Longest subsequence such that adjacent elements have at least one common digit
Last Updated :
11 Jul, 2025
Given an array arr[], the task is to find the length of the longest sub-sequence such that adjacent elements of the subsequence have at least one digit in common.
Examples:
Input: arr[] = [1, 12, 44, 29, 33, 96, 89]
Output: 5
Explanation: The longest sub-sequence is [1 12 29 96 89]
Input: arr[] = [12, 23, 45, 43, 36, 97]
Output: 4
Explanation: The longest sub-sequence is [12 23 43 36]
Using recursion - O(2^n) Time O(n) Space
In this approach, we recursively explore all possible subsequences from the given array. For each subsequence, we check whether every pair of adjacent elements shares at least one common digit. This is done by maintaining a prevDigit array that tracks the digits encountered in the previous subsequence element. If a common digit is found between the current element and the last chosen element, we update the subsequence length. Ultimately, the recursion helps identify the longest subsequence that satisfies the condition where each adjacent pair has at least one digit in common.
C++
// C++ program to find max length of the subsequence
// such that each adjacent element of the subsequence has at
// least one digit in common.
#include <bits/stdc++.h>
using namespace std;
int longestSubsequence(int curr, vector<int> &arr, vector<int> prevDigit) {
// Base case: If we've processed all elements, return 0
if (curr >= arr.size())
return 0;
// Recurse by skipping the current number
int res = longestSubsequence(curr + 1, arr, prevDigit);
bool hasCommonDigit = false;
// Convert the number to string to check its digits
string numStr = to_string(arr[curr]);
// Check if any digit in the current number matches
// a previously used digit
for (auto digitChar : numStr) {
int digit = digitChar - '0';
if (prevDigit[digit] == 1) {
hasCommonDigit = true;
break;
}
}
// If there's a common digit, consider this number as
// part of the subsequence
if (hasCommonDigit) {
// Reset prevDigit to reflect the current
// number's digits
for (int i = 0; i < 10; i++) {
prevDigit[i] = 0;
}
// Mark the digits of the current number as used
for (auto digitChar : numStr) {
prevDigit[digitChar - '0'] = 1;
}
// Recurse, adding the current number to the
// subsequence
res = max(res, 1 + longestSubsequence(curr + 1, arr, prevDigit));
}
return res;
}
int main() {
vector<int> arr = {1, 12, 44, 29, 33, 96, 89};
vector<int> prevDigit(10, 1);
int res = longestSubsequence(0, arr, prevDigit);
cout << res << endl;
return 0;
}
// Java program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common.
import java.util.*;
class GfG {
static int
longestSubsequence(int curr, int[] arr,
int[] prevDigit) {
// Base case: If we've processed all elements,
// return 0
if (curr >= arr.length)
return 0;
// Recurse by skipping the current number
int res = longestSubsequence(
curr + 1, arr, prevDigit);
boolean hasCommonDigit = false;
// Convert the number to string to check its digits
String numStr = Integer.toString(arr[curr]);
// Check if any digit in the current number matches
// a previously used digit
for (char digitChar : numStr.toCharArray()) {
int digit = digitChar - '0';
if (prevDigit[digit] == 1) {
hasCommonDigit = true;
break;
}
}
// If there's a common digit, consider this number
// as part of the subsequence
if (hasCommonDigit) {
// Reset prevDigit to reflect the current
// number's digits
Arrays.fill(prevDigit, 0);
// Mark the digits of the current number as used
for (char digitChar : numStr.toCharArray()) {
prevDigit[digitChar - '0'] = 1;
}
// Recurse, adding the current number to the
// subsequence
res = Math.max(
res,
1 + longestSubsequence(
curr + 1, arr, prevDigit));
}
return res;
}
public static void main(String[] args) {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int[] prevDigit = new int[10];
Arrays.fill(
prevDigit,
1);
int res = longestSubsequence(
0, arr, prevDigit);
System.out.println(res);
