Longest subsequence such that adjacent elements have at least one common digit

Given an array arr[] of N integers, the task is to find the length of the longest sub-sequence such that adjacent elements of the sub-sequence have at least one digit in common.

Examples:

Input: arr[] = {1, 12, 44, 29, 33, 96, 89}
Output: 5
The longest sub-sequence is {1 12 29 96 89}

Input: arr[] = {12, 23, 45, 43, 36, 97}
Output: 4
The longest sub-sequence is {12 23 43 36}

Approach: The idea is to store the length of longest sub-sequence for each digit present in an element of the array.

  • dp[i][d] represents the length of the longest sub-sequence up to ith element if digit d is the common digit.
  • Declare a cnt array and set cnt[d] = 1 for each digit present in current element.
  • Consider each digit d as common digit and find maximum length sub-sequence ending at arr[i] as dp[i][d] = max(dp[j][d]+1) (0<=j<i).
  • Also find a local maximum max(dp[i][d]) for current element.
  • After finding local maximum update dp[i][d] for all digits present in the current element to a local maximum.
  • This is required as local maximum represents maximum length sub-sequence for an element having digit d.
  • E.g. Consider arr[] = {24, 49, 96}.
    The local maximum for 49 is 2 obtain from digit 4.
    This local maximum will be used in finding the local maximum for 96 with common digit 9.
    For that it is required for all digits in 49, dp[i][d] should be set to local maximum.

    Below is the implementation of the above approach:

    C++

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    // C++ program to find maximum length subsequence
    // such that adjacent elements have at least
    // one common digit
    #include <bits/stdc++.h>
    using namespace std;
      
    // Returns length of maximum length subsequence
    int findSubsequence(int arr[], int n)
    {
      
        // To store the length of the
        // maximum length subsequence
        int len = 1;
      
        // To store current element arr[i]
        int tmp;
      
        int i, j, d;
      
        // To store the length of the sub-sequence
        // ending at index i and having common digit d
        int dp[n][10];
      
        memset(dp, 0, sizeof(dp));
      
        // To store digits present in current element
        int cnt[10];
      
        // To store length of maximum length subsequence
        // ending at index i
        int locMax;
      
        // For first element maximum length is 1 for
        // each digit
        tmp = arr[0];
        while (tmp > 0) {
            dp[0][tmp % 10] = 1;
            tmp /= 10;
        }
      
        // Find digits of each element, then find length
        // of subsequence for each digit and then find
        // local maximum
        for (i = 1; i < n; i++) {
            tmp = arr[i];
            locMax = 1;
            memset(cnt, 0, sizeof(cnt));
      
            // Find digits in current element
            while (tmp > 0) {
                cnt[tmp % 10] = 1;
                tmp /= 10;
            }
      
            // For each digit present find length of
            // subsequence and find local maximum
            for (d = 0; d <= 9; d++) {
                if (cnt[d]) {
                    dp[i][d] = 1;
                    for (j = 0; j < i; j++) {
                        dp[i][d] = max(dp[i][d], dp[j][d] + 1);
                        locMax = max(dp[i][d], locMax);
                    }
                }
            }
      
            // Update value of dp[i][d] for each digit
            // present in current element to local maximum
            // found.
            for (d = 0; d <= 9; d++) {
                if (cnt[d]) {
                    dp[i][d] = locMax;
                }
            }
      
            // Update maximum length with local maximum
            len = max(len, locMax);
        }
      
        return len;
    }
      
    // Driver code
    int main()
    {
        int arr[] = { 1, 12, 44, 29, 33, 96, 89 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        cout << findSubsequence(arr, n);
      
        return 0;
    }

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    Java

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    // Java program to find maximum length subsequence
    // such that adjacent elements have at least
    // one common digit
      
    class GFG
    {
      
    // Returns length of maximum length subsequence
    static int findSubsequence(int arr[], int n)
    {
      
        // To store the length of the
        // maximum length subsequence
        int len = 1;
      
        // To store current element arr[i]
        int tmp;
      
        int i, j, d;
      
        // To store the length of the sub-sequence
        // ending at index i and having common digit d
        int[][] dp = new int[n][10];
      
