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    <p>Okay, I think I have a simple solution.<br> I'm going to pretend A,B are strings of binary numbers only. </p> <pre><code>A= 011010101001001101 B= 010 </code></pre> <p>Apply an XNOR filter on A, with B. Result will be <strong>C</strong>.<br> Assume origin of B is at its center.<br> (<em>This means look for B in A. If you find a match, put a 1 in C to that location.</em>) </p> <pre><code>A= 011010101001001101 C= 000010101001000000 </code></pre> <p>Simply, C shows the locations of where A contains B.<br> You could remove all the characters in the locations shown by C, for example, and that would solve your problem. But it wouldn't be the optimum solution. </p> <p>But finding the optimum solution is easy.<br> Okay, now the problem is a lot simpler. I can explain it like these algorithm problems: </p> <ul> <li>C shows the locations of paintings. Each guard can protect a range of X( X=length(B)/2+1 ). Place the minimum number of guards such that all paintings are protected. </li> <li>Remove from C the minimum number of 1's. For each 1 removed, all 1's in a neighbourhood of lenth(B)/2+1 are removed as well. </li> </ul> <p>This is no new problem, and there are existing solutions to it. Apply one such solution.<br> And this will yield the following. <strong>D</strong> shows the locations of 1's removed:</p> <pre><code>C= 000010101001000000 D= 000000100001000000 </code></pre> <p>Count the number of 1's in D -> 2. This means You can remove 2 elements from A so B is not a substring of A.</p>
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    1. PORemoving minimal letters from a string 'A' to remove all instances of string 'B'
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    1. COThis doesn't work at all. You're making the assumption that when you remove a character that the remaining characters will never form a new B. Consider: B=01000 A=010001000, C becomes 001001000 which suggests you should only need to remove one character (the 0 that's shared between the two). However that results in A'=01001000 which still has a B at the end.
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    2. CO@Nate That's correct. I didn't consider that could be the case.
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    3. CO@Nate Now that I look at the problem and solution again, in your counter-example, C=001000100, and D=001000100. This means A'=0100100, and has no substring B. Still, I understand how my solution is flawed. The solution needs to repeated until there are no substrings, and the solution at that point may not be optimal. I'm going to propose a new solution tonight.
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