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diff --git a/AP1403 - Algorithms/src/main/java/Exercises.java b/AP1403 - Algorithms/src/main/java/Exercises.java
index 15a2133..de31de2 100644
--- a/AP1403 - Algorithms/src/main/java/Exercises.java	
+++ b/AP1403 - Algorithms/src/main/java/Exercises.java	
@@ -1,64 +1,98 @@
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
 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) {
-        // todo
+        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) {
-        // todo
-        return null;
+    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++;
+
+            if (top <= bottom) {
+                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
+    public static int[][] intPartitions(int n) {
+        List<int[]> parts = new ArrayList<>();
+        int[] partition = new int[n];
+        int index = 0;
 
-        e.g. 1 -> all partitions of integer 3 are:
-        3
-        2, 1
-        1, 1, 1
+        partition[index] = n;
 
-        e.g. 2 -> for number 4 goes as:
-        4
-        3, 1
-        2, 2
-        2, 1, 1
-        1, 1, 1, 1
+        while (true) {
+            int[] validParts = Arrays.copyOf(partition, index + 1);
+            parts.add(validParts);
 
-        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
+            int remaining = 0;
+            while (index >= 0 && partition[index] == 1) {
+                remaining += partition[index];
+                index--;
+            }
 
-        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 (index < 0) {
+                break;
+            }
 
-        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) {
-        // todo
-        return null;
+            partition[index]--;
+            remaining++;
+
+            while (remaining > partition[index]) {
+                partition[index + 1] = partition[index];
+                remaining -= partition[index];
+                index++;
+            }
+
+            partition[index + 1] = remaining;
+            index++;
+        }
+
+        int[][] result = new int[parts.size()][];
+        for (int i = 0; i < parts.size(); i++) {
+            result[i] = parts.get(i);
+        }
+
+        return result;
     }
 
     public static void main(String[] args) {
         // you can test your code here
     }
 }
+