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February 7, 2016 23:07
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| Print concentric rectangular pattern in a 2d matrix. | |
| Let us show you some examples to clarify what we mean. | |
| Example 1: | |
| Input: A = 4. | |
| Output: | |
| 4 4 4 4 4 4 4 | |
| 4 3 3 3 3 3 4 | |
| 4 3 2 2 2 3 4 | |
| 4 3 2 1 2 3 4 | |
| 4 3 2 2 2 3 4 | |
| 4 3 3 3 3 3 4 | |
| 4 4 4 4 4 4 4 | |
| Example 2: | |
| Input: A = 3. | |
| Output: | |
| 3 3 3 3 3 | |
| 3 2 2 2 3 | |
| 3 2 1 2 3 | |
| 3 2 2 2 3 | |
| 3 3 3 3 3 | |
| The outermost rectangle is formed by A, then the next outermost is formed by A-1 and so on. | |
| You will be given A as an argument to the function you need to implement, and you need to return a 2D array. | |
| public class Solution { | |
| public ArrayList<ArrayList<Integer>> prettyPrint(int a) { | |
| ArrayList<ArrayList<Integer>> sol = new ArrayList<ArrayList<Integer>>(); | |
| Integer size = 2 * a - 1; | |
| for (int i = 0; i < size; i += 1) { | |
| sol.add(new ArrayList<Integer>()); | |
| for (int j = 0; j < size; j += 1) { | |
| sol.get(i).add(a); | |
| } | |
| } | |
| //top half and middle | |
| for (int i = 1; i < (size - 1) / 2 + 1; i += 1) { | |
| int stepDowns = i; | |
| int deficit = 0; | |
| Boolean goUp = false; | |
| //left side and middle | |
| for (int j = 1; j <= (size - 1) /2; j += 1) { | |
| if (stepDowns > 0) { | |
| deficit += 1; | |
| stepDowns -= 1; | |
| sol.get(i).set(j, a - deficit); | |
| } else { | |
| sol.get(i).set(j, a - deficit); | |
| } | |
| } | |
| stepDowns = i; | |
| deficit = 0; | |
| //right side | |
| for (int j = size - 2; j > (size - 1) /2; j -= 1) { | |
| if (stepDowns > 0) { | |
| deficit += 1; | |
| stepDowns -= 1; | |
| sol.get(i).set(j, a - deficit); | |
| } else { | |
| sol.get(i).set(j, a - deficit); | |
| } | |
| } | |
| } | |
| int i = 0; | |
| //well I'll be darned...the bottom two rows are the first two! | |
| for (int j = size - 1; j > (size - 1) /2; j -= 1) { | |
| sol.set(j, sol.get(i)); | |
| i += 1; | |
| } | |
| return sol; | |
| //all 3's | |
| } | |
| } |
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