Recull totes les monedes en un nombre mínim de passos
Donades moltes piles de monedes que estan disposades adjacents. Hem de recollir totes aquestes monedes en el nombre mínim de passos on en un pas podem recollir una línia horitzontal de monedes o una línia vertical de monedes i les monedes recollides han de ser contínues.
Exemples:
Input : height[] = [2 1 2 5 1] Each value of this array corresponds to the height of stack that is we are given five stack of coins where in first stack 2 coins are there then in second stack 1 coin is there and so on. Output : 4 We can collect all above coins in 4 steps which are shown in below diagram. Each step is shown by different color. First we have collected last horizontal line of coins after which stacks remains as [1 0 1 4 0] after that another horizontal line of coins is collected from stack 3 and 4 then a vertical line from stack 4 and at the end a horizontal line from stack 1. Total steps are 4.
Podem resoldre aquest problema utilitzant el mètode de dividir i conquerir. Podem veure que sempre és beneficiós eliminar les línies horitzontals per sota. Suposem que estem treballant en piles des de l'índex l fins a l'índex r en un pas de recursivitat cada vegada que escollim l'alçada mínima, eliminem aquestes moltes línies horitzontals després de les quals la pila es dividirà en dues parts l fins al mínim i el mínim +1 fins a r i trucarem recursivament en aquests subbarrays. Una altra cosa és que també podem recollir monedes mitjançant línies verticals, de manera que triarem el mínim entre el resultat de les trucades recursives i (r - l) perquè utilitzant (r - l) les línies verticals sempre podem recollir totes les monedes.
Com cada vegada que anomenem cada subbarray i trobar el mínim d'aquest temps total, la complexitat de la solució serà O(N 2 )
// C++ program to find minimum number of // steps to collect stack of coins #include using namespace std ; // recursive method to collect coins from // height array l to r with height h already // collected int minStepsRecur ( int height [] int l int r int h ) { // if l is more than r no steps needed if ( l >= r ) return 0 ; // loop over heights to get minimum height // index int m = l ; for ( int i = l ; i < r ; i ++ ) if ( height [ i ] < height [ m ]) m = i ; /* choose minimum from 1) collecting coins using all vertical lines (total r - l) 2) collecting coins using lower horizontal lines and recursively on left and right segments */ return min ( r - l minStepsRecur ( height l m height [ m ]) + minStepsRecur ( height m + 1 r height [ m ]) + height [ m ] - h ); } // method returns minimum number of step to // collect coin from stack with height in // height[] array int minSteps ( int height [] int N ) { return minStepsRecur ( height 0 N 0 ); } // Driver code to test above methods int main () { int height [] = { 2 1 2 5 1 }; int N = sizeof ( height ) / sizeof ( int ); cout < < minSteps ( height N ) < < endl ; return 0 ; }
Java // Java Code to Collect all coins in // minimum number of steps import java.util.* ; class GFG { // recursive method to collect coins from // height array l to r with height h already // collected public static int minStepsRecur ( int height [] int l int r int h ) { // if l is more than r no steps needed if ( l >= r ) return 0 ; // loop over heights to get minimum height // index int m = l ; for ( int i = l ; i < r ; i ++ ) if ( height [ i ] < height [ m ] ) m = i ; /* choose minimum from 1) collecting coins using all vertical lines (total r - l) 2) collecting coins using lower horizontal lines and recursively on left and right segments */ return Math . min ( r - l minStepsRecur ( height l m height [ m ] ) + minStepsRecur ( height m + 1 r height [ m ] ) + height [ m ] - h ); } // method returns minimum number of step to // collect coin from stack with height in // height[] array public static int minSteps ( int height [] int N ) { return minStepsRecur ( height 0 N 0 ); } /* Driver program to test above function */ public static void main ( String [] args ) { int height [] = { 2 1 2 5 1 }; int N = height . length ; System . out . println ( minSteps ( height N )); } } // This code is contributed by Arnav Kr. Mandal.
