Longest subarray such that adjacent elements have at least one common digit | Set 1
Last Updated :
11 Jul, 2025
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
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][10] 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 the worst case.
Steps to solve this problem:
1. Create an array hash. hash is an 2-dimensional array with n rows and 10 columns, where each row represents the presence of each digit (0 to 9) in the corresponding index of the input array a.
2. Initialize hash to all zeros.
3. Loop through the elements of the input array a. For each element, extract the digits from the number by repeatedly dividing the number by 10 until it becomes zero, and mark the corresponding digit in the corresponding row of hash as 1.
4. Initialize longest to the minimum integer value and count to 0.
5. Loop through the elements of the input array a and check for every two consecutive elements. If they have at least one digit in common, increment the count by 1. If they don't have any digit in common, update longest to the maximum of longest and count + 1, and reset count to 0.
6. After the loop, update longest to the maximum of longest and count + 1.
7. Return longest.
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 <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 a[], int n)
{
// remembers the occurrence of digits in
// i-th index number
int hash[n][10];
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[0]);
// function call
cout << longestSubarray(a, n);
return 0;
}
// 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
# 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# 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
// 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(&$a, $n)
{
// remembers the occurrence of
// digits in i-th index number
$hash = array_fill(0, $n,
array_fill(0, 10, NULL));
// marks the presence of digit in
// i-th index number
for ($i = 0; $i < $n; $i++)
{
$num = $a[$i];
while ($num)
{
// marks the digit
$hash[$i][$num % 10] = 1;
$num = intval($num / 10);
}
}
// counts the longest Subarray
$longest = PHP_INT_MIN;
// counts the subarray
$count = 0;
// check for all adjacent elements
for ($i = 0; $i < $n - 1; $i++)
{
for ($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
$a = array(11, 22, 33, 44, 54, 56, 63 );
$n = sizeof($a);
// function call
echo longestSubarray($a, $n);
// This code is contributed by ChitraNayal
?>
<script>
// Javascript 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(a,n)
{
// remembers the occurrence of digits in
// i-th index number
let hash = new Array(n);
for(let i=0;i<n;i++)
{
hash[i]=new Array(10);
for(let j=0;j<10;j++)
{
hash[i][j]=0;
}
}
// marks the presence of digit in i-th
// index number
for (let i = 0; i < n; i++) {
let num = a[i];
while (num != 0) {
// marks the digit
hash[i][num % 10] = 1;
num = Math.floor(num/ 10);
}
}
// counts the longest Subarray
let longest = Number.MIN_VALUE;
// counts the subarray
let count = 0;
// check for all adjacent elements
for (let i = 0; i < n - 1; i++) {
let 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
let a=[11, 22, 33, 44, 54, 56, 63];
let n = a.length;
// function call
document.write(longestSubarray(a, n));
// This code is contributed by rag2127
</script>
Time Complexity: O(n*10)
Longest subarray such that adjacent elements have at least one common digit | Set – 2
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