}
}
# Python program to find the maximum length of the subsequence
# such that each adjacent element of the subsequence has at least
# one digit in common.
def longestSubsequence(curr, arr, prevDigit):
# Base case: If we've processed all elements,
# return 0
if curr >= len(arr):
return 0
# Recurse by skipping the current number
res = longestSubsequence(curr + 1, arr, prevDigit)
hasCommonDigit = False
# Convert the number to string to check its digits
numStr = str(arr[curr])
# Check if any digit in the current number matches
# a previously used digit
for digitChar in numStr:
digit = int(digitChar)
if prevDigit[digit] == 1:
hasCommonDigit = True
break
# If there's a common digit, consider this number
# as part of the subsequence
if hasCommonDigit:
# Reset prevDigit to reflect the current number's digits
prevDigit = [0] * 10
# Mark the digits of the current number as used
for digitChar in numStr:
prevDigit[int(digitChar)] = 1
# Recurse, adding the current number to the subsequence
res = max(
res, 1 + longestSubsequence(curr + 1, arr, prevDigit))
return res
if __name__ == "__main__":
arr = [1, 12, 44, 29, 33, 96, 89]
prevDigit = [1] * 10
res = longestSubsequence(0, arr, prevDigit)
print(res)
// C# program to find the maximum length of the subsequence
// such that each adjacent element of the subsequence has at
// least one digit in common.
using System;
class GfG {
static int
longestSubsequence(int curr, int[] arr,
int[] prevDigit) {
// Base case: If we've processed all elements,
// return 0
if (curr >= arr.Length)
return 0;
// Recurse by skipping the current number
int res = longestSubsequence(
curr + 1, arr, prevDigit);
bool hasCommonDigit = false;
// Convert the number to string to check its digits
string numStr = arr[curr].ToString();
// Check if any digit in the current number matches
// a previously used digit
foreach(char digitChar in numStr) {
int digit = digitChar - '0';
if (prevDigit[digit] == 1) {
hasCommonDigit = true;
break;
}
}
// If there's a common digit, consider this number
// as part of the subsequence
if (hasCommonDigit) {
// Reset prevDigit to reflect the current
// number's digits
Array.Clear(prevDigit, 0, prevDigit.Length);
// Mark the digits of the current number as used
foreach(char digitChar in numStr) {
prevDigit[digitChar - '0'] = 1;
}
// Recurse, adding the current number to the
// subsequence
res = Math.Max(
res,
1
+ longestSubsequence(
curr + 1, arr, prevDigit));
}
return res;
}
static void Main(string[] args) {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int[] prevDigit = new int[10];
Array.Fill(
prevDigit,
1);
int res = longestSubsequence(
0, arr, prevDigit);
Console.WriteLine(res);
}
}
// JavaScript program to find the maximum length of the
// subsequence such that each adjacent element of the
// subsequence has at least one digit in common.
function longestSubsequenceWithCommonDigit(curr, arr,
prevDigit) {
// Base case: If we've processed all elements, return 0
if (curr >= arr.length)
return 0;
// Recurse by skipping the current number
let res = longestSubsequenceWithCommonDigit(
curr + 1, arr, prevDigit);
let hasCommonDigit = false;
// Convert the number to string to check its digits
let numStr = arr[curr].toString();
// Check if any digit in the current number matches a
// previously used digit
for (let digitChar of numStr) {
let digit = parseInt(digitChar);
if (prevDigit[digit] === 1) {
hasCommonDigit = true;
break;
}
}
// If there's a common digit, consider this number as
// part of the subsequence
if (hasCommonDigit) {
// Reset prevDigit to reflect the current number's
// digits
prevDigit.fill(0);
// Mark the digits of the current number as used
for (let digitChar of numStr) {
prevDigit[parseInt(digitChar)] = 1;
}
// Recurse, adding the current number to the
// subsequence
res = Math.max(
res, 1
+ longestSubsequenceWithCommonDigit(
curr + 1, arr, prevDigit));
}
return res;
}
let arr = [ 1, 12, 44, 29, 33, 96, 89 ];
let prevDigit = new Array(10).fill(
1);
let res
= longestSubsequenceWithCommonDigit(0, arr, prevDigit);
console.log(res);
[Expected Approach - 1] Using Tabulation - O(n*n) Time O(n) Space
This approach is similar to finding the Longest Increasing Subsequence (LIS). Here, we need to find subsequences where each adjacent pair of numbers shares at least one common digit.