Python 3 # Python 3 program to find # minimum number of steps # to collect stack of coins # recursive method to collect # coins from height array l to # r with height h already # collected def minStepsRecur ( height l r h ): # if l is more than r # no steps needed if l >= r : return 0 ; # loop over heights to # get minimum height index m = l for i in range ( l r ): if height [ i ] < height [ m ]: m = i # choose minimum from # 1) collecting coins using # all vertical lines (total r - l) # 2) collecting coins using # lower horizontal lines and # recursively on left and # right segments return min ( r - l minStepsRecur ( height l m height [ m ]) + minStepsRecur ( height m + 1 r height [ m ]) + height [ m ] - h ) # method returns minimum number # of step to collect coin from # stack with height in height[] array def minSteps ( height N ): return minStepsRecur ( height 0 N 0 ) # Driver code height = [ 2 1 2 5 1 ] N = len ( height ) print ( minSteps ( height N )) # This code is contributed # by ChitraNayal
C# // C# Code to Collect all coins in // minimum number of steps using System ; class GFG { // recursive method to collect coins from // height array l to r with height h already // collected public static int minStepsRecur ( int [] height int l int r int h ) { // if l is more than r no steps needed if ( l >= r ) return 0 ; // loop over heights to // get minimum height index int m = l ; for ( int i = l ; i < r ; i ++ ) if ( height [ i ] < height [ m ]) m = i ; /* choose minimum from 1) collecting coins using all vertical lines (total r - l) 2) collecting coins using lower horizontal lines and recursively on left and right segments */ return Math . Min ( r - l minStepsRecur ( height l m height [ m ]) + minStepsRecur ( height m + 1 r height [ m ]) + height [ m ] - h ); } // method returns minimum number of step to // collect coin from stack with height in // height[] array public static int minSteps ( int [] height int N ) { return minStepsRecur ( height 0 N 0 ); } /* Driver program to test above function */ public static void Main () { int [] height = { 2 1 2 5 1 }; int N = height . Length ; Console . Write ( minSteps ( height N )); } } // This code is contributed by nitin mittal
PHP // PHP program to find minimum number of // steps to collect stack of coins // recursive method to collect // coins from height array l to // r with height h already // collected function minStepsRecur ( $height $l $r $h ) { // if l is more than r // no steps needed if ( $l >= $r ) return 0 ; // loop over heights to // get minimum height // index $m = $l ; for ( $i = $l ; $i < $r ; $i ++ ) if ( $height [ $i ] < $height [ $m ]) $m = $i ; /* choose minimum from 1) collecting coins using all vertical lines (total r - l) 2) collecting coins using lower horizontal lines and recursively on left and right segments */ return min ( $r - $l minStepsRecur ( $height $l $m $height [ $m ]) + minStepsRecur ( $height $m + 1 $r $height [ $m ]) + $height [ $m ] - $h ); } // method returns minimum number of step to // collect coin from stack with height in // height[] array function minSteps ( $height $N ) { return minStepsRecur ( $height 0 $N 0 ); } // Driver Code $height = array ( 2 1 2 5 1 ); $N = sizeof ( $height ); echo minSteps ( $height $N ) ; // This code is contributed by nitin mittal. ?>
JavaScript < script > // Javascript Code to Collect all coins in // minimum number of steps // recursive method to collect coins from // height array l to r with height h already // collected function minStepsRecur ( height l r h ) { // if l is more than r no steps needed if ( l >= r ) return 0 ; // loop over heights to get minimum height // index let m = l ; for ( let i = l ; i < r ; i ++ ) if ( height [ i ] < height [ m ]) m = i ; /* choose minimum from 1) collecting coins using all vertical lines (total r - l) 2) collecting coins using lower horizontal lines and recursively on left and right segments */ return Math . min ( r - l minStepsRecur ( height l m height [ m ]) + minStepsRecur ( height m + 1 r height [ m ]) + height [ m ] - h ); } // method returns minimum number of step to // collect coin from stack with height in // height[] array function minSteps ( height N ) { return minStepsRecur ( height 0 N 0 ); } /* Driver program to test above function */ let height = [ 2 1 2 5 1 ]; let N = height . length ; document . write ( minSteps ( height N )); // This code is contributed by avanitrachhadiya2155 < /script>
Sortida:
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Complexitat temporal: La complexitat temporal d'aquest algorisme és O(N^2) on N és el nombre d'elements de la matriu d'alçada.
Complexitat espacial: La complexitat espacial d'aquest algorisme és O(N) a causa de les trucades recursives que es fan a la matriu d'alçada.
