Find the length of the longest valid number chain in an Array

Given an array A[] of N numbers, the task is to find the length of the longest valid number that can be formed by connecting one or more numbers from the array such that, while connecting two numbers the last digit of the previous number is the same as the first digit of the next number.

Examples:

Input: A = [100, 234, 416, 654, 412, 298, 820, 177]
Output: 12
Explanation: The valid numbers are 234416654(connecting a1, a2, a3) and 234416654412 connecting a[1], a[2], a[3], a[4]. The longest valid number is 234416654412, which has a length of 12.

Input: A = [81, 24, 478, 905, 331, 138, 721, 565]
Output: 5
Explanation: The valid numbers are 81138(connecting a0, a4), and 565(a7). Maximum length is 5.

Approach: This can be solved with the following idea:

The approach is to use dynamic programming to find the length of the longest valid number that can be formed by connecting one or more numbers from the given array. The approach uses a 2D array dp to store previously computed results, where dp[i][j] represents the length of the longest valid number that ends with the digit i and starts with the digit j. The algorithm iterates through the array in reverse order and for each number, it computes the length of the longest valid number that can be formed by connecting the number to a previously processed number. The computed results are stored in a temporary array v. Finally, the algorithm updates the dp array with the computed results and finds the maximum valid length. The time complexity of this algorithm is O(n^2), where n is the length of the array.

Below are the steps involved in the implementation of the code:

• Create a 2D array of size 10x10 named 'dp' to store previously computed results.
• Iterate through the array 'arr' in reverse order, starting from the last element and moving toward the first element.
• For each element of 'arr', convert it to a string 'numString', and get its last digit 'lastDigit'.
• Create a temporary array 'v' of size 10 to store the results for each digit 'd'.
• For each digit 'd' from 0 to 9, if there is a previously computed result for the pair of digits 'lastDigit' and 'd' in the 'dp' array, update 'v[d]' to be the sum of the length of 'numString' and the previously computed result.
• Update 'v[lastDigit]' to be the maximum between its current value and the length of 'numString'.
• Get the first digit 'firstDigit' of 'numString'.
• For each digit 'd' from 0 to 9, update the value in 'dp[firstDigit][d]' to be the maximum between its current value and 'v[d]'.
• Update the maximum valid length 'maxValidLength' to be the maximum between its current value and 'dp[firstDigit][firstDigit]'.
• Return 'maxValidLength'.

Below is the implementation of the above approach:

// C++ code for the above approach:
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;

int findLongestValidNumberLength(int arr[], int size)
{
    int maxValidLength = 0;

    // Create a 2D array to store previous results
    int dp[10][10] = { 0 };

    // Iterate through the array in reverse order
    for (int i = size - 1; i >= 0; --i) {

        // Convert the number to a string
        string numString = to_string(arr[i]);

        // Get the last digit of the number
        int lastDigit = numString.back() - '0';

        // Create a temporary array to store the results
        int v[10] = { 0 };
        for (int d = 0; d < 10; ++d) {

            // If there is a previously computed result for
            // the last digit and the current digit, update
            // the temporary array
            if (dp[lastDigit][d] > 0) {
                v[d]
                    = numString.length() + dp[lastDigit][d];
            }
        }

        // Update the temporary array to include the current
        // number
        v[lastDigit]
            = max(v[lastDigit],
                  static_cast<int>(numString.length()));

        // Get the first digit of the number
        int firstDigit = numString.front() - '0';
        for (int d = 0; d < 10; ++d) {

            // Update the 2D array with the computed results
            dp[firstDigit][d]
                = max(dp[firstDigit][d], v[d]);
        }

        // Update the maximum valid length if necessary
        maxValidLength = max(maxValidLength,
                             dp[firstDigit][firstDigit]);
    }

    return maxValidLength;
}

// Driver code
int main()
{
    // Sample inputs
    int arr1[] = { 100, 234, 416, 654, 412, 298, 820, 177 };
    int size1 = sizeof(arr1) / sizeof(arr1[0]);

    int arr2[] = { 81, 24, 478, 905, 331, 138, 721, 565 };
    int size2 = sizeof(arr2) / sizeof(arr2[0]);

    int arr3[] = { 1111, 2222, 3333, 4444, 5555 };
    int size3 = sizeof(arr3) / sizeof(arr3[0]);