We use a dp array where dp[i] stores the length of the longest subsequence that ends at the ith index. Initially, each number can form a subsequence of length 1 on its own, so we start by setting dp[i] = 1. For each number at index i, we look at all previous numbers (from j = 0 to i-1). If there is any common digit between the numbers at i and j, we can extend the subsequence ending at j by including the number at i. We then update dp[i] using:
- dp[i] = max(dp[i], dp[j]+1)
The final result is the maximum value in the dp array, which gives the length of the longest subsequence where each adjacent pair of numbers shares at least one common digit.
C++
// C++ program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common.
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int longestSubsequence(vector<int> &arr) {
int n = arr.size();
// Creating DP array of size n
vector<int> dp(n);
// Max length of the subsequence such that
// each adjacent elements of the subsequence have at
// least one digit in common.
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Each number can form a subsequence of
// length 1 by itself
dp[i] = 1;
// Bool array to store the presence of digits
// in the current number
vector<bool> digitsOfCurrentNumber(10, false);
int tem = arr[i];
// Extracting and marking digits of current 'i'th element
while (tem) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
tem /= 10;
}
// Updating DP array by checking for common digits
// with previous elements
for (int j = 0; j < i; j++) {
tem = arr[j];
// Extracting digits of 'j'th element and checking for
// common digit with 'i'th element
while (tem) {
int digit = tem % 10;
// If a common digit is found, we can add the current
// element to the subsequence
if (digitsOfCurrentNumber[digit]) {
dp[i] = max(dp[i], dp[j] + 1);
break;
// No need to check further digits if a
// common one is found
}
tem /= 10;
}
}
// Updating the maximum length of the
// subsequence found so far
maxLength = max(maxLength, dp[i]);
}
return maxLength;
}
int main() {
vector<int> arr = {1, 12, 44, 29, 33, 96, 89};
int res = longestSubsequence(arr);
cout << res << endl;
return 0;
}
// Java program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common.
import java.util.*;
class GfG {
static int
longestSubsequence(int[] arr) {
int n = arr.length;
// Creating DP array of size n
int[] dp = new int[n];
// Max length of the subsequence such that each
// adjacent elements of the subsequence have at
// least one digit in common.
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Each number can form a subsequence of length
// 1 by itself
dp[i] = 1;
// Bool array to store the presence of digits in
// the current number
boolean[] digitsOfCurrentNumber
= new boolean[10];
int tem = arr[i];
// Extracting and marking digits of current
// 'i'th element
while (tem > 0) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
tem /= 10;
}
// Updating DP array by checking for common
// digits with previous elements
for (int j = 0; j < i; j++) {
tem = arr[j];
// Extracting digits of 'j'th element and
// checking for common digit with 'i'th
// element
while (tem > 0) {
int digit = tem % 10;
// If a common digit is found, we can
// add the current element to the
// subsequence
if (digitsOfCurrentNumber[digit]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
break;
// No need to check further digits
// if a common one is found
}
tem /= 10;
}
}
// Updating the maximum length of the
// subsequence found so far
maxLength = Math.max(maxLength, dp[i]);
}
return maxLength;
}
public static void main(String[] args) {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int res = longestSubsequence(arr);
System.out.println(res);
