Passaggi minimi per raggiungere la fine dell'array sotto vincoli
Dato un array contenente numeri a una cifra, presupponendo che ci troviamo al primo indice, dobbiamo raggiungere la fine dell'array utilizzando un numero minimo di passaggi in cui in un passaggio possiamo saltare agli indici vicini o saltare a una posizione con lo stesso valore.
In altre parole, se siamo all'indice i, in un solo passaggio puoi raggiungere arr[i-1] o arr[i+1] o arr[K] tale che arr[K] = arr[i] (il valore di arr[K] è lo stesso di arr[i])
Esempi:
Input : arr[] = {5 4 2 5 0} Output : 2 Explanation : Total 2 step required. We start from 5(0) in first step jump to next 5 and in second step we move to value 0 (End of arr[]). Input : arr[] = [0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7] Output : 5 Explanation : Total 5 step required. 0(0) -> 0(12) -> 6(11) -> 6(6) -> 7(7) -> (18) (inside parenthesis indices are shown) Questo problema può essere risolto utilizzando BFS . Possiamo considerare l'array dato come un grafico non ponderato in cui ogni vertice ha due bordi per gli elementi dell'array successivo e precedente e più bordi per gli elementi dell'array con gli stessi valori. Ora per la lavorazione veloce del terzo tipo di bordi ne teniamo 10 vettori che memorizzano tutti gli indici in cui sono presenti le cifre da 0 a 9. Nell'esempio precedente il vettore corrispondente a 0 memorizzerà [0 12] 2 indici in cui si è verificato 0 nell'array specificato.
Viene utilizzato un altro array booleano in modo da non visitare lo stesso indice più di una volta. Poiché utilizziamo BFS e i proventi BFS livello per livello, sono garantiti passaggi minimi ottimali.
Attuazione:
C++ // C++ program to find minimum jumps to reach end // of array #include using namespace std ; // Method returns minimum step to reach end of array int getMinStepToReachEnd ( int arr [] int N ) { // visit boolean array checks whether current index // is previously visited bool visit [ N ]; // distance array stores distance of current // index from starting index int distance [ N ]; // digit vector stores indices where a // particular number resides vector < int > digit [ 10 ]; // In starting all index are unvisited memset ( visit false sizeof ( visit )); // storing indices of each number in digit vector for ( int i = 1 ; i < N ; i ++ ) digit [ arr [ i ]]. push_back ( i ); // for starting index distance will be zero distance [ 0 ] = 0 ; visit [ 0 ] = true ; // Creating a queue and inserting index 0. queue < int > q ; q . push ( 0 ); // loop until queue in not empty while ( ! q . empty ()) { // Get an item from queue q. int idx = q . front (); q . pop (); // If we reached to last index break from loop if ( idx == N -1 ) break ; // Find value of dequeued index int d = arr [ idx ]; // looping for all indices with value as d. for ( int i = 0 ; i < digit [ d ]. size (); i ++ ) { int nextidx = digit [ d ][ i ]; if ( ! visit [ nextidx ]) { visit [ nextidx ] = true ; q . push ( nextidx ); // update the distance of this nextidx distance [ nextidx ] = distance [ idx ] + 1 ; } } // clear all indices for digit d because all // of them are processed digit [ d ]. clear (); // checking condition for previous index if ( idx -1 >= 0 && ! visit [ idx - 1 ]) { visit [ idx - 1 ] = true ; q . push ( idx - 1 ); distance [ idx - 1 ] = distance [ idx ] + 1 ; } // checking condition for next index if ( idx + 1 < N && ! visit [ idx + 1 ]) { visit [ idx + 1 ] = true ; q . push ( idx + 1 ); distance [ idx + 1 ] = distance [ idx ] + 1 ; } } // N-1th position has the final result return distance [ N - 1 ]; } // driver code to test above methods int main () { int arr [] = { 0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7 }; int N = sizeof ( arr ) / sizeof ( int ); cout < < getMinStepToReachEnd ( arr N ); return 0 ; }
