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9c0d7dd402 |
@@ -8,7 +8,7 @@ Follow these steps to set up your project:
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### 0. Watch the introductory video
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We have recorded a special video, explaining the exact steps that needs to be taken to complete this assignment.
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Here is the [link](https://drive.meshcomp.ir/d/c9138977a329485585ae/)
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Here is the [link]()
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**Make sure to watch the video before starting the assignment.**
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@@ -226,7 +226,7 @@ Before starting, ensure you have:
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## Getting Help
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- **Mentor contact**
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- **Mentor contact:** Available during office hours and via [specify platform]
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- **Git documentation:** https://docs.meshcomp.ir/git
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---
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@@ -7,17 +7,6 @@ public class BonusExercises {
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/*
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complete the method below, so it will validate an email address
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1. Must have exactly one @ (no more, no less)
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2. Split into local-part and domain (before @ and after @)
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3. Local-part rules:
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- Can't be empty
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- Can't start or end with dot
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- Can't have two dots in a row
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4. Domain rules:
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- Can't be empty
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- Can't start or end with hyphen
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- Can't have underscores
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- Each segment (between dots) must follow same hyphen rules
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*/
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public boolean validateEmail(String email) {
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String regex = ""; // todo
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@@ -28,16 +17,10 @@ public class BonusExercises {
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}
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/*
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This method should find and return the first date in a string.
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Supported formats:
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- American: MM/DD/YYYY (e.g., 12/09/2023)
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- British: DD/MM/YYYY (e.g., 12/09/2023 — same pattern, context matters)
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- ISO: YYYY-MM-DD (e.g., 2024-07-15)
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- Slash variant: YYYY/MM/DD (e.g., 2025/01/01)
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If no match for a date is found in the string, return null.
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*/
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this method should find a date in string
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note that it should be in british or american format
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if there's no match for a date, return null
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*/
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public String findDate(String string) {
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// todo
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return null;
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@@ -19,47 +19,41 @@ public class MainExercises
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the output has to be a two-dimensional array of characters, so don't just print the triangle!
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*/
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public char[][] generateTriangle(int n) {
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char[][] triangle = new char[n][];
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for(int i = 0; i < n; i++)
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{
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triangle[i] = new char[i+1];
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for(int j = 0; j <= i; j++)
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{
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if(i == 0 || j == 0 || i == j || i == n-1)
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triangle[i][j] = '*';
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else
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triangle[i][j] = ' ';
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}
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// todo
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return null;
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}
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return triangle;
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}
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/*
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SPIRAL TRAVERSAL OF A RECTANGULAR MATRIX
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given a matrix of random integers, you should do spiral traversal in it
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e.g. if the matrix is as shown below:
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1 2 3
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4 5 6
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7 8 9
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then the spiral traversal of that is:
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{1, 2, 3, 6, 9, 8, 7, 4, 5}
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Given a rectangular matrix (2D array) of integers, this method traverses
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it in a spiral order (clockwise from outside to inside) and returns
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the elements as a 1D array.
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so you should walk in that matrix in a curl and then add the numbers in order you've seen them in a 1D array
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*/
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public int[] spiralTraversal(int[][] values, int rows, int cols) {
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EXAMPLE:
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Input matrix:
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1 2 3
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4 5 6
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7 8 9
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Spiral order: start at top-left (1), go right →, then down ↓,
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then left ←, then up ↑, then repeat inward.
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Result: {1, 2, 3, 6, 9, 8, 7, 4, 5}
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so you should walk in that matrix in a curl and then add the numbers in order you've seen them in a 1D array
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RECTANGULAR MATRIX ASSUMPTION:
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This method assumes the input matrix is RECTANGULAR (all rows have
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the same number of columns). In Java, we can verify this because
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2D arrays might be jagged (rows of different lengths).
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IMPORTANT: In Java, we do NOT need to pass rows and cols!
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The 2D array 'matrix' knows its own dimensions:
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- Number of rows: matrix.length
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- Number of columns: matrix[0].length (if rectangular)
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*/
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public int[] spiralTraversal(int[][] matrix) {
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// todo
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return null;
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}
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/*
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@@ -78,18 +72,14 @@ public class MainExercises
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2, 1, 1
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1, 1, 1, 1
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Note: As you can see in the examples, we want to generate distinct partitions,
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which means 1,2 and 2,1 are not different — they count as the same combination.
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note: as you can see in examples, we want to generate distinct summations, which means 1, 2 and 2, 1 are no different
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you should generate all partitions of the input number and
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You should generate all partitions of the input number.
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hint: you can measure the size and order of arrays by finding the pattern of partitions and their number
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trust me, that one's fun and easy :)
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Hint: You can determine the size and order of the arrays by finding the pattern
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of partitions and their count. Trust me, this one's fun and easy :)
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If you're familiar with Lists and ArrayLists, you can also edit the method's
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body to use them instead of arrays.
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if you're familiar with lists and arraylists, you can also edit method's body to use them instead of array
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*/
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public int[][] intPartitions(int n) {
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// todo
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return null;
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@@ -16,8 +16,10 @@ public class MainTestTraversal {
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int[][] nums = {
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{1, 2, 3, 4, 5}
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};
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int rows = 1;
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int cols = 5;
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int[] expected = {1, 2, 3, 4, 5};
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assertArrayEquals(expected, ex.spiralTraversal(nums));
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assertArrayEquals(expected, ex.spiralTraversal(nums, rows, cols));
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}
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@Test
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@@ -28,8 +30,10 @@ public class MainTestTraversal {
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{7, 8, 9},
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{10, 11, 12}
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};
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int rows = 4;
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int cols = 3;
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int[] expected = {1, 2, 3, 6, 9, 12, 11, 10, 7, 4, 5, 8};
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assertArrayEquals(expected, ex.spiralTraversal(nums));
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assertArrayEquals(expected, ex.spiralTraversal(nums, rows, cols));
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}
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@Test
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@@ -38,8 +42,10 @@ public class MainTestTraversal {
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{1, 2, 3, 4},
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{5, 6, 7, 8}
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};
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int rows = 2;
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int cols = 4;
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int[] expected = {1, 2, 3, 4, 8, 7, 6, 5};
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assertArrayEquals(expected, ex.spiralTraversal(nums));
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assertArrayEquals(expected, ex.spiralTraversal(nums, rows, cols));
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}
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@Test
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@@ -51,8 +57,10 @@ public class MainTestTraversal {
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{19, 20, 21, 22, 23, 24},
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{25, 26, 27, 28, 29, 30}
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};
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int rows = 5;
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int cols = 6;
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int[] expected = {1, 2, 3, 4, 5, 6, 12, 18, 24, 30, 29, 28, 27, 26, 25, 19, 13, 7, 8, 9, 10, 11, 17, 23, 22, 21,
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20, 14, 15, 16};
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assertArrayEquals(expected, ex.spiralTraversal(nums));
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assertArrayEquals(expected, ex.spiralTraversal(nums, rows, cols));
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}
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}
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Reference in New Issue
Block a user