<prompt> Andika suluhisho katika Java. Kwa kuwa na kamba mbili a na b, kurudi urefu wa ndogo ya muda mrefu zaidi isiyo ya kawaida kati ya a na b. Kama ndogo ya muda mrefu zaidi isiyo ya kawaida haipo, kurudi -1. ndogo ya muda mrefu kati ya kamba mbili ni kamba ambayo ni ndogo ya moja lakini si nyingine. ndogo ya kamba s ni kamba ambayo inaweza kupatikana baada ya kufuta idadi yoyote ya herufi kutoka s. - Kwa mfano, abc ni ndogo ya aebdc kwa sababu unaweza kufuta herufi zilizowekwa chini katika aebdc kupata abc.  <response>: Here is a possible solution in Java:public int findLUSlengthString a, String b if a.equalsb return -1; else return Math.maxa.length, b.length; Explanation:- First, we check if the two strings are equal using the equals method. If they are equal, then there is no uncommon subsequence and we return -1.- If the strings are not equal, then the longest uncommon subsequence will be the longer of the two strings. We use the Math.max method to determine the maximum length between a and b.- The method returns the length of the longest uncommon subsequence as an integer. 