}
}
# Python program to find max length of the subsequence
# such that each adjacent element of the subsequence has at
# least one digit in common.
def longestSubsequence(arr):
n = len(arr)
# Creating DP array of size n
dp = [0] * n
# Max length of the subsequence such that each adjacent element
# of the subsequence has at least one digit in common.
maxLength = 0
for i in range(n):
# Each number can form a subsequence of length 1 by itself
dp[i] = 1
# Bool array to store the presence of digits in the current number
digitsOfCurrentNumber = [False] * 10
tem = arr[i]
# Extracting and marking digits of current 'i'th element
while tem > 0:
digit = tem % 10
digitsOfCurrentNumber[digit] = True
tem //= 10
# Updating DP array by checking for common
# digits with previous elements
for j in range(i):
tem = arr[j]
# Extracting digits of 'j'th element and checking for common
# digit with 'i'th element
while tem > 0:
digit = tem % 10
# If a common digit is found, we can add the current element
# to the subsequence
if digitsOfCurrentNumber[digit]:
dp[i] = max(dp[i], dp[j] + 1)
break
tem //= 10
# Updating the maximum length of the subsequence found so far
maxLength = max(maxLength, dp[i])
return maxLength
if __name__ == "__main__":
arr = [1, 12, 44, 29, 33, 96, 89]
res = longestSubsequence(arr)
print(res)
// C# program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common.
using System;
using System.Linq;
class GfG {
static int longestSubsequence(int[] arr) {
int n = arr.Length;
// Creating DP array of size n
int[] dp = new int[n];
// Max length of the subsequence such that each
// adjacent element of the subsequence has at least
// one digit in common.
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Each number can form a subsequence of length
// 1 by itself
dp[i] = 1;
// Bool array to store the presence of digits in
// the current number
bool[] digitsOfCurrentNumber = new bool[10];
int tem = arr[i];
// Extracting and marking digits of current
// 'i'th element
while (tem > 0) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
tem /= 10;
}
// Updating DP array by checking for common
// digits with previous elements
for (int j = 0; j < i; j++) {
tem = arr[j];
// Extracting digits of 'j'th element and
// checking for common digit with 'i'th
// element
while (tem > 0) {
int digit = tem % 10;
// If a common digit is found, we can
// add the current element to the
// subsequence
if (digitsOfCurrentNumber[digit]) {
dp[i] = Math.Max(dp[i], dp[j] + 1);
break;
}
tem /= 10;
}
}
// Updating the maximum length of the
// subsequence found so far
maxLength = Math.Max(maxLength, dp[i]);
}
return maxLength;
}
static void Main() {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int res = longestSubsequence(arr);
Console.WriteLine(res);
}
}
// JavaScript program to find the max length of the
// subsequence such that each adjacent element of the
// subsequence has at least one digit in common.
function longestSubsequence(arr) {
const n = arr.length;
// Creating DP array of size n
const dp = new Array(n).fill(0);
// Max length of the subsequence such that each adjacent
// elements of the subsequence have at least one digit
// in common.
let maxLength = 0;
for (let i = 0; i < n; i++) {
// Each number can form a subsequence of length 1 by
// itself
dp[i] = 1;
// Bool array to store the presence of digits in the
// current number
const digitsOfCurrentNumber
= new Array(10).fill(false);
let temp = arr[i];
// Extracting and marking digits of the current
// 'i'th element
while (temp > 0) {
const digit = temp % 10;
digitsOfCurrentNumber[digit] = true;
temp = Math.floor(temp / 10);
}
// Updating DP array by checking for common digits
// with previous elements
for (let j = 0; j < i; j++) {
temp = arr[j];
// Extracting digits of 'j'th element and
// checking for common digit with 'i'th element
while (temp > 0) {
const digit = temp % 10;
// If a common digit is found, we can add
// the current element to the subsequence
if (digitsOfCurrentNumber[digit]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
break;
}
temp = Math.floor(temp / 10);
}
}
// Updating the maximum length of the subsequence
// found so far
maxLength = Math.max(maxLength, dp[i]);
}
return maxLength;
}
const arr = [ 1, 12, 44, 29, 33, 96, 89 ];
const res = longestSubsequence(arr);
console.log(res);
[Expected Approach - 2] Using Tabulation - O(n) Time O(n) Space
This apporach is similar to the above one, but here for every i, we are not iterating from 0 to i-1. The state we have used is dp[i][j] denotes the length of the longest subsequence till i th index having the last element-containing j as a digit.
