Givet mange stakke af mønter, som er arrangeret ved siden af. Vi er nødt til at samle alle disse mønter i det mindste antal trin, hvor vi i et trin kan samle en vandret linje af mønter eller lodret linje af mønter, og indsamlede mønter skal være kontinuerlige.
Eksempler:
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.
btree og b tree
Vi kan løse dette problem ved at bruge del og hersk metoden. Vi kan se, at det altid er en fordel at fjerne vandrette linjer nedefra. Antag, at vi arbejder på stakke fra l-indeks til r-indeks i et rekursionstrin, hver gang vi vælger minimumshøjde, fjern de mange vandrette linjer, hvorefter stakken vil blive opdelt i to dele l til minimum og minimum +1 til r, og vi kalder rekursivt i disse underarrays. En anden ting er, at vi også kan samle mønter ved hjælp af lodrette linjer, så vi vil vælge minimum mellem resultatet af rekursive kald og (r - l), fordi vi ved at bruge (r - l) lodrette linjer altid kan samle alle mønter.
Da hver gang vi kalder hver undergruppe og finder minimum af den samlede tidskompleksitet af løsningen vil være 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>
Produktion:
4
Tidskompleksitet: Tidskompleksiteten af denne algoritme er O(N^2), hvor N er antallet af elementer i højdearrayet.
Rumkompleksitet: Rumkompleksiteten af denne algoritme er O(N) på grund af de rekursive kald, der foretages på højdearrayet.