Impression de la sous-séquence bitonique la plus longue
Le problème de la sous-séquence bitonique la plus longue consiste à trouver la sous-séquence la plus longue d’une séquence donnée telle qu’elle soit d’abord croissante puis décroissante. Une séquence triée par ordre croissant est considérée comme Bitonic avec la partie décroissante comme vide. De la même manière, une séquence d'ordre décroissant est considérée comme Bitonic avec la partie croissante comme vide. Exemples :
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]
Dans précédent article dont nous avons discuté sur le problème de la sous-séquence bitonique la plus longue. Cependant, le message ne couvrait que le code lié à la recherche de la somme maximale d'une sous-séquence croissante, mais pas à la construction d'une sous-séquence. Dans cet article, nous verrons comment construire la sous-séquence bitonique la plus longue elle-même. Soit arr[0..n-1] le tableau d'entrée. Nous définissons le vecteur LIS tel que LIS[i] est lui-même un vecteur qui stocke la plus longue sous-séquence croissante de arr[0..i] qui se termine par arr[i]. Par conséquent, pour un index i, LIS[i] peut être écrit récursivement sous la forme -
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 Nous définissons également un vecteur LDS tel que LDS[i] est lui-même un vecteur qui stocke la plus longue sous-séquence décroissante de arr[i..n] qui commence par arr[i]. Par conséquent, pour un index i, LDS[i] peut être écrit récursivement sous la forme -
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 Par exemple pour le tableau [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
Par conséquent, la sous-séquence bitonique la plus longue peut être
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]
Vous trouverez ci-dessous la mise en œuvre de l’idée ci-dessus –
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 );
Sortir:
1 11 10 5 2 1
Complexité temporelle de la solution de programmation dynamique ci-dessus est O (n 2 ). Espace auxiliaire utilisé par le programme est O(n 2 ).
