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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 <bits/stdc++.h>
using namespace std;
  
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
int longestSubarray(int arr[], int n)
{
    int i, d;
  
    // To mark presence of digit in current
    // element.
    int hash[2][10];
    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[0];
    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] && 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
int main()
{
    int arr[] = { 11, 22, 33, 44, 54, 56, 63 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << longestSubarray(arr, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to print the length of the 
// longest subarray such that adjacent elements 
// of the subarray have at least one digit in 
// common 
class GFG
{
  
// Function to print the longest subarray 
// such that adjacent elements have at least 
// one digit in common 
static 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[0]; 
    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 Code 
public 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to print the length of the
# longest subarray such that adjacent elements
# of the subarray have at least one digit in
# common
import math
  
# Function to print the longest subarray
# such that adjacent elements have at least
# one digit in common
def 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[0];
    while (tmp > 0): 
        HASH1[0][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 Code
arr = [ 11, 22, 33, 44, 54, 56, 63 ];
n = len(arr);
  
print(longestSubarray(arr, n));
  
# This code is contributed by chandan_jnu

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to print the length of the 
// longest subarray such that adjacent elements 
// of the subarray have at least one digit in 
// common
using System;
  
class GFG
{
  
// Function to print the longest subarray 
// such that adjacent elements have at least 
// one digit in common 
static 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[0]; 
    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 Code 
static 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
  
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
function longestSubarray($arr, $n)
{
    $i = $d = 0;
  
    // To mark presence of digit in current
    // element.
    $hash = array_fill(0, 2, array_fill(0, 10, 0));
  
    // To store current row.
    $currRow = 0;
  
    // To store maximum length subarray length.
    $maxLen = 1;
  
    // To store current subarray length.
    $len = 0;
  
    // To store current array element.
    $tmp = 0;
  
    // Mark the presence of digits
    // of first element.
    $tmp = $arr[0];
    while ($tmp > 0) 
    {
        $hash[0][$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
?>

chevron_right


Output:

4

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



My Personal Notes arrow_drop_up

A Programmer and A Machine learning Enthusiast

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.