<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:
    """
Consider a dp array, where dp[i] denotes the minimum cost to make the first i characters in the string equal.

We can set dp[0] = 0 as base case.

Now, if we encounter an index i such that s[i] != s[i-1] then we can either flip the prefix with cost i or the suffix with cost n-i. We choose the minimum of these two.

We have essentially made the first i+1 characters equal, while maintaining the cost required to make all characters of the remaining suffix equal.

We thus have the relation,
dp[i] = dp[i-1] + min(i, n-i) if s[i] != s[i-1]
dp[i] = dp[i-1] otherwise
    """
    
    """
    "010101"
     [0, 1, 0, 0, 0, 0] 11
    [0, 1, 3, 0, 0, 0]  000
    [0, 1, 3, 6, 0, 0]   1111 or 000[000] both works
    [0, 1, 3, 6, 8, 0]   cheaper to flip back 111[11]
    [0, 1, 3, 6, 8, 9]    flip back
    
    """
    def minimumCost(self, s: str) -> int:
        #let dp[0][i] is the cost to 
        n = len(s)
        dp = [0]*n #dp[i] is the minimum cost to make all char equal at i 
        for i in range(1, n):
            dp[i] = dp[i-1]
            if s[i] != s[i-1]:
                dp[i] += min(i, n-i) # cost of i as to flip the part before i
        return dp[n-1]
167