Find the array element having maximum frequency of the digit K
Last Updated :
19 Sep, 2023
Given an array arr[] of size N and an integer K, the task is to find an array element that contains the digit K a maximum number of times. If more than one solutions exist, then print any one of them. Otherwise, print -1.
Examples:
Input: arr[] = {3, 77, 343, 456}, K = 3
Output: 343
Explanation:
Frequency of 3 in array elements: 1, 0, 2, 0
343 has maximum frequency i.e. 2.
Input: arr[] = {1, 1111, 111, 11}, K = 1
Output: 1111
Explanation:
Frequency of 1 in array elements: 1, 4, 3, 2
1111 has maximum frequency i.e. 4.
Approach: The idea is to traverse the array and for every individual array element, count the occurrences of the digit K in it. Follow the steps below to solve the problem:
- Initialize the max frequency of digit K, say maxfreq, as 0.
- Traverse the given array from the start element till the end.
- For every traversed element, find the frequency of digit K in that element. If it is greater than maxfreq, then update maxfreq and store that element.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of
// digits, k in the given number n
int countFreq(int N, int K)
{
// Stores count of occurrence
// of digit K in N
int count = 0;
// Iterate over digits of N
while (N > 0) {
// If current digit is k
if (N % 10 == K) {
// Update count
count++;
}
// Update N
N = N / 10;
}
return count;
}
// Utility function to find an array element
// having maximum frequency of digit k
int findElementUtil(int arr[], int N, int K)
{
// Stores frequency of
// digit K in arr[i]
int c = 0;
// Stores maximum frequency of
// digit K in the array
int max = 0;
// Stores an array element having
// maximum frequency of digit k
int ele;
// Initialize max
max = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// Count the frequency of
// digit k in arr[i]
c = countFreq(arr[i], K);
// Update max with maximum
// frequency found so far
if (c > max) {
max = c;
// Update ele
ele = arr[i];
}
}
// If there is no array element
// having digit k in it
if (max == 0)
return -1;
else
return ele;
}
// Function to find an array element
// having maximum frequency of digit k
void findElement(int arr[], int K, int N)
{
// Stores an array element having
// maximum frequency of digit k
int ele = findElementUtil(arr, N, K);
// If there is no element found
// having digit k in it
if (ele == -1)
cout << "-1";
else
// Print the element having max
// frequency of digit k
cout << ele;
}
// Driver Code
int main()
{
// The digit whose max
// occurrence has to be found
int K = 1;
// Given array
int arr[] = { 1, 1111, 111, 11};
// Size of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
findElement(arr, K, N);
return 0;
}
// Java program for the above approach
public class GFG {
// Function to find the count of
// digits, k in the given number n
static int countFreq(int N, int K)
{
// Stores count of occurrence
// of digit K in N
int count = 0;
// Iterate over digits of N
while (N > 0)
{
// If current digit is k
if (N % 10 == K)
{
// Update count
count++;
}
// Update N
N = N / 10;
}
return count;
}
// Utility function to find an array element
// having maximum frequency of digit k
static int findElementUtil(int arr[], int N, int K)
{
// Stores frequency of
// digit K in arr[i]
int c;
// Stores maximum frequency of
// digit K in the array
int max;
// Stores an array element having
// maximum frequency of digit k
int ele = 0;
// Initialize max
max = 0;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Count the frequency of
// digit k in arr[i]
c = countFreq(arr[i], K);
// Update max with maximum
// frequency found so far
if (c > max)
{
max = c;
// Update ele
ele = arr[i];
}
}
// If there is no array element
// having digit k in it
if (max == 0)
return -1;
else
return ele;
}
// Function to find an array element
// having maximum frequency of digit k
static void findElement(int arr[], int N, int K)
{
// Stores an array element having
