Udskrivning af Maksimal Sum Stigende Efterfølger
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Produktion
Problemet med den maksimale sum-forøgende delsekvens er at finde den maksimale sum-undersekvens af en given sekvens, således at alle elementer i undersekvensen sorteres i stigende rækkefølge.
Eksempler:
Input: [1 101 2 3 100 4 5]
Output: [1 2 3 100]
Input: [3 4 5 10]
Output: [3 4 5 10]
Input: [10 5 4 3]
Output: [10]
Input: [3 2 6 4 5 1]
Output: [3 4 5]I tidligere indlæg har vi diskuteret problemet med den maksimale sum stigende efterfølger. Indlægget dækkede dog kun kode, der var relateret til at finde den maksimale sum af stigende efterfølger, men ikke til konstruktionen af efterfølgen. I dette indlæg vil vi diskutere, hvordan man konstruerer selve Maksimal Sum Stigende Subsequence.
Lad arr[0..n-1] være input-arrayet. Vi definerer vektor L sådan, at L[i] i sig selv er en vektor, der lagrer Maksimal Sum Stigende Subsequence af arr[0..i], der ender med arr[i]. Derfor kan L[i] for indeks i skrives rekursivt som
L[0] = {arr[0]}
L[i] = {MaxSum(L[j])} + arr[i] where j < i and arr[j] < arr[i]
= arr[i] if there is no j such that arr[j] < arr[i]
For eksempel for array [3 2 6 4 5 1]L[0]: 3
L[1]: 2
L[2]: 3 6
L[3]: 3 4
L[4]: 3 4 5
L[5]: 1C++
Nedenfor er implementeringen af ovenstående idé –Java/* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ #include#include using namespace std ; // Utility function to calculate sum of all // vector elements int findSum ( vector < int > arr ) { int sum = 0 ; for ( int i : arr ) sum += i ; return sum ; } // Function to construct Maximum Sum Increasing // Subsequence void printMaxSumIS ( int arr [] int n ) { // L[i] - The Maximum Sum Increasing // Subsequence that ends with arr[i] vector < vector < int > > L ( n ); // L[0] is equal to arr[0] L [ 0 ]. push_back ( arr [ 0 ]); // start from index 1 for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { /* L[i] = {MaxSum(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if (( arr [ i ] > arr [ j ]) && ( findSum ( L [ i ]) < findSum ( L [ j ]))) L [ i ] = L [ j ]; } // L[i] ends with arr[i] L [ i ]. push_back ( arr [ i ]); // L[i] now stores Maximum Sum Increasing // Subsequence of arr[0..i] that ends with // arr[i] } vector < int > res = L [ 0 ]; // find max for ( vector < int > x : L ) if ( findSum ( x ) > findSum ( res )) res = x ; // max will contain result for ( int i : res ) cout < < i < < ' ' ; cout < < endl ; } // Driver Code int main () { int arr [] = { 3 2 6 4 5 1 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); // construct and print Max Sum IS of arr printMaxSumIS ( arr n ); return 0 ; } Python/* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ import java.util.* ; class GFG { // Utility function to calculate sum of all // vector elements static int findSum ( Vector < Integer > arr ) { int sum = 0 ; for ( int i : arr ) sum += i ; return sum ; } // Function to construct Maximum Sum Increasing // Subsequence static void printMaxSumIs ( int [] arr int n ) { // L[i] - The Maximum Sum Increasing // Subsequence that ends with arr[i] @SuppressWarnings ( 'unchecked' ) Vector < Integer >[] L = new Vector [ n ] ; for ( int i = 0 ; i < n ; i ++ ) L [ i ] = new Vector <> (); // L[0] is equal to arr[0] L [ 0 ] . add ( arr [ 0 ] ); // start from index 1 for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { /* * L[i] = {MaxSum(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if (( arr [ i ] > arr [ j ] ) && ( findSum ( L [ i ] ) < findSum ( L [ j ] ))) { for ( int k : L [ j ] ) if ( ! L [ i ] . contains ( k )) L [ i ] . add ( k ); } } // L[i] ends with arr[i] L [ i ] . add ( arr [ i ] ); // L[i] now stores Maximum Sum Increasing // Subsequence of arr[0..i] that ends with // arr[i] } Vector < Integer > res = new