Base Case:
dp[0][j] = 1 for all j present in 0th element
Recurrence Relation:
For each element arr[i]:
1. Identify the digit present in arr[i]: Let digitsOfCurrentNumber be a boolean array where digitsOfCurrentNumber[d] = 1 if digit d is present in arr[i-1], and otherwise 0.
2. Maximize the subsequence length for each digit j: For each digit that appears in arr[i], update currentMax:
- currentMax = max(currentMax, dp[i-1][d]+1),
- This means that if digit d appeared in the current element, the subsequence can extend from the previous subsequence ending with d, by adding the current element.
3. Update the DP table: For each j from 0 to 9, update the dp table, if digit j present in current element:
- if current j is present in arr[i], dp[i][j] = currentMax
- otherwise, dp[i][j] = dp[i-1][j]
The answer will be maximum value present in last row of dp table.
C++
// C++ program to find max length of the subsequence
// such that each adjacent element of the subsequence has
// at least one digit in common.
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int longestSubsequence(vector<int> &arr) {
int n = arr.size();
// DP matrix: dp[i][j] denotes the length of the longest
// subsequence till i-th element
// where the last element contains the digit 'j'
vector<vector<int>> dp(n + 1, vector<int>(10, 0));
// Iterate through each element in the array
for (int i = 1; i <= n; i++) {
int currentMax = 0;
int tem = arr[i - 1];
// Track the digits present in the current number
vector<bool> digitsOfCurrentNumber(10, false);
// Extract digits from current element and
// maximize currentMax
while (tem) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
currentMax = max(currentMax, dp[i - 1][digit] + 1);
tem /= 10;
}
// Update DP matrix based on the common digits
for (int j = 0; j < 10; j++) {
if (digitsOfCurrentNumber[j]) {
dp[i][j] = currentMax;
}
else {
dp[i][j] = dp[i - 1][j];
}
}
}
// Find the maximum subsequence length from the
// last row of dp
int maxLength = 0;
for (int i = 0; i < 10; i++) {
maxLength = max(maxLength, dp[n][i]);
}
return maxLength;
}
int main() {
vector<int> arr = {1, 12, 44, 29, 33, 96, 89};
int res = longestSubsequence(arr);
cout << res << endl;
return 0;
}
// Java program to find max length of the subsequence
// such that each adjacent element of the subsequence
// has at least one digit in common.
import java.util.*;
class GfG {
static int longestSubsequence(int[] arr) {
int n = arr.length;
// DP matrix: dp[i][j] denotes the length of the
// longest subsequence till i-th element where the
// last element contains the digit 'j'
int[][] dp = new int[n + 1][10];
// Iterate through each element in the array
for (int i = 1; i <= n; i++) {
int currentMax = 0;
int tem = arr[i - 1];
// Track the digits present in the current
// number
boolean[] digitsOfCurrentNumber
= new boolean[10];
// Extract digits from current element and
// maximize currentMax
while (tem > 0) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
currentMax = Math.max(currentMax,
dp[i - 1][digit] + 1);
tem /= 10;
}
// Update DP matrix based on the common digits
for (int j = 0; j < 10; j++) {
if (digitsOfCurrentNumber[j]) {
dp[i][j] = currentMax;
}
else {
dp[i][j] = dp[i - 1][j];
}
}
}
// Find the maximum subsequence length from the last
// row of dp
int maxLength = 0;
for (int i = 0; i < 10; i++) {
maxLength = Math.max(maxLength, dp[n][i]);
}
return maxLength;
}
public static void main(String[] args) {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int res = longestSubsequence(arr);
System.out.println(res);
