Developer2

Exercises.java

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+import java.util.ArrayList;
+import java.util.List;
+import java.util.Scanner;
+public class Exercises {
+
+    /*
+        there is an array of positive integers as input of function and another integer for the target value
+        all the algorithm should do is to find those two integers in array which their multiplication is the target
+        then it should return an array of their indices
+        e.g. {1, 2, 3, 4} with target of 8 -> {1, 3}
+
+        note: you should return the indices in ascending order and every array's solution is unique
+    */
+    public int[] productIndices(int[] values, int target) {
+        for (int i = 0; i < values.length; i++) {
+            for (int j = i + 1; j < values.length; j++) {
+                if (values[i] * values[j] == target) {
+                    return new int[]{i, j};
+                }
+            }
+        }
+        return null;
+    }
+
+    /*
+        given a matrix of random integers, you should do spiral traversal in it
+        e.g. if the matrix is as shown below:
+        1 2 3
+        4 5 6
+        7 8 9
+        then the spiral traversal of that is:
+        {1, 2, 3, 6, 9, 8, 7, 4, 5}
+
+        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
+    */
+    public int[] spiralTraversal(int[][] values, int rows, int cols) {
+        int[] result = new int[rows * cols];
+        int index = 0;
+        int top = 0, bottom = rows - 1, left = 0, right = cols - 1;
+
+        while (top <= bottom && left <= right) {
+            for (int i = left; i <= right; i++) {
+                result[index++] = values[top][i];
+            }
+            top++;
+
+            for (int i = top; i <= bottom; i++) {
+                result[index++] = values[i][right];
+            }
+            right--;
+
+            if (top <= bottom) {
+                for (int i = right; i >= left; i--) {
+                    result[index++] = values[bottom][i];
+                }
+                bottom--;
+            }
+
+            if (left <= right) {
+                for (int i = bottom; i >= top; i--) {
+                    result[index++] = values[i][left];
+                }
+                left++;
+            }
+        }
+
+        return result;
+    }
+
+    /*
+        integer partitioning is a combinatorics problem in discreet maths
+        the problem is to generate sum numbers which their summation is the input number
+
+        e.g. 1 -> all partitions of integer 3 are:
+        3
+        2, 1
+        1, 1, 1
+
+        e.g. 2 -> for number 4 goes as:
+        4
+        3, 1
+        2, 2
+        2, 1, 1
+        1, 1, 1, 1
+
+        note: as you can see in examples, we want to generate distinct summations, which means 1, 2 and 2, 1 are no different
+        you should generate all partitions of the input number and
+
+        hint: you can measure the size and order of arrays by finding the pattern of partitions and their number
+        trust me, that one's fun and easy :)
+
+        if you're familiar with lists and arraylists, you can also edit method's body to use them instead of array
+    */
+    public int[][] intPartitions(int n) {
+        ArrayList<int[]> partitionsList = new ArrayList<>();
+        int[] p = new int[n];
+        int k = 0;
+        p[k] = n;
+
+        while (true) {
+            int[] currentPartition = new int[k + 1];
+            System.arraycopy(p, 0, currentPartition, 0, k + 1);
+            partitionsList.add(currentPartition);
+
+            int a = 0;
+            while (k >= 0 && p[k] == 1) {
+                a += p[k];
+                k--;
+            }
+
+            if (k < 0) {
+                int[][] result = new int[partitionsList.size()][];
+                for (int i = 0; i < partitionsList.size(); i++) {
+                    result[i] = partitionsList.get(i);
+                }
+                return result;
+            }
+
+            p[k]--;
+            a++;
+
+            while (a > p[k]) {
+                p[k + 1] = p[k];
+                a = a - p[k];
+                k++;
+            }
+
+            p[k + 1] = a;
+            k++;
+        }
+    }
+
+    public static void main(String[] args) {
+        Scanner c = new Scanner(System.in);
+        int[] a = new int[4];
+        for (int i = 0; i < 4; i++) {
+            a[i] = c.nextInt();
+        }
+        int b = c.nextInt();
+        Exercises e = new Exercises();
+        int[] indices = e.productIndices(a, b);
+        if (indices != null) {
+            for (int i = 0; i < indices.length; i++) {
+                System.out.print(indices[i] + " ");
+            }
+            System.out.println();
+        } else {
+            System.out.println("No such pair found.");
+        }
+        int rows = c.nextInt();
+        int cols = c.nextInt();
+        int[][] m = new int[rows][cols];
+        for (int i = 0; i < rows; i++) {
+            for (int j = 0; j < cols; j++) {
+                m[i][j] = c.nextInt();
+            }
+        }
+        int[] s = e.spiralTraversal(m, rows, cols);
+        for (int i = 0; i < s.length; i++) {
+            System.out.print(s[i] + " ");
+        }
+        System.out.println();
+
+        int n = c.nextInt();
+        int[][] partitions = e.intPartitions(n);
+        for (int[] partition : partitions) {
+            for (int i = 0; i < partition.length; i++) {
+                System.out.print(partition[i] + " ");
+            }
+            System.out.println();
+        }
+    }
+}
+// good
+
+