      
        // To store digits present in current element
        int[] cnt = new int[10];
      
        // To store length of maximum length subsequence
        // ending at index i
        int locMax;
      
        // For first element maximum length is 1 for
        // each digit
        tmp = arr[0];
        while (tmp > 0
        {
            dp[0][tmp % 10] = 1;
            tmp /= 10;
        }
      
        // Find digits of each element, then find length
        // of subsequence for each digit and then find
        // local maximum
        for (i = 1; i < n; i++) 
        {
            tmp = arr[i];
            locMax = 1;
            for (int x = 0; x < 10; x++)
            cnt[x]=0;
      
            // Find digits in current element
            while (tmp > 0
            {
                cnt[tmp % 10] = 1;
                tmp /= 10;
            }
      
            // For each digit present find length of
            // subsequence and find local maximum
            for (d = 0; d <= 9; d++)
            {
                if (cnt[d] > 0)
                {
                    dp[i][d] = 1;
                    for (j = 0; j < i; j++) 
                    {
                        dp[i][d] = Math.max(dp[i][d], dp[j][d] + 1);
                        locMax = Math.max(dp[i][d], locMax);
                    }
                }
            }
      
            // Update value of dp[i][d] for each digit
            // present in current element to local maximum
            // found.
            for (d = 0; d <= 9; d++)
            {
                if (cnt[d] > 0)
                {
                    dp[i][d] = locMax;
                }
            }
      
            // Update maximum length with local maximum
            len = Math.max(len, locMax);
        }
      
        return len;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int arr[] = { 1, 12, 44, 29, 33, 96, 89 };
        int n = arr.length;
      
        System.out.println(findSubsequence(arr, n));
    }
    }
      
    // This code is contributed by mits

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    Python3

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    # Python3 program to find maximum 
    # Length subsequence such that 
    # adjacent elements have at least
    # one common digit
      
    # Returns Length of maximum 
    # Length subsequence
    def findSubsequence(arr, n):
      
        # To store the Length of the
        # maximum Length subsequence
        Len = 1
      
        # To store current element arr[i]
        tmp = 0
      
        i, j, d = 0, 0, 0
      
        # To store the Length of the sub-sequence
        # ending at index i and having common digit d
        dp = [[0 for i in range(10)] 
                 for i in range(n)]
      
        # To store digits present in current element
        cnt = [0 for i in range(10)]
      
        # To store Length of maximum 
        # Length subsequence ending at index i
        locMax = 0
      
        # For first element maximum 
        # Length is 1 for each digit
        tmp = arr[0]
        while (tmp > 0):
            dp[0][tmp % 10] = 1
            tmp //= 10
      
        # Find digits of each element, 
        # then find Length of subsequence 
        # for each digit and then find
        # local maximum
        for i in range(1, n):
            tmp = arr[i]
            locMax = 1
            cnt = [0 for i in range(10)]
      
            # Find digits in current element
            while (tmp > 0):
                cnt[tmp % 10] = 1
                tmp //= 10
      
            # For each digit present find Length of
            # subsequence and find local maximum
            for d in range(10):
                if (cnt[d]):
                    dp[i][d] = 1
                    for j in range(i):
                        dp[i][d] = max(dp[i][d], 
                                       dp[j][d] + 1)
                        locMax = max(dp[i][d], locMax)
      
            # Update value of dp[i][d] for each digit
            # present in current element to local 
            # maximum found.
            for d in range(10):
                if (cnt[d]):
                    dp[i][d] = locMax
      
            # Update maximum Length 
            # with local maximum
            Len = max(Len, locMax)
        return Len
      
    # Driver code
    arr = [1, 12, 44, 29, 33, 96, 89]
    n = len(arr)
      
    print(findSubsequence(arr, n))
      
    # This code is contributed 
    # by mohit kumar

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    C#

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    // C# program to find maximum length subsequence
    // such that adjacent elements have at least
    // one common digit
    using System;
      
    class GFG
    {
      
    // Returns length of maximum length subsequence
    static int findSubsequence(int []arr, int n)
    {
      
        // To store the length of the
        // maximum length subsequence
        int len = 1;
      