Java // Java program to find minimum jumps // to reach end of array import java.util.* ; class GFG { // Method returns minimum step // to reach end of array static int getMinStepToReachEnd ( int arr [] int N ) { // visit boolean array checks whether // current index is previously visited boolean [] visit = new boolean [ N ] ; // distance array stores distance of // current index from starting index int [] distance = new int [ N ] ; // digit vector stores indices where a // particular number resides Vector < Integer > [] digit = new Vector [ 10 ] ; for ( int i = 0 ; i < 10 ; i ++ ) digit [ i ] = new Vector <> (); // In starting all index are unvisited for ( int i = 0 ; i < N ; i ++ ) visit [ i ] = false ; // storing indices of each number // in digit vector for ( int i = 1 ; i < N ; i ++ ) digit [ arr [ i ]] . add ( i ); // for starting index distance will be zero distance [ 0 ] = 0 ; visit [ 0 ] = true ; // Creating a queue and inserting index 0. Queue < Integer > q = new LinkedList <> (); q . add ( 0 ); // loop until queue in not empty while ( ! q . isEmpty ()) { // Get an item from queue q. int idx = q . peek (); q . remove (); // If we reached to last // index break from loop if ( idx == N - 1 ) break ; // Find value of dequeued index int d = arr [ idx ] ; // looping for all indices with value as d. for ( int i = 0 ; i < digit [ d ] . size (); i ++ ) { int nextidx = digit [ d ] . get ( i ); if ( ! visit [ nextidx ] ) { visit [ nextidx ] = true ; q . add ( nextidx ); // update the distance of this nextidx distance [ nextidx ] = distance [ idx ] + 1 ; } } // clear all indices for digit d // because all of them are processed digit [ d ] . clear (); // checking condition for previous index if ( idx - 1 >= 0 && ! visit [ idx - 1 ] ) { visit [ idx - 1 ] = true ; q . add ( idx - 1 ); distance [ idx - 1 ] = distance [ idx ] + 1 ; } // checking condition for next index if ( idx + 1 < N && ! visit [ idx + 1 ] ) { visit [ idx + 1 ] = true ; q . add ( idx + 1 ); distance [ idx + 1 ] = distance [ idx ] + 1 ; } } // N-1th position has the final result return distance [ N - 1 ] ; } // Driver Code public static void main ( String [] args ) { int arr [] = { 0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7 }; int N = arr . length ; System . out . println ( getMinStepToReachEnd ( arr N )); } } // This code is contributed by 29AjayKumar
Python3 # Python 3 program to find minimum jumps to reach end# of array # Method returns minimum step to reach end of array def getMinStepToReachEnd ( arr N ): # visit boolean array checks whether current index # is previously visited visit = [ False for i in range ( N )] # distance array stores distance of current # index from starting index distance = [ 0 for i in range ( N )] # digit vector stores indices where a # particular number resides digit = [[ 0 for i in range ( N )] for j in range ( 10 )] # storing indices of each number in digit vector for i in range ( 1 N ): digit [ arr [ i ]] . append ( i ) # for starting index distance will be zero distance [ 0 ] = 0 visit [ 0 ] = True # Creating a queue and inserting index 0. q = [] q . append ( 0 ) # loop until queue in not empty while ( len ( q ) > 0 ): # Get an item from queue q. idx = q [ 0 ] q . remove ( q [ 0 ]) # If we reached to last index break from loop if ( idx == N - 1 ): break # Find value of dequeued index d = arr [ idx ] # looping for all indices with value as d. for i in range ( len ( digit [ d ])): nextidx = digit [ d ][ i ] if ( visit [ nextidx ] == False ): visit [ nextidx ] = True q . append ( nextidx ) # update the distance of this nextidx distance [ nextidx ] = distance [ idx ] + 1 # clear all indices for digit d because all # of them are processed # checking condition for previous index if ( idx - 1 >= 0 and visit [ idx - 1 ] == False ): visit [ idx - 1 ] = True q . append ( idx - 1 ) distance [ idx - 1 ] = distance [ idx ] + 1 # checking condition for next index if ( idx + 1 < N and visit [ idx + 1 ] == False ): visit [ idx + 1 ] = True q . append ( idx + 1 ) distance [ idx + 1 ] = distance [ idx ] + 1 # N-1th position has the final result return distance [ N - 1 ] # driver code to test above methods if __name__ == '__main__' : arr = [ 0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7 ] N = len ( arr ) print ( getMinStepToReachEnd ( arr N )) # This code is contributed by # Surendra_Gangwar