    // Function calls
    cout << findLongestValidNumberLength(arr1, size1)
         << endl;
    cout << findLongestValidNumberLength(arr2, size2)
         << endl;
    cout << findLongestValidNumberLength(arr3, size3)
         << endl;

    return 0;
}

// This code is contributed by akshitaguprzj3

// Java code for the above approach:
class GFG {
    public static int
    findLongestValidNumberLength(int[] arr)
    {
        int maxValidLength = 0;

        // Create a 2D array to store
        // previous results
        int[][] dp = new int[10][10];

        // Iterate through the array in
        // reverse order
        for (int i = arr.length - 1; i >= 0; --i) {

            // Convert the number
            // to a string
            String numString = String.valueOf(arr[i]);

            // Get the last digit
            // of the number
            int lastDigit
                = numString.charAt(numString.length() - 1)
                  - '0';

            // Create a temporary array
            // to store the results
            int[] v = new int[10];
            for (int d = 0; d < 10; ++d) {

                // If there is a previously
                // computed result for the
                // last digit and the current
                // digit, update the
                // temporary array
                if (dp[lastDigit][d] > 0) {
                    v[d] = numString.length()
                           + dp[lastDigit][d];
                }
            }
            // Update the temporary array
            // to include the current
            // number
            v[lastDigit] = Math.max(v[lastDigit],
                                    numString.length());

            // Get the first digit
            // of the number
            int firstDigit = numString.charAt(0) - '0';
            for (int d = 0; d < 10; ++d) {

                // Update the 2D array with
                // the computed results
                dp[firstDigit][d]
                    = Math.max(dp[firstDigit][d], v[d]);
            }

            // Update the maximum valid
            // length if necessary
            maxValidLength = Math.max(
                maxValidLength, dp[firstDigit][firstDigit]);
        }
        return maxValidLength;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Sample inputs
        int[] arr1
            = { 100, 234, 416, 654, 412, 298, 820, 177 };
        int[] arr2
            = { 81, 24, 478, 905, 331, 138, 721, 565 };
        int[] arr3 = { 1111, 2222, 3333, 4444, 5555 };

        // Function call
        System.out.println(
            findLongestValidNumberLength(arr1));
        System.out.println(
            findLongestValidNumberLength(arr2));
        System.out.println(
            findLongestValidNumberLength(arr3));
    }
}

# Python3 code for the above approach:
def findLongestValidNumberLength(arr):
    size = len(arr)
    maxValidLength = 0

    # Create a 2D list to store previous results
    dp = [[0] * 10 for _ in range(10)]

    # Iterate through the list in reverse order
    for i in range(size - 1, -1, -1):

        # Convert the number to a string
        numString = str(arr[i])

        # Get the last digit of the number
        lastDigit = int(numString[-1])

        # Create a temporary list to store the results
        v = [0] * 10
        for d in range(10):

            # If there is a previously computed result for
            # the last digit and the current digit, update
            # the temporary list
            if dp[lastDigit][d] > 0:
                v[d] = len(numString) + dp[lastDigit][d]

        # Update the temporary list to include the current
        # number
        v[lastDigit] = max(v[lastDigit], len(numString))

        # Get the first digit of the number
        firstDigit = int(numString[0])
        for d in range(10):

            # Update the 2D list with the computed results
            dp[firstDigit][d] = max(dp[firstDigit][d], v[d])

        # Update the maximum valid length if necessary
        maxValidLength = max(maxValidLength, dp[firstDigit][firstDigit])

    return maxValidLength


# Driver code
arr1 = [100, 234, 416, 654, 412, 298, 820, 177]
arr2 = [81, 24, 478, 905, 331, 138, 721, 565]
arr3 = [1111, 2222, 3333, 4444, 5555]

# Function calls
print(findLongestValidNumberLength(arr1))
print(findLongestValidNumberLength(arr2))
print(findLongestValidNumberLength(arr3))