        // To store current element arr[i]
        int tmp;
      
        int i, j, d;
      
        // To store the length of the sub-sequence
        // ending at index i and having common digit d
        int[,] dp = new int[n, 10];
      
      
        // To store digits present in current element
        int[] cnt = new int[10];
      
        // To store length of maximum length subsequence
        // ending at index i
        int locMax;
      
        // For first element maximum length is 1 for
        // each digit
        tmp = arr[0];
        while (tmp > 0) 
        {
            dp[0, tmp % 10] = 1;
            tmp /= 10;
        }
      
        // Find digits of each element, then find length
        // of subsequence for each digit and then find
        // local maximum
        for (i = 1; i < n; i++) 
        {
            tmp = arr[i];
            locMax = 1;
            for (int x = 0; x < 10; x++)
                cnt[x] = 0;
      
            // Find digits in current element
            while (tmp > 0) 
            {
                cnt[tmp % 10] = 1;
                tmp /= 10;
            }
      
            // For each digit present find length of
            // subsequence and find local maximum
            for (d = 0; d <= 9; d++)
            {
                if (cnt[d] > 0)
                {
                    dp[i, d] = 1;
                    for (j = 0; j < i; j++) 
                    {
                        dp[i, d] = Math.Max(dp[i, d], dp[j, d] + 1);
                        locMax = Math.Max(dp[i, d], locMax);
                    }
                }
            }
      
            // Update value of dp[i,d] for each digit
            // present in current element to local maximum
            // found.
            for (d = 0; d <= 9; d++)
            {
                if (cnt[d] > 0)
                {
                    dp[i, d] = locMax;
                }
            }
      
            // Update maximum length with local maximum
            len = Math.Max(len, locMax);
        }
      
        return len;
    }
      
    // Driver code
    public static void Main() 
    {
        int []arr = { 1, 12, 44, 29, 33, 96, 89 };
        int n = arr.Length;
      
        Console.WriteLine(findSubsequence(arr, n));
    }
    }
      
    // This code is contributed by mits

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    Output:

    5
    

    Time Complexity: O(n2)
    Auxiliary Space: O(n)

    The auxiliary space required for above solution can be further reduced. Observe that for each digit d present in arr[i], it is required to find maximum length sub-sequence upto that digit irrespective of the fact that at which element the sub-sequence end. This reduces auxiliary space required to O(1). For each arr[i], find local maximum and update dig[d] for each digit d in arr[i] to local maximum.

    Below is the implementation of the above approach:

    C++

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    // C++ program to find maximum length subsequence
    // such that adjacent elements have at least
    // one common digit
    #include <bits/stdc++.h>
    using namespace std;
      
    // Returns length of maximum length subsequence
    int findSubsequence(int arr[], int n)
    {
      
        // To store length of maximum length subsequence
        int len = 1;
      
        // To store current element arr[i]
        int tmp;
      
        int i, j, d;
      
        // To store length of subsequence
        // having common digit d
        int dp[10];
      
        memset(dp, 0, sizeof(dp));
      
        // To store digits present in current element
        int cnt[10];
      
        // To store local maximum for current element
        int locMax;
      
        // For first element maximum length is 1 for
        // each digit
        tmp = arr[0];
        while (tmp > 0) {
            dp[tmp % 10] = 1;
            tmp /= 10;
        }
      
        // Find digits of each element, then find length
        // of subsequence for each digit and then find
        // local maximum
        for (i = 1; i < n; i++) {
            tmp = arr[i];
            locMax = 1;
            memset(cnt, 0, sizeof(cnt));
      
            // Find digits in current element
            while (tmp > 0) {
                cnt[tmp % 10] = 1;
                tmp /= 10;
            }
      
            // For each digit present find length of
            // subsequence and find local maximum
            for (d = 0; d <= 9; d++) {
                if (cnt[d]) {
                    dp[d]++;
                    locMax = max(locMax, dp[d]);
                }
            }
      
            // Update value of dp[d] for each digit
            // present in current element to local maximum
            // found
            for (d = 0; d <= 9; d++) {
                if (cnt[d]) {
                    dp[d] = locMax;
                }
            }
      
