All tests passing
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2025mohseni
@@ -1,103 +1,75 @@
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public class MainExercises
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{
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/*
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you should create a triangle with "*" and return a two-dimensional array of characters based on that
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the triangle's area is empty, which means some characters should be " "
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import java.util.*;
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example 1, input = 3:
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*
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**
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***
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public class MainExercises {
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example 2, input = 5:
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*
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**
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* *
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* *
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*****
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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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// اصلاح متد برای ایجاد مثلث به صورت آرایه دو بعدی Jagged
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public char[][] generateTriangle(int n) {
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// todo
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return null;
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char[][] triangle = new char[n][];
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for (int i = 0; i < n; i++) {
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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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// شرط قرارگیری ستاره: ستون اول، ستون آخر یا ردیف آخر
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if (j == 0 || j == i || 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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}
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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 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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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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// متد پیمایش مارپیچی ماتریس
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public int[] spiralTraversal(int[][] matrix) {
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// todo
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return null;
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if (matrix == null || matrix.length == 0) return new int[0];
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int rows = matrix.length;
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int cols = matrix[0].length;
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int[] result = new int[rows * cols];
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int top = 0, bottom = rows - 1, left = 0, right = cols - 1;
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int index = 0;
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while (top <= bottom && left <= right) {
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for (int i = left; i <= right; i++) result[index++] = matrix[top][i];
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top++;
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for (int i = top; i <= bottom; i++) result[index++] = matrix[i][right];
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right--;
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if (top <= bottom) {
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for (int i = right; i >= left; i--) result[index++] = matrix[bottom][i];
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bottom--;
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}
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if (left <= right) {
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for (int i = bottom; i >= top; i--) result[index++] = matrix[i][left];
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left++;
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}
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}
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return result;
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}
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/*
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integer partitioning is a combinatorics problem in discreet maths
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the problem is to generate sum numbers which their summation is the input number
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e.g. 1 -> all partitions of integer 3 are:
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3
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2, 1
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1, 1, 1
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e.g. 2 -> for number 4 goes as:
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4
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3, 1
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2, 2
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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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You should generate all partitions of the input number.
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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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*/
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// متد افراز عدد (Integer Partitioning)
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public int[][] intPartitions(int n) {
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// todo
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return null;
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List<List<Integer>> results = new ArrayList<>();
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generatePartitions(n, n, new ArrayList<>(), results);
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int[][] res = new int[results.size()][];
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for (int i = 0; i < results.size(); i++) {
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res[i] = results.get(i).stream().mapToInt(Integer::intValue).toArray();
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}
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return res;
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}
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private void generatePartitions(int target, int max, List<Integer> current, List<List<Integer>> results) {
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if (target == 0) {
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results.add(new ArrayList<>(current));
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return;
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}
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for (int i = Math.min(target, max); i >= 1; i--) {
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current.add(i);
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generatePartitions(target - i, i, current, results);
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current.remove(current.size() - 1);
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}
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}
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public static void main()
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{
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public static void main(String[] args) {
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// تست دستی در صورت نیاز
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}
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}
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