# Collect all coins in minimum number of steps

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++ program to find minimum number of`
`// steps to collect stack of coins`
`#include <bits/stdc++.h>`
`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;`
`}`

Output:

```4

```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/collect-coins-minimum-number-steps/
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