit is same as d
        if ($x % 10 == $d)
        $count++;
        $x = (int)($x / 10);
    }
    return $count;
}

// Returns the max occurring
// digit in x
function maxOccurring($x)
{
// Handle negative number
if ($x < 0)
    $x = -$x;

$result = 0; // Initialize result
             // which is a digit
$max_count = 1; // Initialize count of result

// Traverse through all digits
for ($d = 0; $d <= 9; $d++)
{
    // Count occurrences of
    // current digit
    $count = countOccurrences($x, $d);

    // Update max_count and result
    // if needed
    if ($count >= $max_count)
    {
        $max_count = $count;
        $result = $d;
    }
}
return $result;
}

// Driver Code
$x = 1223355;
echo "Max occurring digit is " .
               maxOccurring($x);

// This code is contributed by mits
?>

Output

Max occurring digit is 5

Time Complexity: O(log₁₀x)
Auxiliary Space: O(1)

Approach#2: using division and modulus operator.

Count the frequency of each digit using division and modulus operator.

Algorithm

1. Initialize a variable “max_freq” to store the maximum frequency of any digit.
2. Initialize a variable “most_frequent_digit” to store the most frequent digit.
3. Repeat the following steps for each digit in the input number:
a. Initialize a variable “count” to 0.
b. Divide the input number by 10 repeatedly until it becomes 0 or less.
c. If the remainder of the division is equal to the current digit, increment “count”.
d. If “count” is greater than “max_freq”, update “max_freq” to “count” and “most_frequent_digit” to the current digit.
4. Return “most_frequent_digit”.

Java

public class GFG {

    // Function to find the most frequent digit in a number
    static int mostFrequentDigit(int x) {
        int maxFreq = 0;
        int mostFrequentDigit = x % 10;
        while (x > 0) {
            int digit = x % 10;
            int count = 0;
            int temp = x;
            while (temp > 0) {
                if (temp % 10 == digit) {
                    count++;
                }
                temp /= 10;
            }
            if (count > maxFreq) {
                maxFreq = count;
                mostFrequentDigit = digit;
            }
            x /= 10;
        }
        return mostFrequentDigit;
    }

    public static void main(String[] args) {
        int x = 1223355;
        System.out.println(mostFrequentDigit(x));
    }
}

Python3

def most_frequent_digit(x):
    max_freq = 0
    most_frequent_digit = x % 10
    while x > 0:
        digit = x % 10
        count = 0
        temp = x
        while temp > 0:
            if temp % 10 == digit:
                count += 1
            temp //= 10
        if count > max_freq:
            max_freq = count
            most_frequent_digit = digit
        x //= 10
    return most_frequent_digit

x = 1223355
print(most_frequent_digit(x))

Javascript

// This function returns the most frequent digit in a given number
function most_frequent_digit(x) {
    let max_freq = 0;
    let most_frequent_digit = x % 10;

    // Loop through each digit of the number
    while (x > 0) {
        let digit = x % 10;
        let count = 0;
        let temp = x;

        // Count the number of occurrences of the current digit in the number
        while (temp > 0) {
            if (temp % 10 == digit) {
                count += 1;
            }
            temp = Math.floor(temp / 10);
        }

        // If the current digit occurs more frequently than the previous most frequent digit, update the most frequent digit
        if (count > max_freq) {
            max_freq = count;
            most_frequent_digit = digit;
        }
        x = Math.floor(x / 10);
    }
    return most_frequent_digit;
}

// Example usage
let x = 1223355;
console.log(most_frequent_digit(x));

Output

5

Time complexity: O(d * log10(x)), where d is the number of digits in the input number.
Auxiliary Space: O(1)