// maximum frequency of digit k
int ele = findElementUtil(arr, N, K);
// If there is no element found
// having digit k in it
if (ele == -1)
System.out.print("-1");
else
// Print the element having max
// frequency of digit k
System.out.print(ele);
}
// Driver Code
public static void main (String[] args)
{
// The digit whose max
// occurrence has to be found
int K = 1;
// Given array
int arr[] = { 1, 1111, 111, 11};
// Size of array
int N = arr.length;
// Function Call
findElement(arr, N, K);
}
};
// This code is contributed by AnkThon
# Python3 program for the above approach
# Function to find the count of
# digits, k in the given number n
def countFreq(N, K):
# Stores count of occurrence
# of digit K in N
count = 0
# Iterate over digits of N
while (N > 0):
# If current digit is k
if (N % 10 == K):
# Update count
count += 1
# Update N
N = N // 10
return count
# Utility function to find an array element
# having maximum frequency of digit k
def findElementUtil(arr, N, K):
# Stores frequency of
# digit K in arr[i]
c = 0
# Stores maximum frequency of
# digit K in the array
max = 0
# Stores an array element having
# maximum frequency of digit k
ele = 0
# Initialize max
# max = 0
# Traverse the array
for i in range(N):
# Count the frequency of
# digit k in arr[i]
c = countFreq(arr[i], K)
# Update max with maximum
# frequency found so far
if (c > max):
max = c
# Update ele
ele = arr[i]
# If there is no array element
# having digit k in it
if (max == 0):
return -1
else:
return ele
# Function to find an array element
# having maximum frequency of digit k
def findElement(arr, N, K):
# Stores an array element having
# maximum frequency of digit k
ele = findElementUtil(arr, N, K)
# If there is no element found
# having digit k in it
if (ele == -1):
print("-1", end = "")
else:
# Print element having max
# frequency of digit k
print(ele)
# Driver Code
if __name__ == '__main__':
# The digit whose max
# occurrence has to be found
K = 1
# Given array
arr = [1, 1111, 111, 11]
# Size of array
N = len(arr)
# Function Call
findElement(arr, N, K)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
class GFG {
// Function to find the count of
// digits, k in the given number n
static int countFreq(int N, int K)
{
// Stores count of occurrence
// of digit K in N
int count = 0;
// Iterate over digits of N
while (N > 0)
{
// If current digit is k
if (N % 10 == K)
{
// Update count
count++;
}
// Update N
N = N / 10;
}
return count;
}
// Utility function to find an array element
// having maximum frequency of digit k
static int findElementUtil(int []arr, int N, int K)
{
// Stores frequency of
// digit K in arr[i]
int c;
// Stores maximum frequency of
// digit K in the array
int max;
// Stores an array element having
// maximum frequency of digit k
int ele = 0;
// Initialize max
max = 0;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Count the frequency of
// digit k in arr[i]
c = countFreq(arr[i], K);
// Update max with maximum
// frequency found so far
if (c > max)
{
max = c;
// Update ele
ele = arr[i];
}
}
// If there is no array element
// having digit k in it
if (max == 0)
return -1;
else
return ele;
}
// Function to find an array element
// having maximum frequency of digit k
static void findElement(int []arr, int N, int K)
{
// Stores an array element having
// maximum frequency of digit k
int ele = findElementUtil(arr, N, K);
// If there is no element found
// having digit k in it
if (ele == -1)
Console.Write("-1");
else
// Print the element having max
// frequency of digit k
Console.Write(ele);
}
// Driver Code
public static void Main(String[] args)
{
// The digit whose max
// occurrence has to be found
int K = 1;
// Given array
int []arr = { 1, 1111, 111, 11};
// Size of array
int N = arr.Length;
// Function Call
findElement(arr, N, K);
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program to implement
// the above approach
// Function to find the count of
// digits, k in the given number n