}
}
# Python program to find max length of the subsequence
# such that each adjacent element of the subsequence has
# at least one digit in common.
def longestSubsequence(arr):
n = len(arr)
# DP matrix: dp[i][j] denotes the length of the
# longest subsequence till i-th element
# where the last element contains the digit 'j'
dp = [[0] * 10 for _ in range(n + 1)]
# Iterate through each element in the array
for i in range(1, n + 1):
currentMax = 0
tem = arr[i - 1]
# Track the digits present in the current number
digitsOfCurrentNumber = [False] * 10
# Extract digits from the current element
# and maximize currentMax
while tem > 0:
digit = tem % 10
digitsOfCurrentNumber[digit] = True
currentMax = max(currentMax, dp[i - 1][digit] + 1)
tem //= 10
# Update DP matrix based on the common digits
for j in range(10):
if digitsOfCurrentNumber[j]:
dp[i][j] = currentMax
else:
dp[i][j] = dp[i - 1][j]
# Find the maximum subsequence length from the
# last row of dp
maxLength = 0
for i in range(10):
maxLength = max(maxLength, dp[n][i])
return maxLength
arr = [1, 12, 44, 29, 33, 96, 89]
res = longestSubsequence(arr)
print(res)
// C# program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common.
using System;
class GfG {
static int longestSubsequence(int[] arr) {
int n = arr.Length;
// DP matrix: dp[i][j] denotes the length of the
// longest subsequence till i-th element where the
// last element contains the digit 'j'
int[, ] dp = new int[n + 1, 10];
// Iterate through each element in the array
for (int i = 1; i <= n; i++) {
int currentMax = 0;
int tem = arr[i - 1];
// Track the digits present in the current
// number
bool[] digitsOfCurrentNumber = new bool[10];
// Extract digits from the current element and
// maximize currentMax
while (tem > 0) {
int digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
currentMax = Math.Max(currentMax,
dp[i - 1, digit] + 1);
tem /= 10;
}
// Update DP matrix based on the common digits
for (int j = 0; j < 10; j++) {
if (digitsOfCurrentNumber[j]) {
dp[i, j] = currentMax;
}
else {
dp[i, j] = dp[i - 1, j];
}
}
}
// Find the maximum subsequence length from the last
// row of dp
int maxLength = 0;
for (int i = 0; i < 10; i++) {
maxLength = Math.Max(maxLength, dp[n, i]);
}
return maxLength;
}
static void Main() {
int[] arr = { 1, 12, 44, 29, 33, 96, 89 };
int res = longestSubsequence(arr);
Console.WriteLine(res);
}
}
// JavaScript program to find max length of the subsequence
// such that each adjacent element of the subsequence has at
// least one digit in common.
function longestSubsequence(arr) {
const n = arr.length;
// DP matrix: dp[i][j] denotes the length of the longest
// subsequence till i-th element where the last element
// contains the digit 'j'
let dp = Array.from({length : n + 1},
() => Array(10).fill(0));
// Iterate through each element in the array
for (let i = 1; i <= n; i++) {
let currentMax = 0;
let tem = arr[i - 1];
// Track the digits present in the current number
let digitsOfCurrentNumber = Array(10).fill(false);
// Extract digits from the current element and
// maximize currentMax
while (tem > 0) {
const digit = tem % 10;
digitsOfCurrentNumber[digit] = true;
currentMax = Math.max(currentMax,
dp[i - 1][digit] + 1);
tem = Math.floor(tem / 10);
}
// Update DP matrix based on the common digits
for (let j = 0; j < 10; j++) {
if (digitsOfCurrentNumber[j]) {
dp[i][j] = currentMax;
}
else {
dp[i][j] = dp[i - 1][j];
}
}
}
// Find the maximum subsequence length from the last row
// of dp
let maxLength = 0;
for (let i = 0; i < 10; i++) {
maxLength = Math.max(maxLength, dp[n][i]);
}
return maxLength;
}
const arr = [ 1, 12, 44, 29, 33, 96, 89 ];
const res = longestSubsequence(arr);
console.log(res);
[Expected Approach - 3] Using Space Optimized DP - O(n) Time O(1) Space
In this approach, we optimize the DP table by reducing its size from n × 10 to 1D array of size 10. We use a 1D DP array where dp[i] denotes the length of longest subsequence having the last element containing i as a digit.