            // Update maximum length with local maximum
            len = max(len, locMax);
        }
      
        return len;
    }
      
    // Driver code
    int main()
    {
        int arr[] = { 1, 12, 44, 29, 33, 96, 89 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        cout << findSubsequence(arr, n);
      
        return 0;
    }

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    Java

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    // Java program to find maximum length subsequence 
    // such that adjacent elements have at least 
    // one common digit 
    import java.util.*;
      
    class GFG 
    {
      
    // Returns length of maximum length subsequence 
    static int findSubsequence(int arr[], int n) 
      
        // To store length of maximum length subsequence 
        int len = 1
      
        // To store current element arr[i] 
        int tmp; 
      
        int i, j, d; 
      
        // To store length of subsequence 
        // having common digit d 
        int dp[] = new int[10]; 
      
      
        // To store digits present in current element 
        int cnt[] = new int[10]; 
      
        // To store local maximum for current element 
        int locMax; 
      
        // For first element maximum length is 1 for 
        // each digit 
        tmp = arr[0]; 
        while (tmp > 0
        
            dp[tmp % 10] = 1
            tmp /= 10
        
      
        // Find digits of each element, then find length 
        // of subsequence for each digit and then find 
        // local maximum 
        for (i = 1; i < n; i++) 
        
            tmp = arr[i]; 
            locMax = 1
                    Arrays.fill(cnt, 0);
      
            // Find digits in current element 
            while (tmp > 0
            
                cnt[tmp % 10] = 1
                tmp /= 10
            
      
            // For each digit present find length of 
            // subsequence and find local maximum 
            for (d = 0; d <= 9; d++) 
            
                if (cnt[d] == 1
                
                    dp[d]++; 
                    locMax = Math.max(locMax, dp[d]); 
                
            
      
            // Update value of dp[d] for each digit 
            // present in current element to local maximum 
            // found 
            for (d = 0; d <= 9; d++) 
            
                if (cnt[d] == 1
                
                    dp[d] = locMax; 
                
            
      
            // Update maximum length with local maximum 
            len = Math.max(len, locMax); 
        
      
        return len; 
      
    // Driver code 
    public static void main(String[] args) 
    {
        int arr[] = { 1, 12, 44, 29, 33, 96, 89 }; 
        int n = arr.length; 
            System.out.print(findSubsequence(arr, n)); 
    }
    }
      
    /* This code contributed by PrinciRaj1992 */

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    Python3

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    # Python3 program to find maximum length
    # subsequence such that adjacent elements 
    # have at least one common digit 
      
    # Returns length of maximum
    # length subsequence 
    def findSubsequence(arr, n) :
      
        # To store length of maximum 
        # length subsequence 
        length = 1
      
        # To store length of subsequence 
        # having common digit d 
        dp = [0] * 10
      
        # For first element maximum length
        # is 1 for each digit 
        tmp = arr[0]; 
        while (tmp > 0) : 
            dp[tmp % 10] = 1
            tmp //= 10
          
      
        # Find digits of each element, then
        # find length of subsequence for each 
        # digit and then find local maximum 
        for i in range(1, n) :
            tmp = arr[i]; 
            locMax = 1;
            cnt = [0] * 10
              
            # Find digits in current element 
            while (tmp > 0) :
                cnt[tmp % 10] = 1
                tmp //= 10
      
            # For each digit present find length of 
            # subsequence and find local maximum 
            for d in range(10) : 
                if (cnt[d]) : 
                    dp[d] += 1
                    locMax = max(locMax, dp[d]); 
      
            # Update value of dp[d] for each digit 
            # present in current element to local 
            # maximum found 
            for d in range(10) : 
                if (cnt[d]) :
                    dp[d] = locMax; 
          
            # Update maximum length with local 
            # maximum 
            length = max(length, locMax); 
      
        return length; 
      
    # Driver code 
    if __name__ == "__main__" :
        arr = [ 1, 12, 44, 29, 33, 96, 89 ]; 
        n = len(arr)
      
        print(findSubsequence(arr, n));
          