Vector <> ( L [ 0 ] ); // res = L[0]; // find max for ( Vector < Integer > x : L ) if ( findSum ( x ) > findSum ( res )) res = x ; // max will contain result for ( int i : res ) System . out . print ( i + ' ' ); System . out . println (); } // Driver Code public static void main ( String [] args ) { int [] arr = { 3 2 6 4 5 1 }; int n = arr . length ; // construct and print Max Sum IS of arr printMaxSumIs ( arr n ); } } // This code is contributed by // sanjeev2552C## Dynamic Programming solution to construct # Maximum Sum Increasing Subsequence */ # Utility function to calculate sum of all # vector elements def findSum ( arr ): summ = 0 for i in arr : summ += i return summ # Function to construct Maximum Sum Increasing # Subsequence def printMaxSumIS ( arr n ): # L[i] - The Maximum Sum Increasing # Subsequence that ends with arr[i] L = [[] for i in range ( n )] # L[0] is equal to arr[0] L [ 0 ] . append ( arr [ 0 ]) # start from index 1 for i in range ( 1 n ): # for every j less than i for j in range ( i ): # L[i] = {MaxSum(L[j])} + arr[i] # where j < i and arr[j] < arr[i] if (( arr [ i ] > arr [ j ]) and ( findSum ( L [ i ]) < findSum ( L [ j ]))): for e in L [ j ]: if e not in L [ i ]: L [ i ] . append ( e ) # L[i] ends with arr[i] L [ i ] . append ( arr [ i ]) # L[i] now stores Maximum Sum Increasing # Subsequence of arr[0..i] that ends with # arr[i] res = L [ 0 ] # find max for x in L : if ( findSum ( x ) > findSum ( res )): res = x # max will contain result for i in res : print ( i end = ' ' ) # Driver Code arr = [ 3 2 6 4 5 1 ] n = len ( arr ) # construct and prMax Sum IS of arr printMaxSumIS ( arr n ) # This code is contributed by Mohit KumarJavaScript/* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ using System ; using System.Collections.Generic ; class GFG { // Utility function to calculate sum of all // vector elements static int findSum ( List < int > arr ) { int sum = 0 ; foreach ( int i in arr ) sum += i ; return sum ; } // Function to construct Maximum Sum Increasing // Subsequence static void printMaxSumIs ( int [] arr int n ) { // L[i] - The Maximum Sum Increasing // Subsequence that ends with arr[i] List < int > [] L = new List < int > [ n ]; for ( int i = 0 ; i < n ; i ++ ) L [ i ] = new List < int > (); // L[0] is equal to arr[0] L [ 0 ]. Add ( arr [ 0 ]); // start from index 1 for ( int i = 1 ; i < n ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { /* * L[i] = {MaxSum(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if (( arr [ i ] > arr [ j ]) && ( findSum ( L [ i ]) < findSum ( L [ j ]))) { foreach ( int k in L [ j ]) if ( ! L [ i ]. Contains ( k )) L [ i ] . Add ( k ); } } // L[i] ends with arr[i] L [ i ]. Add ( arr [ i ]); // L[i] now stores Maximum Sum Increasing // Subsequence of arr[0..i] that ends with // arr[i] } List < int > res = new List < int > ( L [ 0 ]); // res = L[0]; // find max foreach ( List < int > x in L ) if ( findSum ( x ) > findSum ( res )) res = x ; // max will contain result foreach ( int i in res ) Console . Write ( i + ' ' ); Console . WriteLine (); } // Driver Code public static void Main ( String [] args ) { int [] arr = { 3 2 6 4 5 1 }; int n = arr . Length ; // construct and print Max Sum IS of arr printMaxSumIs ( arr n ); } } // This code is contributed by PrinciRaj1992
' ); } // Driver Code let arr = [ 3 2 6 4 5 1 ]; let n = arr . length ; // construct and print Max Sum IS of arr printMaxSumIs ( arr n ); // This code is contributed by unknown2108 < /script>
Produktion3 4 5
Vi kan optimere ovenstående DP-løsning ved at fjerne findSum()-funktionen. I stedet kan vi opretholde en anden vektor/array for at gemme summen af den maksimale sum stigende undersekvens, der ender med arr[i].Tidskompleksitet af ovenstående dynamisk programmeringsløsning er O(n 2 ).
