Longest subsequence such that adjacent elements have at least one common digit
Given an array arr[] of N integers, the task is to find the length of the longest sub-sequence such that adjacent elements of the sub-sequence have at least one digit in common.
Examples:
Input: arr[] = {1, 12, 44, 29, 33, 96, 89}
Output: 5
The longest sub-sequence is {1 12 29 96 89}
Input: arr[] = {12, 23, 45, 43, 36, 97}
Output: 4
The longest sub-sequence is {12 23 43 36}
Approach: The idea is to store the length of longest sub-sequence for each digit present in an element of the array.
- dp[i][d] represents the length of the longest sub-sequence up to ith element if digit d is the common digit.
- Declare a cnt array and set cnt[d] = 1 for each digit present in current element.
- Consider each digit d as common digit and find maximum length sub-sequence ending at arr[i] as dp[i][d] = max(dp[j][d]+1) (0<=j<i).
- Also find a local maximum max(dp[i][d]) for current element.
- After finding local maximum update dp[i][d] for all digits present in the current element to a local maximum.
- This is required as local maximum represents maximum length sub-sequence for an element having digit d.
E.g. Consider arr[] = {24, 49, 96}.
The local maximum for 49 is 2 obtain from digit 4.
This local maximum will be used in finding the local maximum for 96 with common digit 9.
For that it is required for all digits in 49, dp[i][d] should be set to local maximum.
Below is the implementation of the above approach:
C++
// C++ program to find maximum length subsequence// such that adjacent elements have at least// one common digit#include <bits/stdc++.h>using namespace std;// Returns length of maximum length subsequenceint findSubsequence(int arr[], int n){ // To store the length of the // maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store the length of the sub-sequence // ending at index i and having common digit d int dp[n][10]; memset(dp, 0, sizeof(dp)); // To store digits present in current element int cnt[10]; // To store length of maximum length subsequence // ending at index i int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[0][tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; memset(cnt, 0, sizeof(cnt)); // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d]) { dp[i][d] = 1; for (j = 0; j < i; j++) { dp[i][d] = max(dp[i][d], dp[j][d] + 1); locMax = max(dp[i][d], locMax); } } } // Update value of dp[i][d] for each digit // present in current element to local maximum // found. for (d = 0; d <= 9; d++) { if (cnt[d]) { dp[i][d] = locMax; } } // Update maximum length with local maximum len = max(len, locMax); } return len;}// Driver codeint main(){ int arr[] = { 1, 12, 44, 29, 33, 96, 89 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findSubsequence(arr, n); return 0;} |
Java
// Java program to find maximum length subsequence// such that adjacent elements have at least// one common digitclass GFG{// Returns length of maximum length subsequencestatic int findSubsequence(int arr[], int n){ // To store the length of the // maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store the length of the sub-sequence // ending at index i and having common digit d int[][] dp = new int[n][10]; // To store digits present in current element int[] cnt = new int[10]; // To store length of maximum length subsequence // ending at index i int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[0][tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; for (int x = 0; x < 10; x++) cnt[x]=0; // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i][d] = 1; for (j = 0; j < i; j++) { dp[i][d] = Math.max(dp[i][d], dp[j][d] + 1); locMax = Math.max(dp[i][d], locMax); } } } // Update value of dp[i][d] for each digit // present in current element to local maximum // found. for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i][d] = locMax; } } // Update maximum length with local maximum len = Math.max(len, locMax); } return len;}// Driver codepublic static void main (String[] args){ int arr[] = { 1, 12, 44, 29, 33, 96, 89 }; int n = arr.length; System.out.println(findSubsequence(arr, n));}}// This code is contributed by mits |
Python3
# Python3 program to find maximum# Length subsequence such that# adjacent elements have at least# one common digit# Returns Length of maximum# Length subsequencedef findSubsequence(arr, n): # To store the Length of the # maximum Length subsequence Len = 1 # To store current element arr[i] tmp = 0 i, j, d = 0, 0, 0 # To store the Length of the sub-sequence # ending at index i and having common digit d dp = [[0 for i in range(10)] for i in range(n)] # To store digits present in current element cnt = [0 for i in range(10)] # To store Length of maximum # Length subsequence ending at index i locMax = 0 # For first element maximum # Length is 1 for each digit tmp = arr[0] while (tmp > 0): dp[0][tmp % 10] = 1 tmp //= 10 # Find digits of each element, # then find Length of subsequence # for each digit and then find # local maximum for i in range(1, n): tmp = arr[i] locMax = 1 cnt = [0 for i in range(10)] # Find digits in current element while (tmp > 0): cnt[tmp % 10] = 1 tmp //= 10 # For each digit present find Length of # subsequence and find local maximum for d in range(10): if (cnt[d]): dp[i][d] = 1 for j in range(i): dp[i][d] = max(dp[i][d], dp[j][d] + 1) locMax = max(dp[i][d], locMax) # Update value of dp[i][d] for each digit # present in current element to local # maximum found. for d in range(10): if (cnt[d]): dp[i][d] = locMax # Update maximum Length # with local maximum Len = max(Len, locMax) return Len# Driver codearr = [1, 12, 44, 29, 33, 96, 89]n = len(arr)print(findSubsequence(arr, n))# This code is contributed# by mohit kumar |
C#
