Stack permutācijas
Mums ir tukša kaudze, un mēs varam veikt push un pop darbības. Mums ir doti divi masīvi a[] un b[] kur a[] apzīmē secību, kādā elementi tiek izstumti uz kaudzītes, un b[] apzīmē secību, kādā elementi tiek izlikti no kaudzes. Atrodiet, vai dotās push un pop secības ir derīgas.
Piemēri:
Ievade: a[] = [1 2 3] b[] = [2 1 3]
Izvade: taisnība
Paskaidrojums: Nospiediet 1. un 2. Tā kā b[] vispirms ir nepieciešams 2 pop 2, pēc tam — nākošais 1. Visbeidzot nospiediet 3 un izspiediet to. Push un pop secība atbilst a[] un b[].Ievade: a[] = [1 2 3] b[] = [3 1 2]
Izvade: viltus
Paskaidrojums: Pēc 1 2 un 3 nospiešanas mēs varam nospiest 3 pēc vajadzības. Bet nākamais elements b[] ir 1, bet kaudzes augšdaļa ir 2. Tā kā 1 ir bloķēts zem 2, šo secību nevar sasniegt.
Satura rādītājs
- [Naive Approach] Izmantojot rindu - O(n) laiks un O(n) telpa
- [Paredzamā pieeja] Push un Pop simulācija — O(n) laiks un O(n) telpa
[Naive Approach] Izmantojot rindu - O(n) laiks un O(n) telpa
Ideja ir simulēt kaudzes darbības, vienlaikus sekojot līdzi atlikušajiem elementiem, kas jāapstrādā astes .
Mēs virzām elementus no a[] secībā un katram elementam pārbaudām, vai tas atbilst b[] priekšpusei (paredzamā pop secība). Ja tas atbilst, mēs to noņemam no b[]; ja nē, mēs to uzspiežam uz kaudzes. Pēc katra nospiešanas mēs arī pārbaudām kaudzes augšdaļu, vai tā sakrīt ar b[] priekšpusi, mēs izlecam no kaudzes un noņemam to no b[]. To atkārtojot, mēs redzam, vai visus b[] elementus var saskaņot. Ja jā, pop secība ir derīga; citādi tā nav.
C++ #include #include #include #include using namespace std ; bool checkPerm ( vector < int >& a vector < int >& b ) { queue < int > q1 ; for ( int i = 0 ; i < a . size (); i ++ ) q1 . push ( a [ i ]); queue < int > q2 ; for ( int i = 0 ; i < b . size (); i ++ ) q2 . push ( b [ i ]); stack < int > st ; // Dequeue all items one by one while ( ! q1 . empty ()) { int ele = q1 . front (); q1 . pop (); if ( ele == q2 . front ()) { // If matches dequeue from output queue q2 . pop (); // Pop from stack while top matches q2 front while ( ! st . empty () && ! q2 . empty () && st . top () == q2 . front ()) { st . pop (); q2 . pop (); } } else { st . push ( ele ); } } return q2 . empty (); } int main () { vector < int > a = { 1 2 3 }; vector < int > b = { 3 2 1 }; if ( checkPerm ( a b )) cout < < 'true' < < endl ; else cout < < 'false' < < endl ; return 0 ; }
Java import java.util.LinkedList ; import java.util.Queue ; import java.util.Stack ; public class GfG { static boolean checkPerm ( int [] a int [] b ) { Queue < Integer > q1 = new LinkedList <> (); for ( int i = 0 ; i < a . length ; i ++ ) q1 . add ( a [ i ] ); Queue < Integer > q2 = new LinkedList <> (); for ( int i = 0 ; i < b . length ; i ++ ) q2 . add ( b [ i ] ); Stack < Integer > st = new Stack <> (); // Dequeue all items one by one while ( ! q1 . isEmpty ()) { int ele = q1 . poll (); if ( ele == q2 . peek ()) { // If matches dequeue from output queue q2 . poll (); // Pop from stack while top matches q2 front while ( ! st . isEmpty () && ! q2 . isEmpty () && st . peek () == q2 . peek ()) { st . pop (); q2 . poll (); } } else { st . push ( ele ); } } return q2 . isEmpty (); } public static void main ( String [] args ) { int [] a = { 1 2 3 }; int [] b = { 3 2 1 }; if ( checkPerm ( a b )) System . out . println ( 'true' ); else System . out . println ( 'false' ); } }
Python from collections import deque def checkPerm ( a b ): q1 = deque ( a ) q2 = deque ( b ) st = [] # Dequeue all items one by one while q1 : ele = q1 . popleft () if ele == q2 [ 0 ]: # If matches dequeue from output queue q2 . popleft () # Pop from stack while top matches q2 front while st and q2 and st [ - 1 ] == q2 [ 0 ]: st . pop () q2 . popleft () else : st . append ( ele ) return not q2 if __name__ == '__main__' : a = [ 1 2 3 ] b = [ 3 2 1 ] if checkPerm ( a b ): print ( 'true' ) else : print ( 'false' )
C# using System ; using System.Collections.Generic ; public class GfG { static bool checkPerm ( int [] a int [] b ) { Queue < int > q1 = new Queue < int > ( a ); Queue < int > q2 = new Queue < int > ( b ); Stack < int > st = new Stack < int > (); // Dequeue all items one by one while ( q1 . Count > 0 ) { int ele = q1 . Dequeue (); if ( ele == q2 . Peek ()) { // If matches dequeue from output queue q2 . Dequeue (); // Pop from stack while top matches q2 front while ( st . Count > 0 && q2 . Count > 0 && st . Peek () == q2 . Peek ()) { st . Pop (); q2 . Dequeue (); } } else { st . Push ( ele ); } } return q2 . Count == 0 ; } public static void Main () { int [] a = { 1 2 3 }; int [] b = { 3 2 1 }; if ( checkPerm ( a b )) Console . WriteLine ( 'true' ); else Console . WriteLine ( 'false' ); } }
