Oracle interview experience Real time questions & tips from candidates to crack your interview
Application Developer
Oracle
3 rounds | 4 Coding
problems
Rejected
Interview preparation journey
Preparation
Duration: 1 month
Topics: DSA, OOPS, Operating Sytem, DBMS, Computer Networks
Tip 1 : Be Strong with DSA
Tip 2 : Be Strong with OOPS and Core CS Fundamentals
Application process
Where: Naukri
Eligibility: 7 CGPA
Tip 1 : One Page Resume
Tip 2 : Highlight your coding profiles
Interview rounds
01
Round
Easy
Video Call
Duration60 minutes
Interview date11 Aug 2021
Coding problem2
DSA + Puzzle
1. Puzzle
(Torch and Bridge)
Puzzle: There are 4 persons (A, B, C and D) who want to cross a bridge in night.
A takes 1 minute to cross the bridge.
B takes 2 minutes to cross the bridge.
C takes 5 minutes to cross the bridge.
D takes 8 minutes to cross the bridge.
There is only one torch with them and the bridge cannot be crossed without the torch. There cannot be more than two persons on the bridge at any time, and when two people cross the bridge together, they must move at the slower
Step 1: A and B cross the bridge. A comes back. Time taken 3 minutes. Now B is on the other side.
Step 2: C and D cross the bridge. B comes back. Time taken 8 + 2 = 10 minutes. Now C and D are on the other side.
Step 3: A and B cross the bridge. Time taken is 2 minutes. All are on the other side.
Total time spent: 3 + 10 + 2 = 15 minutes.
2. Spiral Matrix
Easy
15m average time
80% success
0/40
Asked in companies
You are given a 2-D array 'MATRIX' of dimensions N x M, of integers. You need to return the spiral path of the matrix.
Example Of Spiral Path:
Problem approach
Algorithm:
Create and initialize variables k – starting row index, m – ending row index, l – starting column index, n – ending column index
Run a loop until all the squares of loops are printed.
In each outer loop traversal print the elements of a square in a clockwise manner.
Print the top row, i.e. Print the elements of the kth row from column index l to n, and increase the count of k.
Print the right column, i.e. Print the last column or n-1th column from row index k to m and decrease the count of n.
Print the bottom row, i.e. if k < m, then print the elements of m-1th row from column n-1 to l and decrease the count of m
Print the left column, i.e. if l < n, then print the elements of lth column from m-1th row to k and increase the count of l.
02
Round
Easy
Video Call
Duration60 minutes
Interview date12 Aug 2021
Coding problem1
Problem Solving
1. Container With Most Water
Moderate
15m average time
90% success
0/80
Asked in companies
Given a sequence of ‘N’ space-separated non-negative integers A[1],A[2],A[3],......A[i]…...A[n]. Where each number of the sequence represents the height of the line drawn at point 'i'. Hence on the cartesian plane, each line is drawn from coordinate ('i',0) to coordinate ('i', 'A[i]'), here ‘i’ ranges from 1 to ‘N’. Find two lines, which, together with the x-axis forms a container, such that the container contains the most area of water.
Note :
1. You can not slant the container i.e. the height of the water is equal to the minimum height of the two lines which define the container.
2. Do not print anything, you just need to return the area of the container with maximum water.
Example
For the above Diagram, the first red marked line is formed between coordinates (2,0) and (2,10), and the second red-marked line is formed between coordinates (5,0) and (5,9). The area of water contained between these two lines is (height* width) = (5-2)* 9 = 27, which is the maximum area contained between any two lines present on the plane. So in this case, we will return 3* 9=27.
Problem approach
class Solution {
public:
int maxArea(vector& height) {
int maxw = 0;
int left = 0;
int right = height.size()-1;
int water = 0;
while(left < right){
water = min(height[left], height[right])*(right-left);
maxw = max(maxw, water);
if(height[left] < height[right])
left++;
else
right--;
}
return maxw;
}
};
03
Round
Easy
Video Call
Duration60 minutes
Interview date20 Aug 2021
Coding problem1
Mangerial Round - Problem Solving
1. Longest Substring Without Repeating Characters
Moderate
20m average time
80% success
0/80
Asked in companies
Given a string 'S' of length 'L', return the length of the longest substring without repeating characters.
Example:
Suppose given input is "abacb", then the length of the longest substring without repeating characters will be 3 ("acb").
Problem approach
Let us talk about the linear time solution now. This solution uses extra space to store the last indexes of already visited characters. The idea is to scan the string from left to right, keep track of the maximum length Non-Repeating Character Substring seen so far in res. When we traverse the string, to know the length of current window we need two indexes.
1) Ending index ( j ) : We consider current index as ending index.
2) Starting index ( i ) : It is same as previous window if current character was not present in the previous window. To check if the current character was present in the previous window or not, we store last index of every character in an array lasIndex[]. If lastIndex[str[j]] + 1 is more than previous start, then we updated the start index i. Else we keep same i.
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Skill covered: Programming
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