POTD- 01/03/2024 | Peak Element | Problem of the Day | GeeksforGeeks

Welcome to the daily solving of our PROBLEM OF THE DAY with Yash Dwivedi. We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Data Structures and Algorithms but will also help you build up problem-solving skills.

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[pankajsonagara]

here array should be sorted right ? if we search for binary search ?

[ThomasShelby-sq]

If we sort array then we have to check only two conditions even or odd length of array.

If even
Ele_pos=Length/2
Else
Ele_pos=( Length/2)+1

And like your code we seprate check for 1 length and 2 length array.

"If i made any mistake please correct me 😊"

[haha-uifp]

Good solution. You could initially set hi=n-2 instead of n-1, for the same reason you said lo was set to lo=1 and not 0. Also IMHO the while condition and the "return -1" are redundant, after all any array of integers does have a peak and thus it should be impossible for the program to ever get to the stage where lo>hi. Might as well use while(true) instead.