Ņemot vērā daudzas monētu kaudzes, kas ir izvietotas blakus. Mums ir jāsavāc visas šīs monētas minimālā soļu skaitā, lai vienā solī varētu savākt vienu horizontālu monētu līniju vai vertikālu monētu līniju, un savāktajām monētām jābūt nepārtrauktām.
Piemēri:
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.
Java sinhronizācija
Mēs varam atrisināt šo problēmu, izmantojot sadali un valdi metodi. Mēs redzam, ka vienmēr ir izdevīgi noņemt horizontālās līnijas no apakšas. Pieņemsim, ka mēs strādājam ar kaudzēm no l indeksa līdz r indeksam rekursijas solī katru reizi, kad izvēlēsimies minimālo augstumu, noņemiet tās daudzās horizontālās līnijas, pēc kurām steka tiks sadalīta divās daļās no l līdz minimumam un minimumam +1 līdz r, un mēs rekursīvi izsauksim šajos apakšblokos. Vēl viena lieta ir tā, ka mēs varam arī vākt monētas, izmantojot vertikālās līnijas, tāpēc mēs izvēlēsimies minimumu starp rekursīvo zvanu rezultātu un (r - l), jo, izmantojot (r - l) vertikālās līnijas, mēs vienmēr varam savākt visas monētas.
Tā kā katru reizi, kad mēs izsaucam katru apakšgrupu un atrodam šīs risinājuma kopējās laika sarežģītības minimumu, būs O(N2)
C++
// C++ program to find minimum number of // steps to collect stack of coins #include
// 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 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# 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 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. ?> <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(heightlrh) { // 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(heightN) { 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>
Izvade:
4
Laika sarežģītība: Šī algoritma laika sarežģītība ir O(N^2), kur N ir elementu skaits augstuma masīvā.
Telpas sarežģītība: Šī algoritma telpas sarežģītība ir O(N) augstuma masīvā veikto rekursīvo izsaukumu dēļ.