# Longest subarray such that adjacent elements have at least one common digit | Set 1

Given an array of N integers, write a program that prints the length of the longest subarray such that adjacent elements of the subarray have at least one digit in common.

Examples:

```Input : 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 : 11 22 33 44 54 56 63
Output : 4
Explanation: Subarray is 44, 54, 56, 63
```

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

A normal approach will be to check for all the subarrays possible. But the time complexity will be O(n2).

An efficient approach will be to create a hash[n] array which marks the occurrence of digits in the i-th index number. We iterate for every element and check if adjacent elements have a digit common in between. If they have a common digit, we keep the count of the length. If the adjacent elements do not have a digit in common, we initialize the count to zero and start counting again for a subarray. Print the maximum count which is obtained while iteration. We use a hash array to minimize the time complexity as the number can be of range 10^18 which will take 18 iterations in worst case.

Given below is the illustration of the 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 common ` `int` `longestSubarray(``int` `a[], ``int` `n) ` `{ ` `    ``// remembers the occurrence of digits in ` `    ``// i-th index number ` `    ``int` `hash[n]; ` `    ``memset``(hash, 0, ``sizeof``(hash)); ` ` `  `    ``// marks the presence of digit in i-th ` `    ``// index number ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `num = a[i]; ` `        ``while` `(num) { ` `            ``// marks the digit ` `            ``hash[i][num % 10] = 1; ` `            ``num /= 10; ` `        ``} ` `    ``} ` ` `  `    ``// counts the longest Subarray ` `    ``int` `longest = INT_MIN; ` `    ``// counts the subarray ` `    ``int` `count = 0; ` ` `  `    ``// check for all adjacent elements ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``int` `j; ` `        ``for` `(j = 0; j < 10; j++) { ` ` `  `            ``// if adjacent elements have digit j  ` `            ``// in them count and break as we have ` `            ``// got at-least one digit ` `            ``if` `(hash[i][j] and hash[i + 1][j]) { ` `                ``count++; ` `                ``break``; ` `            ``} ` `        ``} ` `        ``// if no digits are common ` `        ``if` `(j == 10) { ` `            ``longest = max(longest, count + 1); ` `            ``count = 0; ` `        ``} ` `    ``} ` ` `  `    ``longest = max(longest, count + 1); ` ` `  `    ``// returns the length of the longest subarray ` `    ``return` `longest; ` `} ` `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 11, 22, 33, 44, 54, 56, 63 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``// function call ` `    ``cout << longestSubarray(a, 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  ` `// common. ` ` `  `class` `GFG { ` ` `  `// function to print the longest subarray ` `// such that adjacent elements have at least ` `// one digit in common ` `    ``static` `int` `longestSubarray(``int` `a[], ``int` `n) { ` `        ``// remembers the occurrence of digits in ` `        ``// i-th index number ` `        ``int` `hash[][] = ``new` `int``[n][``10``]; ` ` `  `        ``// marks the presence of digit in i-th ` `        ``// index number ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``int` `num = a[i]; ` `            ``while` `(num != ``0``) { ` `                ``// marks the digit ` `                ``hash[i][num % ``10``] = ``1``; ` `                ``num /= ``10``; ` `            ``} ` `        ``} ` ` `  `        ``// counts the longest Subarray ` `        ``int` `longest = Integer.MIN_VALUE; ` `        ``// counts the subarray ` `        ``int` `count = ``0``; ` ` `  `        ``// check for all adjacent elements ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `            ``int` `j; ` `            ``for` `(j = ``0``; j < ``10``; j++) { ` ` `  `                ``// if adjacent elements have digit j  ` `                ``// in them count and break as we have ` `                ``// got at-least one digit ` `                ``if` `(hash[i][j] == ``1` `& hash[i + ``1``][j] == ``1``) { ` `                    ``count++; ` `                    ``break``; ` `                ``} ` `            ``} ` `            ``// if no digits are common ` `            ``if` `(j == ``10``) { ` `                ``longest = Math.max(longest, count + ``1``); ` `                ``count = ``0``; ` `            ``} ` `        ``} ` ` `  `        ``longest = Math.max(longest, count + ``1``); ` ` `  `        ``// returns the length of the longest subarray ` `        ``return` `longest; ` `    ``} ` `// Driver Code ` ` `  `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `a[] = {``11``, ``22``, ``33``, ``44``, ``54``, ``56``, ``63``}; ` ` `  `        ``int` `n = a.length; ` `        ``// function call ` `        ``System.out.println(longestSubarray(a, n)); ` ` `  `    ``} ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python 3 program to print the length of the ` `# longest subarray such that adjacent elements ` `# of the subarray have at least one digit in  ` `# common. ` `import` `sys ` ` `  `# function to print the longest subarray ` `# such that adjacent elements have at least ` `# one digit in common ` `def` `longestSubarray(a, n): ` `     `  `    ``# remembers the occurrence of digits  ` `    ``# in i-th index number ` `    ``hash` `=` `[[``0` `for` `i ``in` `range``(``10``)] ` `               ``for` `j ``in` `range``(n)] ` ` `  `    ``# marks the presence of digit in ` `    ``# i-th index number ` `    ``for` `i ``in` `range``(n): ` `        ``num ``=` `a[i] ` `        ``while` `(num): ` `             `  `            ``# marks the digit ` `            ``hash``[i][num ``%` `10``] ``=` `1` `            ``num ``=` `int``(num ``/` `10``) ` `     `  `    ``# counts the longest Subarray ` `    ``longest ``=` `-``sys.maxsize``-``1` `     `  `    ``# counts the subarray ` `    ``count ``=` `0` ` `  `    ``# check for all adjacent elements ` `    ``for` `i ``in` `range``(n ``-` `1``): ` `        ``for` `j ``in` `range``(``10``): ` `             `  `            ``# if adjacent elements have digit j  ` `            ``# in them count and break as we have ` `            ``# got at-least one digit ` `            ``if` `(``hash``[i][j] ``and` `hash``[i ``+` `1``][j]): ` `                ``count ``+``=` `1` `                ``break` `         `  `        ``# if no digits are common ` `        ``if` `(j ``=``=` `10``): ` `            ``longest ``=` `max``(longest, count ``+` `1``) ` `            ``count ``=` `0` `     `  `    ``longest ``=` `max``(longest, count ``+` `1``) ` ` `  `    ``# returns the length of the longest  ` `    ``# subarray ` `    ``return` `longest ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``a ``=` `[``11``, ``22``, ``33``, ``44``, ``54``, ``56``, ``63``] ` ` `  `    ``n ``=` `len``(a) ` `     `  `    ``# function call ` `    ``print``(longestSubarray(a, n)) ` `     `  `# This code is contributed by ` `# Sanjit_Prasad `