// C# program to find maximum length subsequence// such that adjacent elements have at least// one common digitusing System;class GFG{// Returns length of maximum length subsequencestatic int findSubsequence(int []arr, int n){ // To store the length of the // maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store the length of the sub-sequence // ending at index i and having common digit d int[,] dp = new int[n, 10]; // To store digits present in current element int[] cnt = new int[10]; // To store length of maximum length subsequence // ending at index i int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[0, tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; for (int x = 0; x < 10; x++) cnt[x] = 0; // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i, d] = 1; for (j = 0; j < i; j++) { dp[i, d] = Math.Max(dp[i, d], dp[j, d] + 1); locMax = Math.Max(dp[i, d], locMax); } } } // Update value of dp[i,d] for each digit // present in current element to local maximum // found. for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i, d] = locMax; } } // Update maximum length with local maximum len = Math.Max(len, locMax); } return len;}// Driver codepublic static void Main(){ int []arr = { 1, 12, 44, 29, 33, 96, 89 }; int n = arr.Length; Console.WriteLine(findSubsequence(arr, n));}}// This code is contributed by mits |
Javascript
<script>// Javascript program to find maximum length subsequence// such that adjacent elements have at least// one common digit// Returns length of maximum length subsequencefunction findSubsequence(arr,n){ // To store the length of the // maximum length subsequence let len = 1; // To store current element arr[i] let tmp; let i, j, d; // To store the length of the sub-sequence // ending at index i and having common digit d let dp = new Array(n); for(let i = 0; i < n; i++) { dp[i] = new Array(10); for(let j = 0; j < 10; j++) { dp[i][j] = 0; } } // To store digits present in current element let cnt = new Array(10); // To store length of maximum length subsequence // ending at index i let locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[0][tmp % 10] = 1; tmp = Math.floor(tmp/10); } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; for (let x = 0; x < 10; x++) cnt[x]=0; // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp = Math.floor(tmp/10); } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i][d] = 1; for (j = 0; j < i; j++) { dp[i][d] = Math.max(dp[i][d], dp[j][d] + 1); locMax = Math.max(dp[i][d], locMax); } } } // Update value of dp[i][d] for each digit // present in current element to local maximum // found. for (d = 0; d <= 9; d++) { if (cnt[d] > 0) { dp[i][d] = locMax; } } // Update maximum length with local maximum len = Math.max(len, locMax); } return len;}// Driver codelet arr=[ 1, 12, 44, 29, 33, 96, 89];let n = arr.length;document.write(findSubsequence(arr, n)); // This code is contributed by rag2127</script> |
Output:
5
Time Complexity: O(n2)
Auxiliary Space: O(n)
The auxiliary space required for above solution can be further reduced. Observe that for each digit d present in arr[i], it is required to find maximum length sub-sequence upto that digit irrespective of the fact that at which element the sub-sequence end. This reduces auxiliary space required to O(1). For each arr[i], find local maximum and update dig[d] for each digit d in arr[i] to local maximum.
Below is the implementation of the above approach:
C++
// C++ program to find maximum length subsequence// such that adjacent elements have at least// one common digit#include <bits/stdc++.h>using namespace std;// Returns length of maximum length subsequenceint findSubsequence(int arr[], int n){ // To store length of maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store length of subsequence // having common digit d int dp[10]; memset(dp, 0, sizeof(dp)); // To store digits present in current element int cnt[10]; // To store local maximum for current element int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; memset(cnt, 0, sizeof(cnt)); // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d]) { dp[d]++; locMax = max(locMax, dp[d]); } } // Update value of dp[d] for each digit // present in current element to local maximum // found for (d = 0; d <= 9; d++) { if (cnt[d]) { dp[d] = locMax; } } // Update maximum length with local maximum len = max(len, locMax); } return len;}// Driver codeint main(){ int arr[] = { 1, 12, 44, 29, 33, 96, 89 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findSubsequence(arr, n); return 0;} |
Java
// Java program to find maximum length subsequence// such that adjacent elements have at least// one common digitimport java.util.*;class GFG{// Returns length of maximum length subsequencestatic int findSubsequence(int arr[], int n){ // To store length of maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store length of subsequence // having common digit d int dp[] = new int[10]; // To store digits present in current element int cnt[] = new int[10]; // To store local maximum for current element int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; Arrays.fill(cnt, 0); // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d]++; locMax = Math.max(locMax, dp[d]); } } // Update value of dp[d] for each digit // present in current element to local maximum // found for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d] = locMax; } } // Update maximum length with local maximum len = Math.max(len, locMax); } return len;}// Driver codepublic static void main(String[] args){ int arr[] = { 1, 12, 44, 29, 33, 96, 89 }; int n = arr.length; System.out.print(findSubsequence(arr, n));}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 