Ques 1: An array of n integers is there in which the range of elements is n, i.e., the difference between maximum and minimum number is n. Find the repeating numbers. Ques 2: An extension of Question 1. Was asked to find number of times each number is repeated. Ques 3: There are n frames of m data element each. The data element in each frame is arranged in increasing order. You are provided m*n space in which you have to arrange all data in increasing order. Ques4:- Given a 2D array containing only 0/1’s and each row is in sorted order. Find the row which contains maximum number of 1s. I was asked to code. Algo which I told was I will search position of first 1 in 1st row using binary search. And mark it. Now note that position check in 2nd row. If there is 1 for that position already found in 1st row, then binary search from 0 to that position else move to row number 3. Similarly continue further.

Question

Accepted Answer

The interview questions involve finding repeating numbers in an array, counting the number of times each number is repeated, arranging data elements in increasing order, and finding the row with the maximum number of 1s in a 2D array.

For Question 1, iterate through the array and use a hash set to track repeating numbers.
For Question 2, use a hash map to count the occurrences of each number.
For Question 3, merge all the frames into a single array and sort it.
For Question 4, use binary search to find the first occurrence of 1 in each row and keep track of the row with the maximum count.

Accepted Answer

This is done by subtracting min-1 from every element.

So in the above example we get min = 5, thus {5-4,5-4,7-4,7-4,9-4} = {1,1,2,2,5}.

Now you can use the following pseudo code:

for(i=0;i

What we are doing is taking an element arr[i], and adding n+1 at the index value equalling that element(possible because elements between 1 and n), e.g. if we have n = 6 and element = 2 then we add 7 at index '2'.This is because by later dividing with n+1 we get the frequency of the given number at the index.

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