Longest subarray such that adjacent elements have at least one common digit | Set – 2

• Last Updated : 21 Apr, 2021

Given an array of N integers, the task is to find the length of the longest subarray such that adjacent elements of the subarray have at least one digit in common.
Examples:

Input : arr[] = {12, 23, 45, 43, 36, 97}
Output : 3
Explanation: The subarray is 45 43 36 which has
4 common in 45, 43 and 3 common in 43, 36.

Input : arr[] = {11, 22, 33, 44, 54, 56, 63}
Output : 4
Explanation: Subarray is 44, 54, 56, 63

The solution discussed in previous post uses O(N) extra space. The problem can be solved using constant space. A hashmap of constant size is used to store whether a digit is present in a given array element or not. To check if adjacent elements have a common digit, only count of digits for two adjacent elements is required. So the number of rows required in hashmap can be reduced to 2. The variable currRow represents current row and 1 – currRow represents previous row in hashmap. If adjacent elements have common digit then increase current length by 1 and compare it with maximum length. Otherwise set current length to 1.
Below is the implementation of above approach:

C++

 // CPP program to print the length of the// longest subarray such that adjacent elements// of the subarray have at least one digit in// common #include using namespace std; // Function to print the longest subarray// such that adjacent elements have at least// one digit in commonint longestSubarray(int arr[], int n){    int i, d;     // To mark presence of digit in current    // element.    int hash;    memset(hash, 0, sizeof(hash));     // To store current row.    int currRow;     // To store maximum length subarray length.    int maxLen = 1;     // To store current subarray length.    int len = 0;     // To store current array element.    int tmp;     // Mark the presence of digits of first element.    tmp = arr;    while (tmp > 0) {        hash[tmp % 10] = 1;        tmp /= 10;    }     currRow = 1;     // Find digits of each element and check if adjacent    // elements have common digit and update len.    for (i = 1; i < n; i++) {        tmp = arr[i];         for (d = 0; d <= 9; d++)            hash[currRow][d] = 0;         // Find all digits in element.        while (tmp > 0) {            hash[currRow][tmp % 10] = 1;            tmp /= 10;        }         // Find common digit in adjacent element.        for (d = 0; d <= 9; d++) {            if (hash[currRow][d] && hash[1 - currRow][d]) {                len++;                break;            }        }         // If no common digit is found a new subarray        // has to start from current element.        if (d == 10) {            len = 1;        }         maxLen = max(maxLen, len);         currRow = 1 - currRow;    }     return maxLen;} // Driver Codeint main(){    int arr[] = { 11, 22, 33, 44, 54, 56, 63 };    int n = sizeof(arr) / sizeof(arr);     cout << longestSubarray(arr, n);     return 0;}

Java

 // Java program to print the length of the// longest subarray such that adjacent elements// of the subarray have at least one digit in// commonclass GFG{ // Function to print the longest subarray// such that adjacent elements have at least// one digit in commonstatic int longestSubarray(int arr[], int n){    int i, d;     // To mark presence of digit in current    // element.    int hash[][] = new int;         for( i = 0; i < 2; i++)        for(int j = 0; j < 10; j++)            hash[i][j] = 0;     // To store current row.    int currRow;     // To store maximum length subarray length.    int maxLen = 1;     // To store current subarray length.    int len = 0;     // To store current array element.    int tmp;     // Mark the presence of digits of first element.    tmp = arr;    while (tmp > 0)    {        hash[tmp % 10] = 1;        tmp /= 10;    }     currRow = 1;     // Find digits of each element and check if adjacent    // elements have common digit and update len.    for (i = 1; i < n; i++)    {        tmp = arr[i];         for (d = 0; d <= 9; d++)            hash[currRow][d] = 0;         // Find all digits in element.        while (tmp > 0)        {            hash[currRow][tmp % 10] = 1;            tmp /= 10;        }         // Find common digit in adjacent element.        for (d = 0; d <= 9; d++)        {            if (hash[currRow][d] != 0 && hash[1 - currRow][d] != 0)            {                len++;                break;            }        }         // If no common digit is found a new subarray        // has to start from current element.        if (d == 10)        {            len = 1;        }         maxLen = Math.max(maxLen, len);         currRow = 1 - currRow;    }     return maxLen;} // Driver Codepublic static void main(String args[]){    int arr[] = { 11, 22, 33, 44, 54, 56, 63 };    int n = arr.length;     System.out.println( longestSubarray(arr, n));}} // This code is contributed by Arnab Kundu

