Given an array arr[] of N positive integers, the task is to print the length of the longest subsequence of the array such that all of its elements are Nude Numbers.
Examples:
Input: arr[] = {34, 34, 2, 2, 3, 333, 221, 32 }
Output: 4
Explanation:
Longest Nude number subsequence is {2, 2, 3, 333} and hence the answer is 4.
Input: arr[] = {456, 44, 104, 133, 39, 325 }
Output: 1
Explanation:
Longest Nude number subsequence is {44} and hence the answer is 1.
Approach: To solve the problem follow the steps given below:
- Traverse the given array and for each element in the array and check if it is a Nude number or not.
- If the element is a Nude Number, it will be included in the resultant longest subsequence. Hence increment the count of elements in the subsequence by 1.
- Print the value of count after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the number
// is a Nude number
bool isNudeNum(int n)
{
// Variable initialization
int copy, length, flag = 0;
copy = n;
string temp;
// Integer 'copy' is converted
// to a string
temp = to_string(copy);
// Total digits in the number
length = temp.length();
// Loop through all digits and check
// if every digit divides n or not
for (int i = 0; i < length; i++) {
int num = temp[i] - '0';
if (num == 0 or n % num != 0) {
// flag is used to keep check
flag = 1;
}
}
// Return true or false as per
// the condition
if (flag == 1)
return false;
else
return true;
}
// Function to find the longest subsequence
// which contain all Nude numbers
int longestNudeSubseq(int arr[], int n)
{
int answer = 0;
// Find the length of longest
// Nude number subsequence
for (int i = 0; i < n; i++) {
if (isNudeNum(arr[i]))
answer++;
}
return answer;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 34, 34, 2, 2, 3,
333, 221, 32 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << longestNudeSubseq(arr, n)
<< endl;
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to check if the number
// is a Nude number
static boolean isNudeNum(int n)
{
// Variable initialization
int copy, length, flag = 0;
copy = n;
String temp;
// Integer 'copy' is converted
// to a String
temp = String.valueOf(copy);
// Total digits in the number
length = temp.length();
// Loop through all digits and check
// if every digit divides n or not
for(int i = 0; i < length; i++)
{
int num = temp.charAt(i) - '0';
if (num == 0 || n % num != 0)
{
// flag is used to keep check
flag = 1;
}
}
// Return true or false as per
// the condition
if (flag == 1)
return false;
else
return true;
}
// Function to find the longest subsequence
// which contain all Nude numbers
static int longestNudeSubseq(int arr[], int n)
{
int answer = 0;
// Find the length of longest
// Nude number subsequence
for(int i = 0; i < n; i++)
{
if (isNudeNum(arr[i]))
answer++;
}
return answer;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 34, 34, 2, 2, 3,
333, 221, 32 };
int n = arr.length;
// Function call
System.out.print(longestNudeSubseq(arr, n) + "\n");
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program for the above approach
# Function to check if the number
# is a Nude number
def isNudeNum(n):
# Variable initialization
flag = 0
copy = n
# Integer 'copy' is converted
# to a string
temp = str(copy)
# Total digits in the number
length = len(temp)
# Loop through all digits and check
# if every digit divides n or not
for i in range(length):
num = ord(temp[i]) - ord('0')
if ((num == 0) or (n % num != 0)):
# flag is used to keep check
flag = 1
# Return true or false as per
# the condition
if (flag == 1):
return False
else:
return True
# Function to find the longest subsequence
# which contain all Nude numbers
def longestNudeSubseq(arr, n):
answer = 0
# Find the length of longest
# Nude number subsequence
for i in range(n):
if (isNudeNum(arr[i])):
answer += 1
return answer
# Driver Code
# Given array arr[]
arr = [ 34, 34, 2, 2, 3,
333, 221, 32 ]
n = len(arr)
# Function call
print(longestNudeSubseq(arr, n))
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if the number
// is a Nude number
static bool isNudeNum(int n)
{
// Variable initialization
int copy, length, flag = 0;
copy = n;
String temp;
// int 'copy' is converted
// to a String
temp = String.Join("", copy);
// Total digits in the number
length = temp.Length;
// Loop through all digits and check
// if every digit divides n or not
for(int i = 0; i < length; i++)
{
int num = temp[i] - '0';
if (num == 0 || n % num != 0)
{
// flag is used to keep check
flag = 1;
}
}
// Return true or false as per
// the condition
if (flag == 1)
return false;
else
return true;
}
// Function to find the longest subsequence
// which contain all Nude numbers
static int longestNudeSubseq(int []arr, int n)
{
int answer = 0;
// Find the length of longest
// Nude number subsequence
for(int i = 0; i < n; i++)
{
if (isNudeNum(arr[i]))
answer++;
}
return answer;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 34, 34, 2, 2, 3,
333, 221, 32 };
int n = arr.Length;
// Function call
Console.Write(longestNudeSubseq(arr, n) + "\n");
}
}
// This code is contributed by amal kumar choubey
Javascript
<script>
// Javascript program for the above approach
// Function to check if the number
// is a Nude number
function isNudeNum(n)
{
// Variable initialization
let copy, length, flag = 0;
copy = n;
let temp;
// Integer 'copy' is converted
// to a string
temp = copy.toString();
// Total digits in the number
length = temp.length;
// Loop through all digits and check
// if every digit divides n or not
for (let i = 0; i < length; i++) {
let num = temp[i].charCodeAt() - '0'.charCodeAt();
if (num == 0 || n % num != 0) {
// flag is used to keep check
flag = 1;
}
}
// Return true or false as per
// the condition
if (flag == 1)
return false;
else
return true;
}
// Function to find the longest subsequence
// which contain all Nude numbers
function longestNudeSubseq(arr, n)
{
let answer = 0;
// Find the length of longest
// Nude number subsequence
for (let i = 0; i < n; i++) {
if (isNudeNum(arr[i]))
answer++;
}
return answer;
}
// Given array arr[]
let arr = [ 34, 34, 2, 2, 3,
333, 221, 32 ];
let n = arr.length;
// Function Call
document.write(longestNudeSubseq(arr, n));
// This code is contributed by divyesh072019.
</script>
Output
4
Time Complexity: O(N*log10N)
Auxiliary Space: O(1)
Approach : Using Set or Hashing
To solve the problem follow the steps given below:
- Firstly, create a set or hash table to store all Nude Numbers
- Iterate over the given array and for each element check if it is a Nude Number
- If it is then increment the count of Nude Numbers encountered.
- After iterating through the array
- The count of Nude Numbers encountered will give the length of the longest Nude Numbers subsequence.
Below is the implementation of the above approach:
Python3
def is_nude_number(num):
# Define a set of Nude Numbers
nude_numbers = {2, 3, 5, 7}
# Check if the number is a Nude Number
while num > 0:
digit = num % 10
if digit not in nude_numbers:
return False
num //= 10
return True
def longest_nude_numbers_subsequence(arr):
max_length = 0
current_length = 0
for num in arr:
if is_nude_number(num):
current_length += 1
max_length = max(max_length, current_length)
else:
current_length = 0
return max_length
# Example usage:
arr1 = [34, 34, 2, 2, 3, 333, 221, 32]
print(longest_nude_numbers_subsequence(arr1))
Output
4
Time Complexity: O(N * log(num)), where N is the number of elements in the array.
Auxiliary Space: O(1)