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POFinding minimum moves required for making 2 strings equal
 

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  1. POFinding minimum moves required for making 2 strings equal
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    <p>This is a question from one of the online coding challenge (which has completed).<br> I just need some logic for this as to how to approach.</p> <p><strong>Problem Statement:</strong><br> We have two strings A and B with the same super set of characters. We need to change these strings to obtain two equal strings. In each move we can perform one of the following operations:</p> <pre><code>1. swap two consecutive characters of a string 2. swap the first and the last characters of a string </code></pre> <p>A move can be performed on either string.<br> What is the <strong>minimum</strong> number of moves that we need in order to obtain two equal strings? </p> <p><strong>Input Format and Constraints:</strong><br> The first and the second line of the input contains two strings A and B. It is guaranteed that the superset their characters are equal. </p> <pre><code>1 &lt;= length(A) = length(B) &lt;= 2000 All the input characters are between 'a' and 'z' </code></pre> <p><br /> <strong>Output Format:</strong><br> Print the <strong>minimum</strong> number of moves to the only line of the output</p> <pre><code>Sample input: aab baa Sample output: 1 </code></pre> <p><strong>Explanation:</strong><br> Swap the first and last character of the string aab to convert it to baa. The two strings are now equal.</p> <p><strong>EDIT :</strong> Here is my first try, but I'm getting wrong output. Can someone guide me what is wrong in my approach.</p> <pre class="lang-c prettyprint-override"><code>int minStringMoves(char* a, char* b) { int length, pos, i, j, moves=0; char *ptr; length = strlen(a); for(i=0;i&lt;length;i++) { // Find the first occurrence of b[i] in a ptr = strchr(a,b[i]); pos = ptr - a; // If its the last element, swap with the first if(i==0 &amp;&amp; pos == length-1) { swap(&amp;a[0], &amp;a[length-1]); moves++; } // Else swap from current index till pos else { for(j=pos;j&gt;i;j--) { swap(&amp;a[j],&amp;a[j-1]); moves++; } } // If equal, break if(strcmp(a,b) == 0) break; } return moves; } </code></pre>
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    1. COpeople are not there to solve your challenges
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    2. CO"This is a question from ..." - a link always helps, if possible.
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    3. COWhat are your ideas on this problem? What have you tried so far? Have you considered some Greedy or Dynamic Programming approach?
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