Collect all coins in minimum number of steps

• Difficulty Level : Hard
• Last Updated : 26 Apr, 2021

Given many stacks of coins which are arranged adjacently. We need to collect all these coins in the minimum number of steps where in one step we can collect one horizontal line of coins or vertical line of coins and collected coins should be continuous.
Examples :

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 can solve this problem using divide and conquer method. We can see that it is always beneficial to remove horizontal lines from below. Suppose we are working on stacks from l index to r index in a recursion step, each time we will choose minimum height, remove those many horizontal lines after which stack will be broken into two parts, l to minimum and minimum +1 till r and we will call recursively in those subarrays. Another thing is we can also collect coins using vertical lines so we will choose minimum between the result of recursive calls and (r – l) because using (r – l) vertical lines we can always collect all coins.
As each time we are calling each subarray and finding minimum of that, total time complexity of the solution will be O(N2

C++

 // 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// collectedint 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[] arrayint minSteps(int height[], int N){    return minStepsRecur(height, 0, N, 0);} // Driver code to test above methodsint 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 stepsimport 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# collecteddef 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[] arraydef minSteps(height, N):    return minStepsRecur(height, 0, N, 0) # Driver codeheight = [ 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 stepsusing 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

 = \$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[] arrayfunction 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



Output:

4

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