Amazon | OA Coding Questions and Solutions | Set - 10 | 2022

Question

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Answer 1

Problem Statement:

John (1) and Jack (2), are friends who construct the wall as per the number of bricks given to them.
They work turn by turn. John works in the increasing order starting from 1 with an increment of 1. Jack places twice the bricks as John places in previous turn. Goal is to find who placed the last brick and how many bricks will be placed in the end.

Example 1:
numberOfBricks: 13
John & Jack will construct the wall
John 1
Total Bricks till now: 1
Jack 1 * 2
Total Bricks till now: 3
John 2
Total Bricks till now: 5
Jack 2 * 2
Total Bricks till now: 9
John 3
Total Bricks till now: 12
Jack 3 * 2
Total Bricks till now: 18
Since total bricks to be placed were 13. But lastly sum became 18, hence lastly Jack has to place on 1 more brick. The correct answer in result array is:
result[0] = 2 // as Jack placed the last brick
result[1] = 1 // only 1 brick was to be placed in the end

Example 2:
numberOfBricks: 10
John & Jack will construct the wall
John 1
Total Bricks till now: 1
Jack 1 * 2
Total Bricks till now: 3
John 2
Total Bricks till now: 5
Jack 2 * 2
Total Bricks till now: 9
John 3
Total Bricks till now: 12
Since total bricks to be placed were 10. But lastly sum became 12, hence lastly John has to place on 1 more brick. The correct answer in result array is:
result[0] = 1 // as John placed the last brick
result[1] = 1 // only 1 brick was to be placed in the end"

My Soln is :-

class solution{
    public int[] bricksAndWinner(int numberOfBricks){
    int[] res = new int[2];
    int x = 1; int john = 1,jack = 2;
    int winner = 1;
    while(numberOfBricks>=0){           
            
            //checking if john has bricks available to use for wall
            if(numberOfBricks>x) {
                numberOfBricks -= x;//Subtracting the number of bricks used by john at x'th time
            }else{
                winner = john; //Declaring John as winner because he placed the last brick.
                break; //as we are satisfied with our results, breaking out of loop.
            }
            
            //checking if jack has bricks available twice of john to use for wall
            if(numberOfBricks>2*x) {
                numberOfBricks -= 2*x; //Subtracting the number of bricks used by jack at x'th time, the number of bricks used by jack is double of john hence 2 * x.
            }else{
                winner = jack; //Declaring Jack as winner because he placed the last brick.
                break; //as we are satisfied with our results, breaking out of loop.
            }
            x++; // incrementing x by 1 in each loop.
        }
        res[0] = winner;
        res[1] = numberOfBricks;
       
        return res;
    }
Time Complexity = O(logN)
Space Complexity = O(1)