#Simangaliso Collin Mavuso collinmavuso98@gmail.com #QUESTION 3 #for this solution we used a simple approach by traversing over each element when we land #on an element with a count over 1 we return that value #NOTE :we can also solve it in other means that consider space complexities but lets keep it simple. def findDuplicate(dl): n=len(dl) for i in range(n): countGreaterThan1=dl.count(dl[i])# for each element get its occurance count if countGreaterThan1 > 1:# if its greater than 1 we have our answers return dl[i] # after weve iterated over all elemnts we by defalut return -1 when no countGreaterThan1 return -1 #Test Cases dl=[1,3,5,4,5,6,5,6456,64,4,46,6,6] cl=[2.3,4,54,2.3] ddl=[0] ccl=[1,2,3,4,5,6,7] print(findDuplicate(dl)) print(findDuplicate(cl)) print(findDuplicate(ddl)) print(findDuplicate(ccl))
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