Given an array of n integers. The problem is to find maximum length of the subsequence with difference between adjacent elements as either 0 or 1.
Examples:
Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output : 5
The subsequence is {5, 6, 7, 6, 5}.
Input : arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output : 4
The subsequence is {-2, -1, -1, 0}.
Source: Expedia Interview Experience | Set 12
The solution to this problem closely resembles the Longest Increasing Subsequence problem. The only difference is that here we have to check whether the absolute difference between the adjacent elements of the subsequence is either 0 or 1.
C++
// C++ implementation to find maximum length // subsequence with difference between adjacent // elements as either 0 or 1 #include <bits/stdc++.h> using namespace std; // function to find maximum length subsequence // with difference between adjacent elements as // either 0 or 1 int maxLenSub(int arr[], int n) { int mls[n], max = 0; // Initialize mls[] values for all indexes for (int i=0; i<n; i++) mls[i] = 1; // Compute optimized maximum length subsequence // values in bottom up manner for (int i=1; i<n; i++) for (int j=0; j<i; j++) if (abs(arr[i] - arr[j]) <= 1 && mls[i] < mls[j] + 1) mls[i] = mls[j] + 1; // Store maximum of all 'mls' values in 'max' for (int i=0; i<n; i++) if (max < mls[i]) max = mls[i]; // required maximum length subsequence return max; } // Driver program to test above int main() { int arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Maximum length subsequence = " << maxLenSub(arr, n); return 0; } |
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Java
// JAVA Code for Maximum length subsequence // with difference between adjacent elements // as either 0 or 1 import java.util.*; class GFG { // function to find maximum length subsequence // with difference between adjacent elements as // either 0 or 1 public static int maxLenSub(int arr[], int n) { int mls[] = new int[n], max = 0; // Initialize mls[] values for all indexes for (int i = 0; i < n; i++) mls[i] = 1; // Compute optimized maximum length // subsequence values in bottom up manner for (int i = 1; i < n; i++) for (int j = 0; j < i; j++) if (Math.abs(arr[i] - arr[j]) <= 1 && mls[i] < mls[j] + 1) mls[i] = mls[j] + 1; // Store maximum of all 'mls' values in 'max' for (int i = 0; i < n; i++) if (max < mls[i]) max = mls[i]; // required maximum length subsequence return max; } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; int n = arr.length; System.out.println("Maximum length subsequence = "+ maxLenSub(arr, n)); } } // This code is contributed by Arnav Kr. Mandal. |
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Python3
# Python implementation to find maximum length # subsequence with difference between adjacent # elements as either 0 or 1 # function to find maximum length subsequence # with difference between adjacent elements as # either 0 or 1 def maxLenSub( arr, n): mls=[] max = 0 #Initialize mls[] values for all indexes for i in range(n): mls.append(1) #Compute optimized maximum length subsequence # values in bottom up manner for i in range(n): for j in range(i): if (abs(arr[i] - arr[j]) <= 1 and mls[i] < mls[j] + 1): mls[i] = mls[j] + 1 # Store maximum of all 'mls' values in 'max' for i in range(n): if (max < mls[i]): max = mls[i] #required maximum length subsequence return max #Driver program to test above arr = [2, 5, 6, 3, 7, 6, 5, 8] n = len(arr) print("Maximum length subsequence = ",maxLenSub(arr, n)) #This code is contributed by "Abhishek Sharma 44" |
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C#
// C# Code for Maximum length subsequence // with difference between adjacent elements // as either 0 or 1 using System; class GFG { // function to find maximum length subsequence // with difference between adjacent elements as // either 0 or 1 public static int maxLenSub(int[] arr, int n) { int[] mls = new int[n]; int max = 0; // Initialize mls[] values for all indexes for (int i = 0; i < n; i++) mls[i] = 1; // Compute optimized maximum length // subsequence values in bottom up manner for (int i = 1; i < n; i++) for (int j = 0; j < i; j++) if (Math.Abs(arr[i] - arr[j]) <= 1 && mls[i] < mls[j] + 1) mls[i] = mls[j] + 1; // Store maximum of all 'mls' values in 'max' for (int i = 0; i < n; i++) if (max < mls[i]) max = mls[i]; // required maximum length subsequence return max; } /* Driver program to test above function */ public static void Main() { int[] arr = { 2, 5, 6, 3, 7, 6, 5, 8 }; int n = arr.Length; Console.Write("Maximum length subsequence = " + maxLenSub(arr, n)); } } // This code is contributed by Sam007 |
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PHP
<?php // PHP implementation to find maximum length // subsequence with difference between adjacent // elements as either 0 or 1 // function to find maximum length subsequence // with difference between adjacent elements as // either 0 or 1 function maxLenSub($arr, $n) { $mls = array(); $max = 0; // Initialize mls[] values // for all indexes for($i = 0; $i < $n; $i++) $mls[$i] = 1; // Compute optimized maximum // length subsequence // values in bottom up manner for ($i = 1; $i < $n; $i++) for ( $j = 0; $j < $i; $j++) if (abs($arr[$i] - $arr[$j]) <= 1 and $mls[$i] < $mls[$j] + 1) $mls[$i] = $mls[$j] + 1; // Store maximum of all // 'mls' values in 'max' for($i = 0; $i < $n; $i++) if ($max < $mls[$i]) $max = $mls[$i]; // required maximum // length subsequence return $max; } // Driver Code $arr = array(2, 5, 6, 3, 7, 6, 5, 8); $n = count($arr); echo "Maximum length subsequence = " , maxLenSub($arr, $n); // This code is contributed by anuj_67. ?> |
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Output:
Maximum length subsequence = 5
Time Complexity: O(n2)
Auxiliary Space: O(n)
Maximum length subsequence with difference between adjacent elements as either 0 or 1 | Set 2
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Improved By : vt_m
