Given an array of n size, the task is to find the longest subsequence such that difference between adjacents is one.
Examples:
Input : arr[] = {10, 9, 4, 5, 4, 8, 6} Output : 3 As longest subsequences with difference 1 are, "10, 9, 8", "4, 5, 4" and "4, 5, 6" Input : arr[] = {1, 2, 3, 2, 3, 7, 2, 1} Output : 7 As longest consecutive sequence is "1, 2, 3, 2, 3, 2, 1"
This problem is based upon the concept of Longest Increasing Subsequence Problem.
Let arr[0..n-1] be the input array and dp[i] be the length of the longest subsequence (with differences one) ending at index i such that arr[i] is the last element of the subsequence. Then, dp[i] can be recursively written as: dp[i] = 1 + max(dp[j]) where 0 < j < i and [arr[j] = arr[i] -1 or arr[j] = arr[i] + 1] dp[i] = 1, if no such j exists. To find the result for a given array, we need to return max(dp[i]) where 0 < i < n.
Following is a Dynamic Programming based implementation. It follows the recursive structure discussed above.
C++
// C++ program to find the longest subsequence such // the difference between adjacent elements of the // subsequence is one. #include<bits/stdc++.h> using namespace std; // Function to find the length of longest subsequence int longestSubseqWithDiffOne( int arr[], int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int dp[n]; for ( int i = 0; i< n; i++) dp[i] = 1; // Start traversing the given array for ( int i=1; i<n; i++) { // Compare with all the previous elements for ( int j=0; j<i; j++) { // If the element is consecutive then // consider this subsequence and update // dp[i] if required. if ((arr[i] == arr[j]+1) || (arr[i] == arr[j]-1)) dp[i] = max(dp[i], dp[j]+1); } } // Longest length will be the maximum value // of dp array. int result = 1; for ( int i = 0 ; i < n ; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code int main() { // Longest subsequence with one difference is // {1, 2, 3, 4, 3, 2} int arr[] = {1, 2, 3, 4, 5, 3, 2}; int n = sizeof (arr)/ sizeof (arr[0]); cout << longestSubseqWithDiffOne(arr, n); return 0; } |
Java
// Java program to find the longest subsequence // such that the difference between adjacent // elements of the subsequence is one. import java.io.*; class GFG { // Function to find the length of longest // subsequence static int longestSubseqWithDiffOne( int arr[], int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int dp[] = new int [n]; for ( int i = 0 ; i< n; i++) dp[i] = 1 ; // Start traversing the given array for ( int i = 1 ; i < n; i++) { // Compare with all the previous // elements for ( int j = 0 ; j < i; j++) { // If the element is consecutive // then consider this subsequence // and update dp[i] if required. if ((arr[i] == arr[j] + 1 ) || (arr[i] == arr[j] - 1 )) dp[i] = Math.max(dp[i], dp[j]+ 1 ); } } // Longest length will be the maximum // value of dp array. int result = 1 ; for ( int i = 0 ; i < n ; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code public static void main(String[] args) { // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} int arr[] = { 1 , 2 , 3 , 4 , 5 , 3 , 2 }; int n = arr.length; System.out.println(longestSubseqWithDiffOne( arr, n)); } } // This code is contributed by Prerna Saini |
Python
# Function to find the length of longest subsequence def longestSubseqWithDiffOne(arr, n): # Initialize the dp[] array with 1 as a # single element will be of 1 length dp = [ 1 for i in range (n)] # Start traversing the given array for i in range (n): # Compare with all the previous elements for j in range (i): # If the element is consecutive then # consider this subsequence and update # dp[i] if required. if ((arr[i] = = arr[j] + 1 ) or (arr[i] = = arr[j] - 1 )): dp[i] = max (dp[i], dp[j] + 1 ) # Longest length will be the maximum value # of dp array. result = 1 for i in range (n): if (result < dp[i]): result = dp[i] return result # Driver code arr = [ 1 , 2 , 3 , 4 , 5 , 3 , 2 ] # Longest subsequence with one difference is # {1, 2, 3, 4, 3, 2} n = len (arr) print longestSubseqWithDiffOne(arr, n) # This code is contributed by Afzal Ansari |
C#
// C# program to find the longest subsequence // such that the difference between adjacent // elements of the subsequence is one. using System; class GFG { // Function to find the length of longest // subsequence static int longestSubseqWithDiffOne( int []arr, int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int []dp = new int [n]; for ( int i = 0; i< n; i++) dp[i] = 1; // Start traversing the given array for ( int i = 1; i < n; i++) { // Compare with all the previous // elements for ( int j = 0; j < i; j++) { // If the element is consecutive // then consider this subsequence // and update dp[i] if required. if ((arr[i] == arr[j] + 1) || (arr[i] == arr[j] - 1)) dp[i] = Math.Max(dp[i], dp[j]+1); } } // Longest length will be the maximum // value of dp array. int result = 1; for ( int i = 0 ; i < n ; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code public static void Main() { // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} int []arr = {1, 2, 3, 4, 5, 3, 2}; int n = arr.Length; Console.Write( longestSubseqWithDiffOne(arr, n)); } } // This code is contributed by nitin mittal. |
PHP
<?php // PHP program to find the longest // subsequence such the difference // between adjacent elements of the // subsequence is one. // Function to find the length of // longest subsequence function longestSubseqWithDiffOne( $arr , $n ) { // Initialize the dp[] // array with 1 as a // single element will // be of 1 length $dp [ $n ] = 0; for ( $i = 0; $i < $n ; $i ++) $dp [ $i ] = 1; // Start traversing the // given array for ( $i = 1; $i < $n ; $i ++) { // Compare with all the // previous elements for ( $j = 0; $j < $i ; $j ++) { // If the element is // consecutive then // consider this // subsequence and // update dp[i] if // required. if (( $arr [ $i ] == $arr [ $j ] + 1) || ( $arr [ $i ] == $arr [ $j ] - 1)) $dp [ $i ] = max( $dp [ $i ], $dp [ $j ] + 1); } } // Longest length will be // the maximum value // of dp array. $result = 1; for ( $i = 0 ; $i < $n ; $i ++) if ( $result < $dp [ $i ]) $result = $dp [ $i ]; return $result ; } // Driver code // Longest subsequence with // one difference is // {1, 2, 3, 4, 3, 2} $arr = array (1, 2, 3, 4, 5, 3, 2); $n = sizeof( $arr ); echo longestSubseqWithDiffOne( $arr , $n ); // This code is contributed by nitin mittal. ?> |
Output:
6
Time Complexity: O(n2)
Auxiliary Space: O(n)
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