Let Σ be a finite alphabet and let L1 and L2 be two languages over Σ. Assume you have access to two routines IsStringInL1(u) and IsStringInL2(u). The former routine decides whether a given string u is in L1 and the latter whether u is in L2. Using these routines as black boxes describe an efficient algorithm that given an arbitrary string w â Σ â decides whether w â (L1 + L2 )â. To evaluate the running time of your solution you can assume that calls to IsStringInL1 () and IsStringInL2 () take constant time. Note that you are not assuming any property of L1 or L2 other than being able to test membership in those languages.

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Print "censored" if userinput contains the word "darn", else print userinput. end with newline. ex: if userinput is "that darn cat.", then output is: censoredex: if userinput is "dang, that was scary! ", then output is: dang, that was scary! note: if the submitted code has an out-of-range access, the system will stop running the code after a few seconds, and report "program end never reached." the system doesn't print the test case that caused the reported message.#include #include using namespace std; int main() {string userinput; getline(cin, userinput); int ispresent = userinput.find("darn"); if (ispresent > 0){cout < < "censored" < < endl; /* your solution goes here */return 0; }