# This code is contributed by rambabuguphka

using System;

class Program {
    static void Main(string[] args)
    {
        int[] arr1
            = { 100, 234, 416, 654, 412, 298, 820, 177 };
        int[] arr2
            = { 81, 24, 478, 905, 331, 138, 721, 565 };
        int[] arr3 = { 1111, 2222, 3333, 4444, 5555 };

        Console.WriteLine(
            findLongestValidNumberLength(arr1));
        Console.WriteLine(
            findLongestValidNumberLength(arr2));
        Console.WriteLine(
            findLongestValidNumberLength(arr3));
        Console.ReadKey();
    }

    public static int
    findLongestValidNumberLength(int[] arr)
    {
        int maxValidLength = 0;

        // Create a 2D array to store
        // previous results
        int[, ] dp = new int[10, 10];

        // Iterate through the array in
        // reverse order
        for (int i = arr.Length - 1; i >= 0; --i) {

            // Convert the number
            // to a string
            string numString = arr[i].ToString();

            // Get the last digit
            // of the number
            int lastDigit
                = numString[numString.Length - 1] - '0';

            // Create a temporary array
            // to store the results
            int[] v = new int[10];
            for (int d = 0; d < 10; ++d) {

                // If there is a previously
                // computed result for the
                // last digit and the current
                // digit, update the
                // temporary array
                if (dp[lastDigit, d] > 0) {
                    v[d] = numString.Length
                           + dp[lastDigit, d];
                }
            }
            // Update the temporary array
            // to include the current
            // number
            v[lastDigit]
                = Math.Max(v[lastDigit], numString.Length);

            // Get the first digit
            // of the number
            int firstDigit = numString[0] - '0';
            for (int d = 0; d < 10; ++d) {

                // Update the 2D array with
                // the computed results
                dp[firstDigit, d]
                    = Math.Max(dp[firstDigit, d], v[d]);
            }

            // Update the maximum valid
            // length if necessary
            maxValidLength = Math.Max(
                maxValidLength, dp[firstDigit, firstDigit]);
        }
        return maxValidLength;
    }
}

// This code is contributed by Tapesh(tapeshdua420)

// JavaScript code for the above approach:
function findLongestValidNumberLength(arr) {
  let size = arr.length;
  let maxValidLength = 0;

  // Create a 2D array to store previous results
  let dp = new Array(10);
  for (let i = 0; i < 10; i++) {
    dp[i] = new Array(10).fill(0);
  }

  // Iterate through the array in reverse order
  for (let i = size - 1; i >= 0; i--) {
    // Convert the number to a string
    let numString = String(arr[i]);

    // Get the last digit of the number
    let lastDigit = parseInt(numString[numString.length - 1]);

    // Create a temporary array to store the results
    let v = new Array(10).fill(0);
    for (let d = 0; d < 10; d++) {
      // If there is a previously computed result for
      // the last digit and the current digit, update
      // the temporary array
      if (dp[lastDigit][d] > 0) {
        v[d] = numString.length + dp[lastDigit][d];
      }
    }

    // Update the temporary array to include the current
    // number
    v[lastDigit] = Math.max(v[lastDigit], numString.length);

    // Get the first digit of the number
    let firstDigit = parseInt(numString[0]);
    for (let d = 0; d < 10; d++) {
      // Update the 2D array with the computed results
      dp[firstDigit][d] = Math.max(dp[firstDigit][d], v[d]);
    }

    // Update the maximum valid length if necessary
    maxValidLength = Math.max(maxValidLength, dp[firstDigit][firstDigit]);
  }

  return maxValidLength;
}

// Driver code
let arr1 = [100, 234, 416, 654, 412, 298, 820, 177];
let arr2 = [81, 24, 478, 905, 331, 138, 721, 565];
let arr3 = [1111, 2222, 3333, 4444, 5555];

// Function calls
console.log(findLongestValidNumberLength(arr1));
console.log(findLongestValidNumberLength(arr2));
console.log(findLongestValidNumberLength(arr3));

// This code is contributed by Tapesh(tapeshdua420)

12
5
4

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