Hjælpeplads brugt af programmet er O(n 2 ).Fremgangsmåde 2: ( Bruger Dynamisk programmering ved hjælp af O(N) mellemrum
Ovenstående tilgang dækkede, hvordan man konstruerer en Maksimal Sum Stigende Subsequence i O(N 2 ) tid og O(N 2 ) plads. I denne tilgang vil vi optimere rumkompleksiteten og konstruere den maksimale sum-forøgende delsekvens i O(N) 2 ) tid og O(N) mellemrum.
- Lad arr[0..n-1] være input-arrayet.
- Vi definerer en vektor af par L, således at L[i] først gemmer den maksimale sum stigende delsekvens af arr[0..i], der slutter med arr[i] og L[i].second gemmer indekset for det foregående element, der blev brugt til at generere summen.
- Da det første element ikke har noget tidligere element, vil dets indeks være -1 i L[0].
F.eks
array = [3 2 6 4 5 1]
L[0]: {3 -1}
L[1]: {2 1}
L[2]: {9 0}
L[3]: {7 0}
L[4]: {12 3}
L[5]: {1 5}Som vi kan se ovenfor er værdien af den Maksimale Sum Stigende Subsequence 12. For at konstruere den faktiske Subsequence vil vi bruge indekset gemt i L[i].second. Trinene til at konstruere undersekvensen er vist nedenfor:
- I et vektorresultat gemmes værdien af det element, hvor den maksimale sum-forøgende delsekvens blev fundet (dvs. ved currIndex = 4). Så i resultatvektoren tilføjer vi arr[currIndex].
- Opdater currIndex til L[currIndex].second og gentag trin 1, indtil currIndex ikke er -1, eller det ikke ændrer sig (dvs. currIndex == previousIndex).
- Vis elementerne i resultatvektoren i omvendt rækkefølge.
Nedenfor er implementeringen af ovenstående idé:
C++14 /* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ #include using namespace std ; // Function to construct and print the Maximum Sum // Increasing Subsequence void constructMaxSumIS ( vector < int > arr int n ) { // L[i] stores the value of Maximum Sum Increasing // Subsequence that ends with arr[i] and the index of // previous element used to construct the Subsequence vector < pair < int int > > L ( n ); int index = 0 ; for ( int i : arr ) { L [ index ] = { i index }; index ++ ; } // Set L[0].second equal to -1 L [ 0 ]. second = -1 ; // start from index 1 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 ] and L [ i ]. first < arr [ i ] + L [ j ]. first ) { L [ i ]. first = arr [ i ] + L [ j ]. first ; L [ i ]. second = j ; } } } int maxi = INT_MIN currIndex track = 0 ; for ( auto p : L ) { if ( p . first > maxi ) { maxi = p . first ; currIndex = track ; } track ++ ; } // Stores the final Subsequence vector < int > result ; // Index of previous element // used to construct the Subsequence int prevoiusIndex ; while ( currIndex >= 0 ) { result . push_back ( arr [ currIndex ]); prevoiusIndex = L [ currIndex ]. second ; if ( currIndex == prevoiusIndex ) break ; currIndex = prevoiusIndex ; } for ( int i = result . size () - 1 ; i >= 0 ; i -- ) cout < < result [ i ] < < ' ' ; } // Driver Code int main () { vector < int > arr = { 1 101 2 3 100 4 5 }; int n = arr . size (); // Function call constructMaxSumIS ( arr n ); return 0 ; }