function countFreq(N, K)
{
// Stores count of occurrence
// of digit K in N
let count = 0;
// Iterate over digits of N
while (N > 0)
{
// If current digit is k
if (N % 10 == K)
{
// Update count
count++;
}
// Update N
N = Math.floor(N / 10);
}
return count;
}
// Utility function to find an array element
// having maximum frequency of digit k
function findElementUtil(arr, N, K)
{
// Stores frequency of
// digit K in arr[i]
let c;
// Stores maximum frequency of
// digit K in the array
let max;
// Stores an array element having
// maximum frequency of digit k
let ele = 0;
// Initialize max
max = 0;
// Traverse the array
for (let i = 0; i < N; i++)
{
// Count the frequency of
// digit k in arr[i]
c = countFreq(arr[i], K);
// Update max with maximum
// frequency found so far
if (c > max)
{
max = c;
// Update ele
ele = arr[i];
}
}
// If there is no array element
// having digit k in it
if (max == 0)
return -1;
else
return ele;
}
// Function to find an array element
// having maximum frequency of digit k
function findElement(arr, N, K)
{
// Stores an array element having
// maximum frequency of digit k
let ele = findElementUtil(arr, N, K);
// If there is no element found
// having digit k in it
if (ele == -1)
document.write("-1");
else
// Print the element having max
// frequency of digit k
document.write(ele);
}
// Driver Code
// The digit whose max
// occurrence has to be found
let K = 3;
// Given array
let arr = [1, 1111, 111, 11];
// Size of array
let N = arr.length;
// Function Call
findElement(arr, N, K};
// This code is contributed by souravghosh0416.
</script>
Time Complexity: O(N * log10(N))
Auxiliary Space: O(1)
Another Approach:
- Start with an array arr[] of size N and an integer K.
- Initialize max_freq as 0 and result as -1.
- Traverse the array from the start to the end.
- For each element, count the number of occurrences of the digit K by iterating over its digits. You can do this by repeatedly dividing the number by 10 and checking the remainder for equality with K. If it's equal, increment a frequency counter.
- Compare the frequency counter with max_freq. If it's greater than max_freq, update max_freq with the frequency counter and result with the current element.
- After the traversal, if the max_freq is still 0, it means that no element contains the digit K, so return -1. Otherwise, return the result.
- Done.
C++
#include <iostream>
#include <vector>
using namespace std;
int findElement(int arr[], int N, int K) {
int max_freq = 0;
int result = -1;
for (int i = 0; i < N; i++) {
int num = arr[i];
int freq = 0;
while (num > 0) {
if (num % 10 == K) {
freq++;
}
num /= 10;
}
if (freq > max_freq) {
max_freq = freq;
result = arr[i];
}
}
return result;
}
int main() {
int arr[] = { 1, 1111, 111, 11 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 1;
int result = findElement(arr, N, K);
if (result == -1) {
cout << "No element contains the digit " << K << endl;
}
else {
cout<<result<<endl;
}
return 0;
}
import java.util.Scanner;
public class Main {
// Function to find the element with the highest frequency of the digit K
static int findElement(int[] arr, int N, int K) {
int maxFreq = 0;
int result = -1;
// Traverse the array and find the frequency of digit K in each element
for (int i = 0; i < N; i++) {
int num = arr[i];
int freq = 0;
while (num > 0) {
if (num % 10 == K) {
freq++;
}
num /= 10;
}
// Update the result if the current element has a higher frequency of digit K
if (freq > maxFreq) {
maxFreq = freq;
result = arr[i];
}
}
return result;
}
// Driver Code
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int arr[] = {1, 1111, 111, 11};
int N = arr.length;
int K = 1;
int result = findElement(arr, N, K);
System.out.println(result);
}
}
# Define a function to find the element with the highest frequency of a given digit
def findElement(arr, N, K):
max_freq = 0
result = -1
for i in range(N):
num = arr[i]
freq = 0