C# // C# program to find minimum jumps // to reach end of array using System ; using System.Collections.Generic ; class GFG { // Method returns minimum step // to reach end of array static int getMinStepToReachEnd ( int [] arr int N ) { // visit boolean array checks whether // current index is previously visited bool [] visit = new bool [ N ]; // distance array stores distance of // current index from starting index int [] distance = new int [ N ]; // digit vector stores indices where a // particular number resides List < int > [] digit = new List < int > [ 10 ]; for ( int i = 0 ; i < 10 ; i ++ ) digit [ i ] = new List < int > (); // In starting all index are unvisited for ( int i = 0 ; i < N ; i ++ ) visit [ i ] = false ; // storing indices of each number // in digit vector for ( int i = 1 ; i < N ; i ++ ) digit [ arr [ i ]]. Add ( i ); // for starting index distance will be zero distance [ 0 ] = 0 ; visit [ 0 ] = true ; // Creating a queue and inserting index 0. Queue < int > q = new Queue < int > (); q . Enqueue ( 0 ); // loop until queue in not empty while ( q . Count != 0 ) { // Get an item from queue q. int idx = q . Peek (); q . Dequeue (); // If we reached to last // index break from loop if ( idx == N - 1 ) break ; // Find value of dequeued index int d = arr [ idx ]; // looping for all indices with value as d. for ( int i = 0 ; i < digit [ d ]. Count ; i ++ ) { int nextidx = digit [ d ][ i ]; if ( ! visit [ nextidx ]) { visit [ nextidx ] = true ; q . Enqueue ( nextidx ); // update the distance of this nextidx distance [ nextidx ] = distance [ idx ] + 1 ; } } // clear all indices for digit d // because all of them are processed digit [ d ]. Clear (); // checking condition for previous index if ( idx - 1 >= 0 && ! visit [ idx - 1 ]) { visit [ idx - 1 ] = true ; q . Enqueue ( idx - 1 ); distance [ idx - 1 ] = distance [ idx ] + 1 ; } // checking condition for next index if ( idx + 1 < N && ! visit [ idx + 1 ]) { visit [ idx + 1 ] = true ; q . Enqueue ( idx + 1 ); distance [ idx + 1 ] = distance [ idx ] + 1 ; } } // N-1th position has the final result return distance [ N - 1 ]; } // Driver Code public static void Main ( String [] args ) { int [] arr = { 0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7 }; int N = arr . Length ; Console . WriteLine ( getMinStepToReachEnd ( arr N )); } } // This code is contributed by PrinciRaj1992
JavaScript < script > // Javascript program to find minimum jumps // to reach end of array // Method returns minimum step // to reach end of array function getMinStepToReachEnd ( arr N ) { // visit boolean array checks whether // current index is previously visited let visit = new Array ( N ); // distance array stores distance of // current index from starting index let distance = new Array ( N ); // digit vector stores indices where a // particular number resides let digit = new Array ( 10 ); for ( let i = 0 ; i < 10 ; i ++ ) digit [ i ] = []; // In starting all index are unvisited for ( let i = 0 ; i < N ; i ++ ) visit [ i ] = false ; // storing indices of each number // in digit vector for ( let i = 1 ; i < N ; i ++ ) digit [ arr [ i ]]. push ( i ); // for starting index distance will be zero distance [ 0 ] = 0 ; visit [ 0 ] = true ; // Creating a queue and inserting index 0. let q = []; q . push ( 0 ); // loop until queue in not empty while ( q . length != 0 ) { // Get an item from queue q. let idx = q . shift (); // If we reached to last // index break from loop if ( idx == N - 1 ) break ; // Find value of dequeued index let d = arr [ idx ]; // looping for all indices with value as d. for ( let i = 0 ; i < digit [ d ]. length ; i ++ ) { let nextidx = digit [ d ][ i ]; if ( ! visit [ nextidx ]) { visit [ nextidx ] = true ; q . push ( nextidx ); // update the distance of this nextidx distance [ nextidx ] = distance [ idx ] + 1 ; } } // clear all indices for digit d // because all of them are processed digit [ d ] = []; // checking condition for previous index if ( idx - 1 >= 0 && ! visit [ idx - 1 ]) { visit [ idx - 1 ] = true ; q . push ( idx - 1 ); distance [ idx - 1 ] = distance [ idx ] + 1 ; } // checking condition for next index if ( idx + 1 < N && ! visit [ idx + 1 ]) { visit [ idx + 1 ] = true ; q . push ( idx + 1 ); distance [ idx + 1 ] = distance [ idx ] + 1 ; } } // N-1th position has the final result return distance [ N - 1 ]; } // Driver Code let arr = [ 0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7 ]; let N = arr . length ; document . write ( getMinStepToReachEnd ( arr N )); // This code is contributed by rag2127 < /script>
Produzione
5
Complessità temporale: O(N) dove N è il numero di elementi dell'array.
Complessità spaziale: O(N) dove N è il numero di elementi dell'array. Stiamo utilizzando un array di distanze e visite di dimensione N e una coda di dimensione N per memorizzare gli indici dell'array.
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