Stampa della sottosequenza bitonica più lunga
Il problema della sottosequenza bitonica più lunga consiste nel trovare la sottosequenza più lunga di una data sequenza tale che sia prima crescente e poi decrescente. Una sequenza ordinata in ordine crescente è considerata bitonica con la parte decrescente vuota. Allo stesso modo la sequenza di ordine decrescente è considerata bitonica con la parte crescente vuota. Esempi:
Input: [1 11 2 10 4 5 2 1] Output: [1 2 10 4 2 1] OR [1 11 10 5 2 1] OR [1 2 4 5 2 1] Input: [12 11 40 5 3 1] Output: [12 11 5 3 1] OR [12 40 5 3 1] Input: [80 60 30 40 20 10] Output: [80 60 30 20 10] OR [80 60 40 20 10]
In precedente post di cui abbiamo discusso sul problema della sottosequenza bitonica più lunga. Tuttavia il post riguardava solo il codice relativo alla ricerca della somma massima di una sottosequenza crescente ma non alla costruzione della sottosequenza. In questo post discuteremo come costruire la stessa sottosequenza bitonica più lunga. Sia arr[0..n-1] l'array di input. Definiamo il vettore LIS in modo tale che LIS[i] sia esso stesso un vettore che memorizza la sottosequenza crescente più lunga di arr[0..i] che termina con arr[i]. Pertanto per un indice i LIS[i] può essere scritto ricorsivamente come -
LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] where j < i and arr[j] < arr[i] = arr[i] if there is no such j Definiamo anche un vettore LDS tale che LDS[i] è esso stesso un vettore che memorizza la sottosequenza decrescente più lunga di arr[i..n] che inizia con arr[i]. Pertanto per un indice i LDS[i] può essere scritto ricorsivamente come -
LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} where j > i and arr[j] < arr[i] = arr[i] if there is no such j Ad esempio per l'array [1 11 2 10 4 5 2 1]
LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1
LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1
Pertanto la sottosequenza bitonica più lunga può essere
LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1]
Di seguito è riportata l'implementazione dell'idea di cui sopra:
C++ /* Dynamic Programming solution to print Longest Bitonic Subsequence */ #include using namespace std ; // Utility function to print Longest Bitonic // Subsequence void print ( vector < int >& arr int size ) { for ( int i = 0 ; i < size ; i ++ ) cout < < arr [ i ] < < ' ' ; } // Function to construct and print Longest // Bitonic Subsequence void printLBS ( int arr [] int n ) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] vector < vector < int >> LIS ( n ); // initialize LIS[0] to arr[0] LIS [ 0 ]. push_back ( arr [ 0 ]); // Compute LIS values from left to right for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { if (( arr [ j ] < arr [ i ]) && ( LIS [ j ]. size () > LIS [ i ]. size ())) LIS [ i ] = LIS [ j ]; } LIS [ i ]. push_back ( arr [ i ]); } /* LIS[i] now stores Maximum Increasing Subsequence of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] vector < vector < int >> LDS ( n ); // initialize LDS[n-1] to arr[n-1] LDS [ n - 1 ]. push_back ( arr [ n - 1 ]); // Compute LDS values from right to left for ( int i = n - 2 ; i >= 0 ; i -- ) { // for every j greater than i for ( int j = n - 1 ; j > i ; j -- ) { if (( arr [ j ] < arr [ i ]) && ( LDS [ j ]. size () > LDS [ i ]. size ())) LDS [ i ] = LDS [ j ]; } LDS [ i ]. push_back ( arr [ i ]); } // reverse as vector as we're inserting at end for ( int i = 0 ; i < n ; i ++ ) reverse ( LDS [ i ]. begin () LDS [ i ]. end ()); /* LDS[i] now stores Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i] */ int max = 0 ; int maxIndex = -1 ; for ( int i = 0 ; i < n ; i ++ ) { // Find maximum value of size of LIS[i] + size // of LDS[i] - 1 if ( LIS [ i ]. size () + LDS [ i ]. size () - 1 > max ) { max = LIS [ i ]. size () + LDS [ i ]. size () - 1 ; maxIndex = i ; } } // print all but last element of LIS[maxIndex] vector print ( LIS [ maxIndex ] LIS [ maxIndex ]. size () - 1 ); // print all elements of LDS[maxIndex] vector print ( LDS [ maxIndex ] LDS [ maxIndex ]. size ()); } // Driver program int main () { int arr [] = { 1 11 2 10 4 5 2 1 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); printLBS ( arr n ); return 0 ; }
Java /* Dynamic Programming solution to print Longest Bitonic Subsequence */ import java.util.* ; class GFG { // Utility function to print Longest Bitonic // Subsequence static void print ( Vector < Integer > arr int size ) { for ( int i = 0 ; i < size ; i ++ ) System . out . print ( arr . elementAt ( i ) + ' ' ); } // Function to construct and print Longest // Bitonic Subsequence static void printLBS ( int [] arr int n ) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] @SuppressWarnings ( 'unchecked' ) Vector < Integer >[] LIS = new Vector [ n ] ; for ( int i = 0 ; i < n ; i ++ ) LIS [ i ] = new Vector <> (); // initialize LIS[0] to arr[0] LIS [ 0 ] . add ( arr [ 0 ] ); // Compute LIS values from left to right for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { if (( arr [ i ] > arr [ j ] ) && LIS [ j ] . size () > LIS [ i ] . size ()) { for ( int k : LIS [ j ] ) if ( ! LIS [ i ] . contains ( k )) LIS [ i ] . add ( k ); } } LIS [ i ] . add ( arr [ i ] ); } /* * LIS[i] now stores Maximum Increasing Subsequence * of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] @SuppressWarnings ( 'unchecked' ) Vector < Integer >[] LDS = new Vector [ n ] ; for ( int i = 0 ; i < n ; i ++ ) LDS [ i ] = new Vector <> (); // initialize LDS[n-1] to arr[n-1] LDS [ n - 1 ] . add ( arr [ n - 1 ] ); // Compute LDS values from right to left for ( int i = n - 2 ; i >= 0 ; i -- ) { // for every j greater than i for ( int j = n - 1 ; j > i ; j -- ) { if ( arr [ j ] < arr [ i ] && LDS [ j ] . size () > LDS [ i ] . size ()) for ( int k : LDS [ j ] ) if ( ! LDS [ i ] . contains ( k )) LDS [ i ] . add ( k ); } LDS [ i ] . add ( arr [ i ] ); } // reverse as vector as we're inserting at end for ( int i = 0 ; i < n ; i ++ ) Collections . reverse ( LDS [ i ] ); /* * LDS[i] now stores Maximum Decreasing Subsequence * of arr[i..n] that starts with arr[i] */ int max = 0 ; int maxIndex = - 1 ; for ( int i = 0 ; i < n ; i ++ ) { // Find maximum value of size of // LIS[i] + size of LDS[i] - 1 if ( LIS [ i ] . size () + LDS [ i ] . size () - 1 > max ) { max = LIS [ i ] . size () + LDS [ i ] . size () - 1 ; maxIndex = i ; } } // print all but last element of LIS[maxIndex] vector print ( LIS [ maxIndex ] LIS [ maxIndex ] . size () - 1 ); // print all elements of LDS[maxIndex] vector print ( LDS [ maxIndex ] LDS [ maxIndex ] . size ()); } // Driver Code public static void main ( String [] args ) { int [] arr = { 1 11 2 10 4 5 2 1 }; int n = arr . length ; printLBS ( arr n ); } } // This code is contributed by // sanjeev2552
