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There is an array of size n which consists of any numbers from 1 to n .Find the numbers in the array which are appear more than once

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1. take sum of array let say asum 2. let esum is expected sum of array if number is not repeated. by math formula esum = n*(n-1)/2 3. repeated element = esum - asum

1.create another array of size n arrNew[n] and initialize all the elements to 1(existing array arrExist[]). 2.loop through the given array n and populate arrNew[ ] as --> arrNew[arrExist[ i ] ]++ 3.loop through new array and the indexes having value > 1 are repeated and indexes having value = 0 are missing.

Khemraj ..u got the question wrong way...let me take a sample input ans output Suppose the array of 6 elements...so the array can be (2,2,2,3,3,1) So ouput (2,3) Now Chetan u r right..but u have taken a lot of space the interviewer said to take a bit array rather than a byte array.so I consume just 1byte for 8 elements whereas in your case u use 8 bytes for 8 elements (probably even more)...so the way to declare size for that array will be ....char *p =Malloc(n/8+1) ...n is number of elements...hope this helps now.there are many more methods on net..

How will ensure that attrition is with acceptable limits of 5 to 8% ?

6 Answers

Where do you want to see your self in 1 year

2 Answers

Why you want to change

2 Answers

Hope you would not be taking or asking for leaves for at least 6 months.?

1 Answer

Technical Questions

1 Answer

What did feel most proud of in your previous job ?

1 Answer

How best you wife and family can describe you and that too in 3 words.

1 Answer

Management questions - What is the current role? How do you manage the team? How do you manage stakeholders? Scenario on how you handle knowledge transfer session?

1 Answer

Plans for the programme you are hired for

1 Answer
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