Java // Dynamic Programming solution to construct // Maximum Sum Increasing Subsequence import java.util.* ; import java.awt.Point ; class GFG { // Function to construct and print the Maximum Sum // Increasing Subsequence static void constructMaxSumIS ( List < Integer > arr int n ) { // L.get(i) stores the value of Maximum Sum Increasing // Subsequence that ends with arr.get(i) and the index of // previous element used to construct the Subsequence List < Point > L = new ArrayList < Point > (); int index = 0 ; for ( int i : arr ) { L . add ( new Point ( i index )); index ++ ; } // Set L[0].second equal to -1 L . set ( 0 new Point ( L . get ( 0 ). x - 1 )); // Start from index 1 for ( int i = 1 ; i < n ; i ++ ) { // For every j less than i for ( int j = 0 ; j < i ; j ++ ) { if ( arr . get ( i ) > arr . get ( j ) && L . get ( i ). x < arr . get ( i ) + L . get ( j ). x ) { L . set ( i new Point ( arr . get ( i ) + L . get ( j ). x j )); } } } int maxi = - 100000000 currIndex = 0 track = 0 ; for ( Point p : L ) { if ( p . x > maxi ) { maxi = p . x ; currIndex = track ; } track ++ ; } // Stores the final Subsequence List < Integer > result = new ArrayList < Integer > (); // Index of previous element // used to construct the Subsequence int prevoiusIndex ; while ( currIndex >= 0 ) { result . add ( arr . get ( currIndex )); prevoiusIndex = L . get ( currIndex ). y ; if ( currIndex == prevoiusIndex ) break ; currIndex = prevoiusIndex ; } for ( int i = result . size () - 1 ; i >= 0 ; i -- ) System . out . print ( result . get ( i ) + ' ' ); } // Driver Code public static void main ( String [] s ) { List < Integer > arr = new ArrayList < Integer > (); arr . add ( 1 ); arr . add ( 101 ); arr . add ( 2 ); arr . add ( 3 ); arr . add ( 100 ); arr . add ( 4 ); arr . add ( 5 ); int n = arr . size (); // Function call constructMaxSumIS ( arr n ); } } // This code is contributed by rutvik_56
Python # Dynamic Programming solution to construct # Maximum Sum Increasing Subsequence import sys # Function to construct and print the Maximum Sum # Increasing Subsequence def constructMaxSumIS ( arr n ) : # L[i] stores the value of Maximum Sum Increasing # Subsequence that ends with arr[i] and the index of # previous element used to construct the Subsequence L = [] index = 0 for i in arr : L . append ([ i index ]) index += 1 # Set L[0].second equal to -1 L [ 0 ][ 1 ] = - 1 # start from index 1 for i in range ( 1 n ) : # for every j less than i for j in range ( i ) : if ( arr [ i ] > arr [ j ] and L [ i ][ 0 ] < arr [ i ] + L [ j ][ 0 ]) : L [ i ][ 0 ] = arr [ i ] + L [ j ][ 0 ] L [ i ][ 1 ] = j maxi currIndex track = - sys . maxsize 0 0 for p in L : if ( p [ 0 ] > maxi ) : maxi = p [ 0 ] currIndex = track track += 1 # Stores the final Subsequence result = [] while ( currIndex >= 0 ) : result . append ( arr [ currIndex ]) prevoiusIndex = L [ currIndex ][ 1 ] if ( currIndex == prevoiusIndex ) : break currIndex = prevoiusIndex for i in range ( len ( result ) - 1 - 1 - 1 ) : print ( result [ i ] end = ' ' ) arr = [ 1 101 2 3 100 4 5 ] n = len ( arr ) # Function call constructMaxSumIS ( arr n ) # This code is contributed by divyeshrabadiya07