For each element, we extract the digits and update the dp array as follows:
We iterate through all digits present in the current element. For each digit d, we calculate the currentVal as the maximum value of dp[d](the longest subsequence ending with d) to ensure that subsequences are extended properly.
- currentVal = max(currentVal, dp[d]) for all d present in current element.
After calculating currentVal, we update the dp[d] for each digit d present in the current element, ensuring the subsequences are extended accordingly.
C++
// C++ program to find max length of the subsequence
// such that each adjacent element of the subsequence
// has at least one digit in common .
#include <iostream>
#include <vector>
using namespace std;
int longestSubsequence(vector<int> &arr) {
// Creating DP array of size 10
// where dp[i] denotes the length of longest
// subsequence having the last element containing 'i' as a digit.
int n = arr.size();
// DP array to track the longest subsequence
// for each digit (0-9)
vector<int> dp(10, 0);
// Initialize the maximum length of subsequence
int maxLength = 0;
for (int i = 0; i < n; i++) {
int currentMax = 0;
int currentElement = arr[i];
// Bool array to store the presence of digits
// in the current element
vector<bool> digitsOfCurrentElement(10, false);
// Extracting digits of the current element and
// maximizing currentMax
while (currentElement) {
int digit = currentElement % 10;
digitsOfCurrentElement[digit] = true;
currentMax = max(currentMax, dp[digit] + 1);
currentElement /= 10;
}
// Update the DP array for all digits present
// in the current element
for (int digit = 0; digit < 10; digit++) {
if (digitsOfCurrentElement[digit]) {
dp[digit] = currentMax;
}
}
// Update the maximum length of subsequence
// found so far
maxLength = max(maxLength, currentMax);
}
return maxLength;
}
int main() {
vector<int> arr = {1, 12, 44, 29, 33, 96, 89};
int res = longestSubsequence(arr);
cout << res << endl;
return 0;
}
// Java program to find max length of the subsequence
// such that each adjacent element of the subsequence has
// at least one digit in common .
import java.util.*;
class GfG {
static int
longestSubsequence(int[] arr) {
// Creating DP array of size 10
// where dp[i] denotes the length of longest
// subsequence having the last element containing
// 'i' as a digit.
int n = arr.length;
// DP array to track the longest subsequence for
// each digit (0-9)
int[] dp = new int[10];
// Initialize the maximum length of subsequence
int maxLength = 0;
for (int i = 0; i < n; i++) {
int currentMax = 0;
int currentElement = arr[i];
// Bool array to store the presence of digits in
// the current element
boolean[] digitsOfCurrentElement
= new boolean[10];
// Extracting digits of the current element and
// maximizing currentMax
while (currentElement > 0) {
int digit = currentElement % 10;
digitsOfCurrentElement[digit] = true;
currentMax
= Math.max(currentMax, dp[digit] + 1);
currentElement /= 10;
}
// Update the DP array for all digits present in
// the current element
for (int digit = 0; digit < 10; digit++) {
if (digitsOfCurrentElement[digit]) {
dp[digit] = currentMax;
}
}
// Update the maximum length of subsequence
// found so far
maxLength = Math.max(maxLength, currentMax);
}
return maxLength;
}
public static void main(String[] args) {
int[] arr = {
1, 12, 44, 29, 33, 96, 89
};
int res = longestSubsequence(arr);
System.out.println(res);
}
}
# Java program to find max length of the subsequence
# such that each adjacent element of the subsequence has
# at least one digit in common .
def longestSubsequence(arr):
