<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> of length <code>n</code> on which you can apply two types of operations:</p>

<ul>
	<li>Choose an index <code>i</code> and invert all characters from&nbsp;index <code>0</code> to index <code>i</code>&nbsp;(both inclusive), with a cost of <code>i + 1</code></li>
	<li>Choose an index <code>i</code> and invert all characters&nbsp;from&nbsp;index <code>i</code> to index <code>n - 1</code>&nbsp;(both inclusive), with a cost of <code>n - i</code></li>
</ul>

<p>Return <em>the <strong>minimum cost </strong>to make all characters of the string <strong>equal</strong></em>.</p>

<p><strong>Invert</strong> a character means&nbsp;if its value is &#39;0&#39; it becomes &#39;1&#39; and vice-versa.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0011&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Apply the second operation with <code>i = 2</code> to obtain <code>s = &quot;0000&quot; for a cost of 2</code>. It can be shown that 2 is the minimum cost to make all characters equal.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;010101&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Apply the first operation with i = 2 to obtain s = &quot;101101&quot; for a cost of 3.
Apply the first operation with i = 1 to obtain s = &quot;011101&quot; for a cost of 2. 
Apply the first operation with i = 0 to obtain s = &quot;111101&quot; for a cost of 1. 
Apply the second operation with i = 4 to obtain s = &quot;111110&quot; for a cost of 2.
Apply the second operation with i = 5 to obtain s = &quot;111111&quot; for a cost of 1. 
The total cost to make all characters equal is 9. It can be shown that 9 is the minimum cost to make all characters equal.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length == n &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code></li>
</ul>

class Solution:
    def minimumCost(self, s: str) -> int:
        n = len(s)

        prefix = [0]
        for i in range(1, n):
            if s[i] == s[i - 1]:
                prefix.append(prefix[-1])
            else:
                prefix.append(prefix[-1] + i)

        suffix = [0 for i in range(n)]
        for i in range(n - 2, -1, -1):
            if s[i] == s[i + 1]:
                suffix[i] = suffix[i + 1]
            else:
                suffix[i] = suffix[i + 1] + (n - 1 - i)

        _min = float('inf')
        for i in range(n):
            _min = min(_min, suffix[i] + prefix[i])
        return _min


# problem:
# given string s
# two operations:
# 1. choose index i and invert all characters from index 0 to index i, with a cost of i + 1, [0, i]
# 2. choose an index i and invert all characters from index i to index n - 1, with a cost of n - i, [i, n - 1]
# return MIN cost to make all characters of string equal

# solution:
# we can do an operation which updates previous items, or next items
# lets split up the dp


# prefix[i] = min number of operations to make [0, i - 1] == s[i]
# base case: prefix[0] = 0
# transition states:
# s[i] == s[i - 1] -> prefix[i] = prefix[i - 1] # because prev string all made equal 
# s[i] != s[i - 1] -> prefix[i] = prefix[i - 1] + (i) # previous string all equal but diff value, flip

# suffix[i] = min number operations to make [i + 1, n - 1] == s[i]
# base case: suffix[n - 1] = 0
# s[i] == s[i + 1] -> suffix[i] = suffix[i + 1] # because front string all made equal 
# s[i] != s[i + 1] -> prefix[i] = suffix[i + 1] + (n - 1 - i) # previous string all equal but diff value, flip

# for each i:
# get total cost to make the rest of the array the same -> prefix[i] + suffix[i]
# find min

26800