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    <p>The first thing to answer is </p> <pre><code>What is a mismatch ? </code></pre> <p>Mismatch is when the characters are out of order. </p> <pre><code>When are characters out of order ? </code></pre> <p>First assumption : The 2 strings are inter-convertible.</p> <p>Second assumption : Both strings are of same length.</p> <p>So an out of order means that for a character <strong>i</strong> in first string, there <strong>certainly</strong> exists another character at position <strong>j</strong> in the second string. </p> <pre><code>When are swaps required ? </code></pre> <p>So for any given index <strong>i</strong>, if the characters in both the strings are not equal. There is a mismatch. How many mismatches ? 2 mismatches. Since the mismatched character from first string will cause a mismatch in the second string at another position <code>j</code>. So the total number of mismatches will always be a multiple of 2. </p> <p>One swap corrects position of 2 characters. So the optimal number of mismatches is half the total number of mismatches.</p> <p>Here is some sample code.</p> <pre><code>public class CharacterMismatch { /** * Returns -1 if not possible * @param first * @param second * @return */ public static int maxSwaps(char first[], char second[]) { if (first.length != second.length) { return -1; } int freqFirst[] = new int[Character.MAX_VALUE], freqSecond[] = new int[Character.MAX_VALUE]; int len = first.length; for (int i = 0; i &lt; len; ++i) { freqFirst[first[i]]++; freqSecond[second[i]]++; } for (int i = 0; i &lt; len; ++i) { if (freqFirst[i] != freqSecond[i]) { return -1; } } //count mismatches. That is the minimal number of swaps required. int swaps = 0; for (int i = 0; i &lt; len; ++i) { if (first[i] != second[i]) { swaps++; } } return swaps / 2; } public static void main(String[] args) { String first[] = { "aabbccdd", "abcc", "abccab" }; String second[] = { "ddbbccaa", "accb", "abcabc"}; for (int i = 0; i &lt; first.length; ++i) { System.out.printf("%s %s =&gt; %d%n", first[i], second[i], maxSwaps(first[i].toCharArray(), second[i].toCharArray())); } } } </code></pre> <p>Returns</p> <pre><code>aabbccdd ddbbccaa =&gt; 2 abcc accb =&gt; 1 abccab abcabc =&gt; 1 </code></pre> <p>Complexity is O(n).</p> <p>There is some boiler plate code which checks whether they are really inter-convertible by checking the frequency of each character in both strings and returns -1 if not possible.</p> <p><strong>Edit:</strong></p> <p>Sorry this works only when the max number of swaps for every mismatched character is at most one. </p>
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    1. POFinding the minimum number of swaps to convert one string to another, where the strings may have repeated characters
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    1. CO"abccab abcabc => 1" I'm not seeing one swap.
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      1. USDavid Eisenstat
    2. CO@DavidEisenstat yes you are right. This does not seem to work for all the cases. Basically it only works when every character is out of place by a maximum of 1 swap. But I think I have a better solution(which looks like O(n lg n)). I will post it tomorrow.
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