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}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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 

Output:

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++

// 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; 
} 

Java

// 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 */

Python3

# 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 

C#

// 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 

PHP

<?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 
?> 

Output:

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

PHP

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