JavaScript function checkPerm ( a b ) { // simulate queue with array let q1 = a ; // simulate queue with array let q2 = b ; let st = []; // pointer for front of q1 let front1 = 0 ; // pointer for front of q2 let front2 = 0 ; while ( front1 < q1 . length ) { let ele = q1 [ front1 ]; front1 ++ ; if ( ele === q2 [ front2 ]) { front2 ++ ; // Pop from stack while top matches q2 front while ( st . length > 0 && st [ st . length - 1 ] === q2 [ front2 ]) { st . pop (); front2 ++ ; } } else { st . push ( ele ); } } return front2 === q2 . length ; } // Driver Code let a = [ 1 2 3 ]; let b = [ 3 2 1 ]; console . log ( checkPerm ( a b ));
Izvade
true
[Paredzamā pieeja] Push un Pop simulācija — O(n) laiks un O(n) telpa
Izmantojot šo pieeju, mēs faktiski neveidojam rindas un nepārveidojam ievades masīvus. Tā vietā mēs tieši simulējam push un pop darbības stekā.
Katrs elements no [] pa vienam tiek uzspiests uz kaudzītes. Pēc katra nospiešanas mēs pārbaudām, vai kaudzes augšdaļa atbilst pašreizējam b[] elementam. Ja tā notiek, mēs to izņemam no steka un virzāmies uz priekšu b[]. Šis process atkārtojas, līdz visi a[] elementi ir nospiesti un pārbaudīti. Ja līdz beigām visi b[] elementi ir veiksmīgi saskaņoti un parādīti, permutācija ir derīga (atgriež true); pretējā gadījumā tas ir nederīgs (atgriež false).
C++ #include #include #include using namespace std ; bool checkPerm ( vector < int >& a vector < int >& b ) { stack < int > st ; int j = 0 ; for ( int i = 0 ; i < a . size (); i ++ ) { // Push top of a[] to stack st . push ( a [ i ]); // Keep popping from stack while it // matches front of the output queue while ( ! st . empty () && st . top () == b [ j ]) { st . pop (); j ++ ; } } return ( j == b . size ()); } int main () { vector < int > a = { 1 2 3 }; vector < int > b = { 2 1 3 }; cout < < ( checkPerm ( a b ) ? 'true' : 'false' ) < < endl ; return 0 ; }
Java import java.util.Stack ; public class GfG { static boolean checkPerm ( int [] a int [] b ) { Stack < Integer > st = new Stack <> (); int j = 0 ; for ( int i = 0 ; i < a . length ; i ++ ) { // Push top of a[] to stack st . push ( a [ i ] ); // Keep popping from stack while it // matches front of the output array while ( ! st . isEmpty () && st . peek (). equals ( b [ j ] )) { st . pop (); j ++ ; } } return ( j == b . length ); } public static void main ( String [] args ) { int [] a = { 1 2 3 }; int [] b = { 2 1 3 }; System . out . println ( checkPerm ( a b ) ? 'true' : 'false' ); } }
Python def checkPerm ( a b ): st = [] j = 0 for i in range ( len ( a )): # Push top of a[] to stack st . append ( a [ i ]) # Keep popping from stack while it # matches front of the output queue while st and st [ - 1 ] == b [ j ]: st . pop () j += 1 return j == len ( b ) if __name__ == '__main__' : a = [ 1 2 3 ] b = [ 2 1 3 ] print ( 'true' if checkPerm ( a b ) else 'false' )
C# using System ; using System.Collections.Generic ; class GfG { static bool checkPerm ( int [] a int [] b ) { Stack < int > stack = new Stack < int > (); int j = 0 ; for ( int i = 0 ; i < a . Length ; i ++ ) { // Push top of a[] to stack stack . Push ( a [ i ]); // Keep popping from stack while it matches b[j] while ( stack . Count > 0 && stack . Peek () == b [ j ]) { stack . Pop (); j ++ ; } } return j == b . Length ; } static void Main () { int [] a = { 1 2 3 }; int [] b = { 2 1 3 }; Console . WriteLine ( checkPerm ( a b ) ? 'true' : 'false' ); } }
JavaScript function checkPerm ( a b ) { const stack = []; let j = 0 ; for ( let i = 0 ; i < a . length ; i ++ ) { // Push top of a[] to stack stack . push ( a [ i ]); // Keep popping from stack while it // matches front of the output queue while ( stack . length > 0 && stack [ stack . length - 1 ] === b [ j ]) { stack . pop (); j ++ ; } } return j === b . length ; } //Driven Code const a = [ 1 2 3 ]; const b = [ 2 1 3 ]; console . log ( checkPerm ( a b ) ? 'true' : 'false' );
Izvade
trueIzveidojiet viktorīnu