    # This code is contributed by Ryuga

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    C#

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    // C# program to find maximum length subsequence 
    // such that adjacent elements have at least 
    // one common digit 
    using System;
      
    class GFG 
    {
      
    // Returns length of maximum length subsequence 
    static int findSubsequence(int []arr, int n) 
      
        // To store length of maximum length subsequence 
        int len = 1; 
      
        // To store current element arr[i] 
        int tmp; 
      
        int i, j, d; 
      
        // To store length of subsequence 
        // having common digit d 
        int []dp = new int[10]; 
      
      
        // To store digits present in current element 
        int []cnt = new int[10]; 
      
        // To store local maximum for current element 
        int locMax; 
      
        // For first element maximum length is 1 for 
        // each digit 
        tmp = arr[0]; 
        while (tmp > 0) 
        
            dp[tmp % 10] = 1; 
            tmp /= 10; 
        
      
        // Find digits of each element, then find length 
        // of subsequence for each digit and then find 
        // local maximum 
        for (i = 1; i < n; i++) 
        
            tmp = arr[i]; 
            locMax = 1; 
            for(int k = 0; k < 10; k++)
            {
                cnt[k] = 0;
            }
            // Find digits in current element 
            while (tmp > 0) 
            
                cnt[tmp % 10] = 1; 
                tmp /= 10; 
            
      
            // For each digit present find length of 
            // subsequence and find local maximum 
            for (d = 0; d <= 9; d++) 
            
                if (cnt[d] == 1) 
                
                    dp[d]++; 
                    locMax = Math.Max(locMax, dp[d]); 
                
            
      
            // Update value of dp[d] for each digit 
            // present in current element to local maximum 
            // found 
            for (d = 0; d <= 9; d++) 
            
                if (cnt[d] == 1) 
                
                    dp[d] = locMax; 
                
            
      
            // Update maximum length with local maximum 
            len = Math.Max(len, locMax); 
        
      
        return len; 
      
    // Driver code 
    public static void Main(String[] args) 
    {
        int []arr = { 1, 12, 44, 29, 33, 96, 89 }; 
        int n = arr.Length; 
            Console.WriteLine(findSubsequence(arr, n)); 
    }
    }
      
    // This code contributed by Rajput-Ji

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    PHP

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    <?php
    // PHP program to find maximum length subsequence
    // such that adjacent elements have at least
    // one common digit
      
    // Returns length of maximum length subsequence
    function findSubsequence($arr, $n)
    {
      
        // To store length of maximum length
        // subsequence
        $len = 1;
      
        // To store length of subsequence
        // having common digit d
        $dp = array_fill(0, 10, NULL);
      
        // For first element maximum length is 1 
        // for each digit
        $tmp = $arr[0];
        while ($tmp > 0)
        {
            $dp[$tmp % 10] = 1;
            $tmp = intval($tmp / 10);
        }
      
        // Find digits of each element, then 
        // find length of subsequence for each 
        // digit and then find local maximum
        for ($i = 1; $i < $n; $i++)
        {
            $tmp = $arr[$i];
            $locMax = 1;
            $cnt = array_fill(0, 10, NULL);
              
            // Find digits in current element
            while ($tmp > 0) 
            {
                $cnt[$tmp % 10] = 1;
                $tmp = intval($tmp / 10);
            }
      
            // For each digit present find length of
            // subsequence and find local maximum
            for ($d = 0; $d <= 9; $d++)
            {
                if ($cnt[$d]) 
                {
                    $dp[$d]++;
                    $locMax = max($locMax, $dp[$d]);
                }
            }
      
            // Update value of dp[d] for each digit
            // present in current element to local 
            // maximum found
            for ($d = 0; $d <= 9; $d++)
            {
                if ($cnt[$d]) 
                {
                    $dp[$d] = $locMax;
                }
            }
      
            // Update maximum length with
            // local maximum
            $len = max($len, $locMax);
        }
      
        return $len;
    }
      
    // Driver code
    $arr = array( 1, 12, 44, 29, 33, 96, 89 );
    $n = sizeof($arr);
    echo findSubsequence($arr, $n);
      
    // This code is contributed by ita_c
    ?>

    chevron_right

    
    

    Output:

    5
    

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



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