## C#

 `     `  `// 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; ` `public` `class` `GFG {  ` ` `  `// function to print the longest subarray  ` `// such that adjacent elements have at least  ` `// one digit in common  ` `    ``static` `int` `longestSubarray(``int` `[]a, ``int` `n) {  ` `        ``// remembers the occurrence of digits in  ` `        ``// i-th index number  ` `        ``int` `[,]hash = ``new` `int``[n,10];  ` ` `  `        ``// marks the presence of digit in i-th  ` `        ``// index number  ` `        ``for` `(``int` `i = 0; i < n; i++) {  ` `            ``int` `num = a[i];  ` `            ``while` `(num != 0) {  ` `                ``// marks the digit  ` `                ``hash[i,num % 10] = 1;  ` `                ``num /= 10;  ` `            ``}  ` `        ``}  ` ` `  `        ``// counts the longest Subarray  ` `        ``int` `longest = ``int``.MinValue;  ` `        ``// counts the subarray  ` `        ``int` `count = 0;  ` ` `  `        ``// check for all adjacent elements  ` `        ``for` `(``int` `i = 0; i < n - 1; i++) {  ` `            ``int` `j;  ` `            ``for` `(j = 0; j < 10; j++) {  ` ` `  `                ``// if adjacent elements have digit j  ` `                ``// in them count and break as we have  ` `                ``// got at-least one digit  ` `                ``if` `(hash[i,j] == 1 & hash[i + 1,j] == 1) {  ` `                    ``count++;  ` `                    ``break``;  ` `                ``}  ` `            ``}  ` `            ``// if no digits are common  ` `            ``if` `(j == 10) {  ` `                ``longest = Math.Max(longest, count + 1);  ` `                ``count = 0;  ` `            ``}  ` `        ``}  ` ` `  `        ``longest = Math.Max(longest, count + 1);  ` ` `  `        ``// returns the length of the longest subarray  ` `        ``return` `longest;  ` `    ``}  ` `// Driver Code  ` ` `  `    ``public` `static` `void` `Main() {  ` `        ``int` `[]a = {11, 22, 33, 44, 54, 56, 63};  ` ` `  `        ``int` `n = a.Length;  ` `        ``// function call  ` `        ``Console.Write(longestSubarray(a, n));  ` ` `  `    ``}  ` `}  ` `// This code is contributed by Rajput-Ji// `

## PHP

 ` `

Output:

```4
```

Time Complexity: O(n*10)

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

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