C# /* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ using System ; using System.Collections.Generic ; class GFG { // Function to construct and print the Maximum Sum // Increasing Subsequence static void constructMaxSumIS ( List < int > arr int n ) { // L[i] stores the value of Maximum Sum Increasing // Subsequence that ends with arr[i] and the index of // previous element used to construct the Subsequence List < Tuple < int int >> L = new List < Tuple < int int >> (); int index = 0 ; foreach ( int i in arr ) { L . Add ( new Tuple < int int > ( i index )); index ++ ; } // Set L[0].second equal to -1 L [ 0 ] = new Tuple < int int > ( L [ 0 ]. Item1 - 1 ); // start from index 1 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 ] && L [ i ]. Item1 < arr [ i ] + L [ j ]. Item1 ) { L [ i ] = new Tuple < int int > ( arr [ i ] + L [ j ]. Item1 j ); } } } int maxi = Int32 . MinValue currIndex = 0 track = 0 ; foreach ( Tuple < int int > p in L ) { if ( p . Item1 > maxi ) { maxi = p . Item1 ; currIndex = track ; } track ++ ; } // Stores the final Subsequence List < int > result = new List < int > (); // Index of previous element // used to construct the Subsequence int prevoiusIndex ; while ( currIndex >= 0 ) { result . Add ( arr [ currIndex ]); prevoiusIndex = L [ currIndex ]. Item2 ; if ( currIndex == prevoiusIndex ) break ; currIndex = prevoiusIndex ; } for ( int i = result . Count - 1 ; i >= 0 ; i -- ) Console . Write ( result [ i ] + ' ' ); } static void Main () { List < int > arr = new List < int > ( new int [] { 1 101 2 3 100 4 5 }); int n = arr . Count ; // Function call constructMaxSumIS ( arr n ); } } // This code is contributed by divyesh072019
JavaScript < script > // Dynamic Programming solution to construct // Maximum Sum Increasing Subsequence // Function to construct and print the Maximum Sum // Increasing Subsequence function constructMaxSumIS ( arr n ){ // L[i] stores the value of Maximum Sum Increasing // Subsequence that ends with arr[i] and the index of // previous element used to construct the Subsequence let L = [] let index = 0 for ( let i of arr ){ L . push ([ i index ]) index += 1 } // Set L[0].second equal to -1 L [ 0 ][ 1 ] = - 1 // start from index 1 for ( let i = 1 ; i < n ; i ++ ){ // for every j less than i for ( let j = 0 ; j < i ; j ++ ){ if ( arr [ i ] > arr [ j ] && L [ i ][ 0 ] < arr [ i ] + L [ j ][ 0 ]){ L [ i ][ 0 ] = arr [ i ] + L [ j ][ 0 ] L [ i ][ 1 ] = j } } } let maxi = Number . MIN_VALUE currIndex = 0 track = 0 for ( let p of L ){ if ( p [ 0 ] > maxi ){ maxi = p [ 0 ] currIndex = track } track += 1 } // Stores the final Subsequence let result = [] while ( currIndex >= 0 ){ result . push ( arr [ currIndex ]) let prevoiusIndex = L [ currIndex ][ 1 ] if ( currIndex == prevoiusIndex ) break currIndex = prevoiusIndex } for ( let i = result . length - 1 ; i >= 0 ; i -- ) document . write ( result [ i ] ' ' ) } let arr = [ 1 101 2 3 100 4 5 ] let n = arr . length // Function call constructMaxSumIS ( arr n ) // This code is contributed by shinjanpatra < /script>
Produktion
1 2 3 100
Tidskompleksitet: PÅ 2 )
Rumkompleksitet: PÅ)