# Creates a DP array of size 10 where dp[i] denotes
# the length of the longest subsequence having the last
# element containing 'i' as a digit.
# Initialize DP array of size 10
dp = [0] * 10
# Initialize the maximum length of subsequence
maxLength = 0
# Iterate through each element in the array
for num in arr:
currentMax = 0
# Bool array to store the presence of digits
# in the current element
digitsOfCurrentElement = [False] * 10
# Extract digits of the current element and
# maximize currentMax
temp = num
while temp > 0:
digit = temp % 10
digitsOfCurrentElement[digit] = True
currentMax = max(currentMax, dp[digit] + 1)
temp //= 10
# Update the DP array for all digits present in
# the current element
for digit in range(10):
if digitsOfCurrentElement[digit]:
dp[digit] = currentMax
# Update the maximum length of subsequence found so far
maxLength = max(maxLength, currentMax)
return maxLength
arr = [1, 12, 44, 29, 33, 96, 89]
res = longestSubsequence(arr)
print(res)
// C# program to find max length of the subsequence such
// that each adjacent element of the subsequence has at
// least one digit in common .
using System;
using System.Collections.Generic;
class GfG {
static int
longestSubsequence(List<int> arr) {
int[] dp = new int[10];
// Initialize the maximum length of subsequence
int maxLength = 0;
// Iterate through each element in the array
foreach(int num in arr) {
int currentMax = 0;
// Bool array to store the presence of digits in
// the current element
bool[] digitsOfCurrentElement = new bool[10];
// Extract digits of the current element and
// maximize currentMax
int temp = num;
while (temp > 0) {
int digit = temp % 10;
digitsOfCurrentElement[digit] = true;
currentMax
= Math.Max(currentMax, dp[digit] + 1);
temp /= 10;
}
// Update the DP array for all digits present in
// the current element
for (int digit = 0; digit < 10; digit++) {
if (digitsOfCurrentElement[digit]) {
dp[digit] = currentMax;
}
}
// Update the maximum length of subsequence
// found so far
maxLength = Math.Max(maxLength, currentMax);
}
return maxLength;
}
static void Main(string[] args) {
List<int> arr
= new List<int>{ 1, 12, 44, 29, 33, 96, 89 };
int res = longestSubsequence(arr);
Console.WriteLine(res);
}
}
// JavaScript program to find max length of the subsequence
// such that each adjacent element of the subsequence has at
// least one digit in common .
function longestSubsequence(arr) {
// Initialize DP array of size 10
let dp = new Array(10).fill(0);
// Initialize the maximum length of subsequence
let maxLength = 0;
// Iterate through each element in the array
for (let num of arr) {
let currentMax = 0;
// Bool array to store the presence of digits
// in the current element
let digitsOfCurrentElement = new Array(10).fill(false);
// Extract digits of the current element and
// maximize currentMax
let temp = num;
while (temp > 0) {
let digit = temp % 10;
digitsOfCurrentElement[digit] = true;
currentMax = Math.max(currentMax, dp[digit] + 1);
temp = Math.floor(temp / 10);
}
// Update the DP array for all digits present in
// the current element
for (let digit = 0; digit < 10; digit++) {
if (digitsOfCurrentElement[digit]) {
dp[digit] = currentMax;
}
}
// Update the maximum length of subsequence found so far
maxLength = Math.max(maxLength, currentMax);
}
return maxLength;
}
const arr = [1, 12, 44, 29, 33, 96, 89];
const res = longestSubsequence(arr);
console.log(res);
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