Finding node with duplicate id and changing the ID - xslt

Question

I need to find duplicate nodes (identified by ID) and if such nodes exist, then I need to update the id of one of them. Will appreciate if someone can let me know how to do it based on xpath or xsl.
example xml:
<music>
<title id="1"/>
<title id="2"/>
<title id="1"/>
</music>
The first and third node have the same ID. So, the id of the third is changed to '3'. I need to change it to the following:
<music>
<title id="1"/>
<title id="2"/>
<title id="3"/>
</music>

Answer 1

Please try the following template:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:template match="music">
<xsl:copy>
<xsl:for-each select="*">
<xsl:element name="{name()}">
<xsl:attribute name="id">
<xsl:choose>
<xsl:when test="preceding::*/#id=current()/#id">
<xsl:value-of select="generate-id()"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="#id"/>
</xsl:otherwise>
</xsl:choose>
</xsl:attribute>
<xsl:apply-templates/>
</xsl:element>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

Answer 2

Usually, the purpose of an ID is to uniquely identify elements. If so, it is insignificant what the actual ID string is - as long as there are no duplicates.
Therefore, the easiest approach to your problem is to number all title elements consistently, as remarked already by #michael.hor257k. This can be done using either position() or xsl:number.
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/music">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="title">
<xsl:copy>
<xsl:attribute name="id">
<xsl:number/>
</xsl:attribute>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Output
<?xml version="1.0" encoding="UTF-8"?>
<music>
<title id="1"/>
<title id="2"/>
<title id="3"/>
</music>