Python3

 # Python3 program to print the length of the# longest subarray such that adjacent elements# of the subarray have at least one digit in# commonimport math # Function to print the longest subarray# such that adjacent elements have at least# one digit in commondef longestSubarray(arr, n):     i = d = 0;     # To mark presence of digit in current    # element.    HASH1 = [[0 for x in range(10)]                for y in range(2)];     # To store current row.    currRow = 0;     # To store maximum length subarray length.    maxLen = 1;     # To store current subarray length.    len1 = 0;     # To store current array element.    tmp = 0;     # Mark the presence of digits    # of first element.    tmp = arr;    while (tmp > 0):        HASH1[tmp % 10] = 1;        tmp = tmp // 10;     currRow = 1;     # Find digits of each element and check    # if adjacent elements have common digit    # and update len.    for i in range(0, n):        tmp = arr[i];         for d in range(0, 10):            HASH1[currRow][d] = 0;         # Find all digits in element.        while (tmp > 0):            HASH1[currRow][tmp % 10] = 1;            tmp = tmp // 10;         # Find common digit in adjacent element.        for d in range(0, 10):            if (HASH1[currRow][d] and                HASH1[1 - currRow][d]):                len1 += 1;                break;         # If no common digit is found a new subarray        # has to start from current element.        if (d == 10):            len1 = 1;         maxLen = max(maxLen, len1);         currRow = 1 - currRow;     return maxLen; # Driver Codearr = [ 11, 22, 33, 44, 54, 56, 63 ];n = len(arr); print(longestSubarray(arr, n)); # This code is contributed by chandan_jnu

C#

 // C# program to print the length of the// longest subarray such that adjacent elements// of the subarray have at least one digit in// commonusing System; class GFG{ // Function to print the longest subarray// such that adjacent elements have at least// one digit in commonstatic int longestSubarray(int []arr, int n){    int i, d;     // To mark presence of digit in current    // element.    int[,] hash = new int[2,10];         for( i = 0; i < 2; i++)        for(int j = 0; j < 10; j++)            hash[i,j] = 0;     // To store current row.    int currRow;     // To store maximum length subarray length.    int maxLen = 1;     // To store current subarray length.    int len = 0;     // To store current array element.    int tmp;     // Mark the presence of digits of first element.    tmp = arr;    while (tmp > 0)    {        hash[0,tmp % 10] = 1;        tmp /= 10;    }     currRow = 1;     // Find digits of each element and check if adjacent    // elements have common digit and update len.    for (i = 1; i < n; i++)    {        tmp = arr[i];         for (d = 0; d <= 9; d++)            hash[currRow,d] = 0;         // Find all digits in element.        while (tmp > 0)        {            hash[currRow,tmp % 10] = 1;            tmp /= 10;        }         // Find common digit in adjacent element.        for (d = 0; d <= 9; d++)        {            if (hash[currRow,d] != 0 &&                hash[1 - currRow,d] != 0)            {                len++;                break;            }        }         // If no common digit is found a new subarray        // has to start from current element.        if (d == 10)        {            len = 1;        }         maxLen = Math.Max(maxLen, len);         currRow = 1 - currRow;    }     return maxLen;} // Driver Codestatic void Main(){    int []arr = { 11, 22, 33, 44, 54, 56, 63 };    int n = arr.Length;     Console.WriteLine( longestSubarray(arr, n));}} // This code is contributed by chandan_jnu

PHP

 0)    {        \$hash[\$tmp % 10] = 1;        \$tmp = (int)(\$tmp / 10);    }     \$currRow = 1;     // Find digits of each element and check    // if adjacent elements have common digit    // and update len.    for (\$i = 1; \$i < \$n; \$i++)    {        \$tmp = \$arr[\$i];         for (\$d = 0; \$d <= 9; \$d++)            \$hash[\$currRow][\$d] = 0;         // Find all digits in element.        while (\$tmp > 0)        {            \$hash[\$currRow][\$tmp % 10] = 1;            \$tmp =(int)(\$tmp/10);        }         // Find common digit in adjacent element.        for (\$d = 0; \$d <= 9; \$d++)        {            if (\$hash[\$currRow][\$d] &&                \$hash[1 - \$currRow][\$d])            {                \$len++;                break;            }        }         // If no common digit is found a new subarray        // has to start from current element.        if (\$d == 10)        {            \$len = 1;        }         \$maxLen = max(\$maxLen, \$len);         \$currRow = 1 - \$currRow;    }     return \$maxLen;} // Driver Code\$arr = array( 11, 22, 33, 44, 54, 56, 63 );\$n = count(\$arr); echo longestSubarray(\$arr, \$n); // This code is contributed by chandan_jnu?>

Javascript


Output:
4

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

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