Find the array element having maximum frequency of the digit K
Given an array arr[] of size N and an integer K, the task is to find an array element which contains the digit K 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, the 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 nint 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 kint 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; // 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 kvoid 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) cout << "-1"; else // Print the element having max // frequency of digit k cout << ele;}// Driver Codeint main(){ // The digit whose max // occurrence has to be found int K = 3; // Given array int arr[] = { 3, 77, 343, 456 }; // Size of array int N = sizeof(arr) / sizeof(arr[0]); // Function Call findElement(arr, K, N); return 0;} |
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Java
// Java program for the above approachclass 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 = 3; // Given array int arr[] = { 3, 77, 343, 456 }; // Size of array int N = arr.length; // Function Call findElement(arr, K, N); }}// This code is contributed by AnkThon |
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Python3
# Python3 program for the above approach# Function to find the count of# digits, k in the given number ndef 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 kdef 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 kdef 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: # Prthe element having max # frequency of digit k print(ele)# Driver Codeif __name__ == '__main__': # The digit whose max # occurrence has to be found K = 3 # Given array arr = [3, 77, 343, 456] # Size of array N = len(arr) # Function Call findElement(arr, K, N)# This code is contributed by mohit kumar 29 |
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C#
// C# program for the above approachusing 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 = 3; // Given array int []arr = { 3, 77, 343, 456 }; // Size of array int N = arr.Length; // Function Call findElement(arr, K, N); }}// This code is contributed by 29AjayKumar |
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Output:
343
Time Complexity: O(N * log10(N))
Auxiliary Space: O(1)
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