# Count the frequency of the digit in the current number
while num > 0:
if num % 10 == K:
freq += 1
num //= 10
# Update the result if the current number has a higher frequency of the digit
if freq > max_freq:
max_freq = freq
result = arr[i]
return result
# Test the function with sample inputs
arr = [1, 1111, 111, 11]
N = len(arr)
K = 1
result = findElement(arr, N, K)
if result == -1:
print("No element contains the digit", K)
else:
print(result)
using System;
namespace FindElementWithMaxDigit
{
class Program
{
// Function to find the element with the highest frequency of the digit K
static int FindElement(int[] arr, int N, int K)
{
int maxFreq = 0;
int result = -1;
// Traverse the array and find the frequency of digit K in each element
for (int i = 0; i < N; i++)
{
int num = arr[i];
int freq = 0;
while (num > 0)
{
if (num % 10 == K)
{
freq++;
}
num /= 10;
}
// Update the result if the current
// element has a higher frequency of digit K
if (freq > maxFreq)
{
maxFreq = freq;
result = arr[i];
}
}
return result;
}
static void Main(string[] args)
{
int[] arr = { 1, 1111, 111, 11 };
int N = arr.Length;
int K = 1;
int result = FindElement(arr, N, K);
if (result == -1)
{
Console.WriteLine($"No element contains the digit {K}");
}
else
{
Console.WriteLine(result);
}
}
}
}
// Define a function to find the element with the highest frequency of a given digit
function findElement(arr, N, K) {
let max_freq = 0;
let result = -1;
for (let i = 0; i < N; i++) {
let num = arr[i];
let freq = 0;
// Count the frequency of the digit in the current number
while (num > 0) {
if (num % 10 == K) {
freq += 1;
}
num = Math.floor(num / 10);
}
// Update the result if the current number has a higher frequency of the digit
if (freq > max_freq) {
max_freq = freq;
result = arr[i];
}
}
return result;
}
// Test the function with sample inputs
let arr = [1, 1111, 111, 11];
let N = arr.length;
let K = 1;
let result = findElement(arr, N, K);
if (result == -1) {
console.log("No element contains the digit", K);
} else {
console.log(result);
}
Time Complexity: O(N*M)
Auxiliary Space: O(1)
Similar Reads
Find elements having at least K times the minimum frequency Given an array A, the task is to find elements with the occurrence at least K times the frequency of the least occurring element in the array. Examples: Input: A = {4, 4, 3, 6, 6, 6, 8, 9}, K = 2Output: {4, 6}Explanation: Frequency Table : {4:2, 3:1, 6:3, 8:1, 9:1}, least occurring elements in nums
6 min read
Find maximum element among the elements with minimum frequency in given Array Given an array arr[] consisting of N integers, the task is to find the maximum element with the minimum frequency. Examples: Input: arr[] = {2, 2, 5, 50, 1}Output: 50Explanation:The element with minimum frequency is {1, 5, 50}. The maximum element among these element is 50. Input: arr[] = {3, 2, 5,
13 min read
Maximum element in an array which is equal to its frequency Given an array of integers arr[] of size N, the task is to find the maximum element in the array whose frequency equals to it's value Examples: Input: arr[] = {3, 2, 2, 3, 4, 3} Output: 3 Frequency of element 2 is 2 Frequency of element 3 is 3 Frequency of element 4 is 1 2 and 3 are elements which h
11 min read
Find element with highest frequency in given nested Array Given an array arr[] of N integers. The task is to create a frequency array freq[] of the given array arr[] and find the maximum element of the frequency array. If two elements have the same frequency in the array freq[], then return the element which has a smaller value. Examples: Input: arr[] = {1
8 min read
Maximum frequency of any array element possible by at most K increments Given an array arr[] of size N and an integer K, the task is to find the maximum possible frequency of any array element by at most K increments. Examples: Input: arr[] = {1, 4, 8, 13}, N = 4, K = 5 Output: 2 Explanation: Incrementing arr[0] twice modifies arr[] to {4, 4, 8, 13}. Maximum frequency =
15+ min read