Koska monet pinot kolikoita, jotka on järjestetty vierekkäin. Meidän on kerättävä kaikki nämä kolikot mahdollisimman vähissä vaiheissa, jolloin yhdessä vaiheessa voimme kerätä yhden vaakasuoran kolikon rivin tai pystysuoran kolikkorivin ja kerättyjen kolikoiden tulee olla jatkuvia.
Esimerkkejä:
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.
päivitys liittymisestä sql:iin
Voimme ratkaista tämän ongelman hajoa ja hallitse -menetelmällä. Näemme, että on aina hyödyllistä poistaa vaakaviivat alhaalta. Oletetaan, että käsittelemme pinoja l-indeksistä r-indeksiin rekursiovaiheessa joka kerta, kun valitsemme minimikorkeuden, poista ne monet vaakasuorat viivat, minkä jälkeen pino jaetaan kahteen osaan l:stä minimiin ja minimiin +1 r-tilaan ja kutsumme rekursiivisesti näissä aliryhmissä. Toinen asia on se, että voimme kerätä kolikoita myös pystysuorien viivojen avulla, joten valitsemme minimin rekursiivisten kutsujen tuloksen ja (r - l) välillä, koska pystysuorilla viivoilla (r - l) voimme aina kerätä kaikki kolikot.
Koska joka kerta kun kutsumme jokaista aliryhmää ja löydämme ratkaisun kokonaisaikamonimutkaisuuden minimin, on 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>
Lähtö:
4
Aika monimutkaisuus: Tämän algoritmin aikamonimutkaisuus on O(N^2), jossa N on korkeustaulukon elementtien lukumäärä.
Avaruuden monimutkaisuus: Tämän algoritmin avaruuden monimutkaisuus on O(N), koska korkeustaulukossa tehdään rekursiivisia kutsuja.