program to find maximum length# subsequence such that adjacent elements# have at least one common digit# Returns length of maximum# length subsequencedef findSubsequence(arr, n) : # To store length of maximum # length subsequence length = 1; # To store length of subsequence # having common digit d dp = [0] * 10; # For first element maximum length # is 1 for each digit tmp = arr[0]; while (tmp > 0) : dp[tmp % 10] = 1; tmp //= 10; # Find digits of each element, then # find length of subsequence for each # digit and then find local maximum for i in range(1, n) : tmp = arr[i]; locMax = 1; cnt = [0] * 10 # Find digits in current element while (tmp > 0) : cnt[tmp % 10] = 1; tmp //= 10; # For each digit present find length of # subsequence and find local maximum for d in range(10) : if (cnt[d]) : dp[d] += 1; locMax = max(locMax, dp[d]); # Update value of dp[d] for each digit # present in current element to local # maximum found for d in range(10) : if (cnt[d]) : dp[d] = locMax; # Update maximum length with local # maximum length = max(length, locMax); return length;# Driver codeif __name__ == "__main__" : arr = [ 1, 12, 44, 29, 33, 96, 89 ]; n = len(arr) print(findSubsequence(arr, n)); # This code is contributed by Ryuga |
C#
// C# program to find maximum length subsequence// such that adjacent elements have at least// one common digitusing System;class GFG{// Returns length of maximum length subsequencestatic int findSubsequence(int []arr, int n){ // To store length of maximum length subsequence int len = 1; // To store current element arr[i] int tmp; int i, j, d; // To store length of subsequence // having common digit d int []dp = new int[10]; // To store digits present in current element int []cnt = new int[10]; // To store local maximum for current element int locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[tmp % 10] = 1; tmp /= 10; } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; for(int k = 0; k < 10; k++) { cnt[k] = 0; } // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp /= 10; } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d]++; locMax = Math.Max(locMax, dp[d]); } } // Update value of dp[d] for each digit // present in current element to local maximum // found for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d] = locMax; } } // Update maximum length with local maximum len = Math.Max(len, locMax); } return len;}// Driver codepublic static void Main(String[] args){ int []arr = { 1, 12, 44, 29, 33, 96, 89 }; int n = arr.Length; Console.WriteLine(findSubsequence(arr, n));}}// This code contributed by Rajput-Ji |
PHP
<?php// PHP program to find maximum length subsequence// such that adjacent elements have at least// one common digit// Returns length of maximum length subsequencefunction findSubsequence($arr, $n){ // To store length of maximum length // subsequence $len = 1; // To store length of subsequence // having common digit d $dp = array_fill(0, 10, NULL); // For first element maximum length is 1 // for each digit $tmp = $arr[0]; while ($tmp > 0) { $dp[$tmp % 10] = 1; $tmp = intval($tmp / 10); } // Find digits of each element, then // find length of subsequence for each // digit and then find local maximum for ($i = 1; $i < $n; $i++) { $tmp = $arr[$i]; $locMax = 1; $cnt = array_fill(0, 10, NULL); // Find digits in current element while ($tmp > 0) { $cnt[$tmp % 10] = 1; $tmp = intval($tmp / 10); } // For each digit present find length of // subsequence and find local maximum for ($d = 0; $d <= 9; $d++) { if ($cnt[$d]) { $dp[$d]++; $locMax = max($locMax, $dp[$d]); } } // Update value of dp[d] for each digit // present in current element to local // maximum found for ($d = 0; $d <= 9; $d++) { if ($cnt[$d]) { $dp[$d] = $locMax; } } // Update maximum length with // local maximum $len = max($len, $locMax); } return $len;}// Driver code$arr = array( 1, 12, 44, 29, 33, 96, 89 );$n = sizeof($arr);echo findSubsequence($arr, $n);// This code is contributed by ita_c?> |
Javascript
<script>// JavaScript program to find maximum length subsequence// such that adjacent elements have at least// one common digit // Returns length of maximum length subsequence function findSubsequence(arr , n) { // To store length of maximum length subsequence var len = 1; // To store current element arr[i] var tmp; var i, j, d; // To store length of subsequence // having common digit d var dp = Array(10).fill(0); // To store digits present in current element var cnt = Array(10).fill(0); // To store local maximum for current element var locMax; // For first element maximum length is 1 for // each digit tmp = arr[0]; while (tmp > 0) { dp[tmp % 10] = 1; tmp = parseInt(tmp/10); } // Find digits of each element, then find length // of subsequence for each digit and then find // local maximum for (var i = 1; i < n; i++) { tmp = arr[i]; locMax = 1; cnt.fill( 0); // Find digits in current element while (tmp > 0) { cnt[tmp % 10] = 1; tmp = parseInt(tmp/10); } // For each digit present find length of // subsequence and find local maximum for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d]++; locMax = Math.max(locMax, dp[d]); } } // Update value of dp[d] for each digit // present in current element to local maximum // found for (d = 0; d <= 9; d++) { if (cnt[d] == 1) { dp[d] = locMax; } } // Update maximum length with local maximum len = Math.max(len, locMax); } return len; } // Driver code var arr = [ 1, 12, 44, 29, 33, 96, 89 ]; var n = arr.length; document.write(findSubsequence(arr, n));// This code contributed by gauravrajput1</script> |
Output:
5
Time Complexity: O(n)
Auxiliary Space: O(1)
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