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Length of longest subsequence in an Array having all elements as Nude Numbers
  • Last Updated : 24 Mar, 2021

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)

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