Skip to content
Related Articles

Related Articles

Length of largest subsequence consisting of a pair of alternating digits
  • Difficulty Level : Medium
  • Last Updated : 19 Aug, 2020

Given a numeric string s consisting of digits 0 to 9, the task is to find the length of the largest subsequence consisting of a pair of alternating digits.

An alternating digits subsequence consisting of two different digits a and b can be represented as “abababababababababab….”.

Examples:

Input: s = “1542745249842”
Output: 6
Explanation: 
The largest substring of alternating digits in the given string is 424242.

Input:  s = “1212312323232”
Output: 9
Explanation:
The largest substring of alternating digits in the given string is 232323232.



Approach: The string consists of only decimal digits i.e., 0-9, thus the sequence can be checked for the presence of all possible subsequences consisting of two alternating digits. For this purpose follow the below approach:

  • Use nested loops from 0 to 9 each, for selecting the ordered pair of digits. When the digits are the same then the string is not traversed. When the digits are different, then the string is traversed and the length of the subsequence consisting of the digits of the ordered pair occurring alternate digits is found.
  • If the maximum length is 1, it implies the second digit in any of the ordered pairs never occurred in the given sequence thus making it a mono-digit sequence. The required kind of subsequence in such a sequence wouldn’t exist thus output 0.
  • If maximum length found is greater than 1, this means at least 2 different digits are present in the given sequence thus output this found length.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to find the length of the
// largest subsequence consisting of
// a pair of alternating digits
void largestSubsequence(string s)
{
    // Variable initialization
    int maxi = 0;
    char prev1;
 
    // Nested loops for iteration
    for (int i = 0; i < 10; i++) {
        for (int j = 0; j < 10; j++) {
 
            // Check if i is not eqaul to j
            if (i != j) {
 
                // Initialize length as 0
                int len = 0;
                prev1 = j + '0';
 
                // Iterate from 0 till the
                // size of the string
                for (int k = 0; k < s.size(); k++) {
 
                    if (s[k] == i + '0'
                        && prev1 == j + '0') {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                    else if (s[k] == j + '0'
                             && prev1 == i + '0') {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                }
 
                // Update maxi
                maxi = max(len, maxi);
            }
        }
    }
 
    // Check if maxi is not equal to
    // 1 the print it otherwise print 0
    if (maxi != 1)
        cout << maxi << endl;
    else
        cout << 0 << endl;
}
 
// Driver Code
int main()
{
    // Given string
    string s = "1542745249842";
 
    // Function call
    largestSubsequence(s);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the length of the
// largest subsequence consisting of
// a pair of alternating digits
static void largestSubsequence(char []s)
{
    // Variable initialization
    int maxi = 0;
    char prev1;
 
    // Nested loops for iteration
    for (int i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
 
            // Check if i is not eqaul to j
            if (i != j)
            {
 
                // Initialize length as 0
                int len = 0;
                prev1 = (char) (j + '0');
 
                // Iterate from 0 till the
                // size of the String
                for (int k = 0; k < s.length; k++)
                {
                    if (s[k] == i + '0' &&
                        prev1 == j + '0')
                    {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                    else if (s[k] == j + '0' &&
                             prev1 == i + '0')
                    {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                }
 
                // Update maxi
                maxi = Math.max(len, maxi);
            }
        }
    }
 
    // Check if maxi is not equal to
    // 1 the print it otherwise print 0
    if (maxi != 1)
        System.out.print(maxi + "\n");
    else
        System.out.print(0 + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
    // Given String
    String s = "1542745249842";
 
    // Function call
    largestSubsequence(s.toCharArray());
}
}
 
// This code is contributed by Rohit_ranjan

Python3




# Python3 program for the above approach
 
# Function to find the length of the
# largest subsequence consisting of
# a pair of alternating digits
def largestSubsequence(s):
     
    # Variable initialization
    maxi = 0
 
    # Nested loops for iteration
    for i in range(10):
        for j in range(10):
 
            # Check if i is not eqaul to j
            if (i != j):
 
                # Initialize length as 0
                lenn = 0
                prev1 = chr(j + ord('0'))
 
                # Iterate from 0 till the
                # size of the string
                for k in range(len(s)):
                    if (s[k] == chr(i + ord('0')) and
                       prev1 == chr(j + ord('0'))):
                        prev1 = s[k]
 
                        # Increment length
                        lenn += 1
                         
                    elif (s[k] == chr(j + ord('0')) and
                         prev1 == chr(i + ord('0'))):
                        prev1 = s[k]
 
                        # Increment lenngth
                        lenn += 1
 
                # Update maxi
                maxi = max(lenn, maxi)
 
    # Check if maxi is not equal to
    # 1 the prit otherwise pr0
    if (maxi != 1):
        print(maxi)
    else:
        print(0)
 
# Driver Code
if __name__ == '__main__':
     
    # Given string
    s = "1542745249842"
 
    # Function call
    largestSubsequence(s)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
class GFG{
 
// Function to find the length of the
// largest subsequence consisting of
// a pair of alternating digits
static void largestSubsequence(char []s)
{
    // Variable initialization
    int maxi = 0;
    char prev1;
 
    // Nested loops for iteration
    for (int i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
 
            // Check if i is not eqaul to j
            if (i != j)
            {
 
                // Initialize length as 0
                int len = 0;
                prev1 = (char) (j + '0');
 
                // Iterate from 0 till the
                // size of the String
                for (int k = 0; k < s.Length; k++)
                {
                    if (s[k] == i + '0' &&
                        prev1 == j + '0')
                    {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                    else if (s[k] == j + '0' &&
                             prev1 == i + '0')
                    {
                        prev1 = s[k];
 
                        // Increment length
                        len++;
                    }
                }
 
                // Update maxi
                maxi = Math.Max(len, maxi);
            }
        }
    }
 
    // Check if maxi is not equal to
    // 1 the print it otherwise print 0
    if (maxi != 1)
        Console.Write(maxi + "\n");
    else
        Console.Write(0 + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given String
    String s = "1542745249842";
 
    // Function call
    largestSubsequence(s.ToCharArray());
}
}
 
// This code is contributed by Rohit_ranjan
Output: 
6





 

Time Complexity: O(10*10*N) 
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :