<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/leetcode_keyboard.png" style="width: 349px; height: 209px;" />
<p>You have a keyboard layout as shown above in the <strong>X-Y</strong> plane, where each English uppercase letter is located at some coordinate.</p>

<ul>
	<li>For example, the letter <code>&#39;A&#39;</code> is located at coordinate <code>(0, 0)</code>, the letter <code>&#39;B&#39;</code> is located at coordinate <code>(0, 1)</code>, the letter <code>&#39;P&#39;</code> is located at coordinate <code>(2, 3)</code> and the letter <code>&#39;Z&#39;</code> is located at coordinate <code>(4, 1)</code>.</li>
</ul>

<p>Given the string <code>word</code>, return <em>the minimum total <strong>distance</strong> to type such string using only two fingers</em>.</p>

<p>The <strong>distance</strong> between coordinates <code>(x<sub>1</sub>, y<sub>1</sub>)</code> and <code>(x<sub>2</sub>, y<sub>2</sub>)</code> is <code>|x<sub>1</sub> - x<sub>2</sub>| + |y<sub>1</sub> - y<sub>2</sub>|</code>.</p>

<p><strong>Note</strong> that the initial positions of your two fingers are considered free so do not count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.</p>

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

<pre>
<strong>Input:</strong> word = &quot;CAKE&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> Using two fingers, one optimal way to type &quot;CAKE&quot; is: 
Finger 1 on letter &#39;C&#39; -&gt; cost = 0 
Finger 1 on letter &#39;A&#39; -&gt; cost = Distance from letter &#39;C&#39; to letter &#39;A&#39; = 2 
Finger 2 on letter &#39;K&#39; -&gt; cost = 0 
Finger 2 on letter &#39;E&#39; -&gt; cost = Distance from letter &#39;K&#39; to letter &#39;E&#39; = 1 
Total distance = 3
</pre>

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

<pre>
<strong>Input:</strong> word = &quot;HAPPY&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Using two fingers, one optimal way to type &quot;HAPPY&quot; is:
Finger 1 on letter &#39;H&#39; -&gt; cost = 0
Finger 1 on letter &#39;A&#39; -&gt; cost = Distance from letter &#39;H&#39; to letter &#39;A&#39; = 2
Finger 2 on letter &#39;P&#39; -&gt; cost = 0
Finger 2 on letter &#39;P&#39; -&gt; cost = Distance from letter &#39;P&#39; to letter &#39;P&#39; = 0
Finger 1 on letter &#39;Y&#39; -&gt; cost = Distance from letter &#39;A&#39; to letter &#39;Y&#39; = 4
Total distance = 6
</pre>

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

<ul>
	<li><code>2 &lt;= word.length &lt;= 300</code></li>
	<li><code>word</code> consists of uppercase English letters.</li>
</ul>

from functools import cache

@cache
def get_char_coords():
    keyboard = [
        "ABCDEF",
        "GHIJKL",
        "MNOPQR",
        "STUVWX",
        "YZ"
    ]
    out = dict()
    for row_idx, row in enumerate(keyboard):
        for col_idx, char in enumerate(row):
            out[char] = [row_idx, col_idx]
    return out

def move_distance(c1, c2):
    char_coordinates = get_char_coords()
    if c1 == 'initial':
        return 0
    x1, y1 = char_coordinates[c1]
    x2, y2 = char_coordinates[c2]

    return abs(x1 - x2) + abs(y1 - y2)

class Solution:
    def minimumDistance(self, word: str) -> int:

        # dynamic program dp(idx, f1, f2) -> min distance to type word[idx:]
        #      if finger1 at f1 and finger2 at f2
        # base case: idx = len(word) -> 0
        # recurrence:
        # let c = word[idx]
        #     try moving finger1 to c and recurse: distance(f1, c) + dp(idx + 1, c, f2)
        #.    try moving finger1 to c and recurse: distance(f2, c) + dp(idx + 1, f1, c)
        # return the minimum of these two
        
        @cache
        def dp(idx, f1='initial', f2='initial'):
            if idx == len(word):
                return 0

            c = word[idx]
            out = float('inf')

            # try moving finger1 to c and recurse
            out = min(out, move_distance(f1, c) + dp(idx + 1, c, f2))
            out = min(out, move_distance(f2, c) + dp(idx + 1, c, f1))

            return out
        
        return dp(0)


        

        
22200