Python3 # Dynamic Programming solution to print Longest # Bitonic Subsequence def _print ( arr : list size : int ): for i in range ( size ): print ( arr [ i ] end = ' ' ) # Function to construct and print Longest # Bitonic Subsequence def printLBS ( arr : list n : int ): # LIS[i] stores the length of the longest # increasing subsequence ending with arr[i] LIS = [ 0 ] * n for i in range ( n ): LIS [ i ] = [] # initialize LIS[0] to arr[0] LIS [ 0 ] . append ( arr [ 0 ]) # Compute LIS values from left to right for i in range ( 1 n ): # for every j less than i for j in range ( i ): if (( arr [ j ] < arr [ i ]) and ( len ( LIS [ j ]) > len ( LIS [ i ]))): LIS [ i ] = LIS [ j ] . copy () LIS [ i ] . append ( arr [ i ]) # LIS[i] now stores Maximum Increasing # Subsequence of arr[0..i] that ends with # arr[i] # LDS[i] stores the length of the longest # decreasing subsequence starting with arr[i] LDS = [ 0 ] * n for i in range ( n ): LDS [ i ] = [] # initialize LDS[n-1] to arr[n-1] LDS [ n - 1 ] . append ( arr [ n - 1 ]) # Compute LDS values from right to left for i in range ( n - 2 - 1 - 1 ): # for every j greater than i for j in range ( n - 1 i - 1 ): if (( arr [ j ] < arr [ i ]) and ( len ( LDS [ j ]) > len ( LDS [ i ]))): LDS [ i ] = LDS [ j ] . copy () LDS [ i ] . append ( arr [ i ]) # reverse as vector as we're inserting at end for i in range ( n ): LDS [ i ] = list ( reversed ( LDS [ i ])) # LDS[i] now stores Maximum Decreasing Subsequence # of arr[i..n] that starts with arr[i] max = 0 maxIndex = - 1 for i in range ( n ): # Find maximum value of size of LIS[i] + size # of LDS[i] - 1 if ( len ( LIS [ i ]) + len ( LDS [ i ]) - 1 > max ): max = len ( LIS [ i ]) + len ( LDS [ i ]) - 1 maxIndex = i # print all but last element of LIS[maxIndex] vector _print ( LIS [ maxIndex ] len ( LIS [ maxIndex ]) - 1 ) # print all elements of LDS[maxIndex] vector _print ( LDS [ maxIndex ] len ( LDS [ maxIndex ])) # Driver Code if __name__ == '__main__' : arr = [ 1 11 2 10 4 5 2 1 ] n = len ( arr ) printLBS ( arr n ) # This code is contributed by # sanjeev2552
C# /* Dynamic Programming solution to print longest Bitonic Subsequence */ using System ; using System.Linq ; using System.Collections.Generic ; class GFG { // Utility function to print longest Bitonic // Subsequence static void print ( List < int > arr int size ) { for ( int i = 0 ; i < size ; i ++ ) Console . Write ( arr [ i ] + ' ' ); } // Function to construct and print longest // Bitonic Subsequence static void printLBS ( int [] arr int n ) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] List < int > [] LIS = new List < int > [ n ]; for ( int i = 0 ; i < n ; i ++ ) LIS [ i ] = new List < int > (); // initialize LIS[0] to arr[0] LIS [ 0 ]. Add ( arr [ 0 ]); // Compute LIS values from left to right for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { if (( arr [ i ] > arr [ j ]) && LIS [ j ]. Count > LIS [ i ]. Count ) { foreach ( int k in LIS [ j ]) if ( ! LIS [ i ]. Contains ( k )) LIS [ i ]. Add ( k ); } } LIS [ i ]. Add ( arr [ i ]); } /* * LIS[i] now stores Maximum Increasing Subsequence * of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] List < int > [] LDS = new List < int > [ n ]; for ( int i = 0 ; i < n ; i ++ ) LDS [ i ] = new List < int > (); // initialize LDS[n-1] to arr[n-1] LDS [ n - 1 ]. Add ( arr [ n - 1 ]); // Compute LDS values from right to left for ( int i = n - 2 ; i >= 0 ; i -- ) { // for every j greater than i for ( int j = n - 1 ; j > i ; j -- ) { if ( arr [ j ] < arr [ i ] && LDS [ j ]. Count > LDS [ i ]. Count ) foreach ( int k in LDS [ j ]) if ( ! LDS [ i ]. Contains ( k )) LDS [ i ]. Add ( k ); } LDS [ i ]. Add ( arr [ i ]); } // reverse as vector as we're inserting at end for ( int i = 0 ; i < n ; i ++ ) LDS [ i ]. Reverse (); /* * LDS[i] now stores Maximum Decreasing Subsequence * of arr[i..n] that starts with arr[i] */ int max = 0 ; int maxIndex = - 1 ; for ( int i = 0 ; i < n ; i ++ ) { // Find maximum value of size of // LIS[i] + size of LDS[i] - 1 if ( LIS [ i ]. Count + LDS [ i ]. Count - 1 > max ) { max = LIS [ i ]. Count + LDS [ i ]. Count - 1 ; maxIndex = i ; } } // print all but last element of LIS[maxIndex] vector print ( LIS [ maxIndex ] LIS [ maxIndex ]. Count - 1 ); // print all elements of LDS[maxIndex] vector print ( LDS [ maxIndex ] LDS [ maxIndex ]. Count ); } // Driver Code public static void Main ( String [] args ) { int [] arr = { 1 11 2 10 4 5 2 1 }; int n = arr . Length ; printLBS ( arr n ); } } // This code is contributed by PrinciRaj1992
JavaScript // Function to print the longest bitonic subsequence function _print ( arr size ) { for ( let i = 0 ; i < size ; i ++ ) { process . stdout . write ( arr [ i ] + ' ' ); } } // Function to construct and print the longest bitonic subsequence function printLBS ( arr n ) { // LIS[i] stores the length of the longest increasing subsequence ending with arr[i] let LIS = new Array ( n ); for ( let i = 0 ; i < n ; i ++ ) { LIS [ i ] = []; } // initialize LIS[0] to arr[0] LIS [ 0 ]. push ( arr [ 0 ]); // Compute LIS values from left to right for ( let i = 1 ; i < n ; i ++ ) { // for every j less than i for ( let j = 0 ; j < i ; j ++ ) { if ( arr [ j ] < arr [ i ] && LIS [ j ]. length > LIS [ i ]. length ) { LIS [ i ] = LIS [ j ]. slice (); } } LIS [ i ]. push ( arr [ i ]); } // LIS[i] now stores the Maximum Increasing Subsequence of arr[0..i] that ends with arr[i] // LDS[i] stores the length of the longest decreasing subsequence starting with arr[i] let LDS = new Array ( n ); for ( let i = 0 ; i < n ; i ++ ) { LDS [ i ] = []; } // initialize LDS[n-1] to arr[n-1] LDS [ n - 1 ]. push ( arr [ n - 1 ]); // Compute LDS values from right to left for ( let i = n - 2 ; i >= 0 ; i -- ) { // for every j greater than i for ( let j = n - 1 ; j > i ; j -- ) { if ( arr [ j ] < arr [ i ] && LDS [ j ]. length > LDS [ i ]. length ) { LDS [ i ] = LDS [ j ]. slice (); } } LDS [ i ]. push ( arr [ i ]); } // reverse the LDS vector as we're inserting at the end for ( let i = 0 ; i < n ; i ++ ) { LDS [ i ]. reverse (); } // LDS[i] now stores the Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i] let max = 0 ; let maxIndex = - 1 ; for ( let i = 0 ; i < n ; i ++ ) { // Find maximum value of size of LIS[i] + size of LDS[i] - 1 if ( LIS [ i ]. length + LDS [ i ]. length - 1 > max ) { max = LIS [ i ]. length + LDS [ i ]. length - 1 ; maxIndex = i ; } } // print all but // print all but last element of LIS[maxIndex] array _print ( LIS [ maxIndex ]. slice ( 0 - 1 ) LIS [ maxIndex ]. length - 1 ); // print all elements of LDS[maxIndex] array _print ( LDS [ maxIndex ] LDS [ maxIndex ]. length ); } // Driver program const arr = [ 1 11 2 10 4 5 2 1 ]; const n = arr . length ; printLBS ( arr n );
Produzione:
1 11 10 5 2 1
Complessità temporale della soluzione di Programmazione Dinamica di cui sopra è O(n 2 ). Spazio ausiliario utilizzato dal programma è O(n 2 ).
