Verzamel alle munten in een minimaal aantal stappen
Gegeven veel stapels munten die naast elkaar zijn gerangschikt. We moeten al deze munten verzamelen in het minimale aantal stappen waarbij we in één stap één horizontale lijn munten of een verticale lijn munten kunnen verzamelen en de verzamelde munten continu moeten zijn.
Voorbeelden:
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.
We kunnen dit probleem oplossen met behulp van de verdeel-en-heers-methode. We kunnen zien dat het altijd nuttig is om horizontale lijnen van onderaf te verwijderen. Stel dat we werken aan stapels van l index tot r index in een recursiestap, elke keer dat we de minimale hoogte kiezen, verwijderen we die vele horizontale lijnen, waarna de stapel in twee delen wordt opgesplitst, l tot minimum en minimum +1 tot r, en we zullen recursief aanroepen in die subarrays. Een ander ding is dat we ook munten kunnen verzamelen met behulp van verticale lijnen, dus we zullen een minimum kiezen tussen het resultaat van recursieve oproepen en (r - l), omdat we met behulp van (r - l) verticale lijnen altijd alle munten kunnen verzamelen.
Elke keer dat we elke subarray aanroepen en het minimum van die totale tijdscomplexiteit van de oplossing vinden, zal 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>
Uitgang:
4
Tijdcomplexiteit: De tijdscomplexiteit van dit algoritme is O(N^2), waarbij N het aantal elementen in de hoogtearray is.
Ruimtecomplexiteit: De ruimtecomplexiteit van dit algoritme is O(N) vanwege de recursieve aanroepen die op de hoogtearray worden gedaan.
