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
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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.
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; } |
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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 |
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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. |
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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 |
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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 ?> |
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Output:
Max occurring digit is 5
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