dsa

- (leetcode) product of array except self
- prefix sum

Author: xy-241

content/Algorithm/Prefix Sum (前缀和).md

@@ -6,7 +6,7 @@ Author Profile:
 tags:
   - dsa
 Creation Date: 2023-10-08T19:28:00
-Last Date: 2024-05-31T14:37:50+08:00
+Last Date: 2024-06-01T13:20:47+08:00
 ---
 ## Abstract
 ---
@@ -15,7 +15,7 @@ Last Date: 2024-05-31T14:37:50+08:00
 
 - Prefix sum is an [[Array]] that has the **same length** as a **given array**
 - The **value at each index** is the **sum** of values from the **first element of the given array** to the **element at the current index**
-- Calculates the **sum of a particular range** of a **given array** efficiently by leveraging on [[Memoization]]
+- [[#Range Sum Query]] enables us to calculates the **sum of a particular range** of a **given array** efficiently by leveraging on [[Memoization]]
 
 ## Range Sum Query
 ---
@@ -28,12 +28,12 @@ Last Date: 2024-05-31T14:37:50+08:00
 
 
 >[!tip]
->$Range[0, j]$ is **same** as $P_j$. So we can just retrieve the value from the prefix sum array with index $j$, instead of using the formula stated in the diagram above, which requires us to have a placeholder at index $0$.
+>$Range[0, j]$ is **same** as $P_j$. So we can just retrieve the value from the **prefix sum array** with index $j$, instead of using the formula stated in the diagram above, which requires us to have a placeholder at index $0$.
 
 
 
 ## Practice Questions
 ---
-- [ ] [[Product of Array Except Self]]
+- [ ] [[Leetcode - Product of Array Except Self]]
 - [ ] [[Romantic Glasses]]
 - [ ] [[Closest Cities]]

content/cp/prefix_sum/Leetcode - Product of Array Except Self.md

@@ -0,0 +1,118 @@
+---
+sthNew: true
+Mastery Level:
+  - 📘
+Time Taken: 15
+Space:
+  - O(1)
+Time: O(n)
+Appears On:
+  - Grind 75
+Brush: 5
+Difficulty:
+  - Medium
+Area:
+  - prefix_sum
+  - dynamic_programming
+Reference 1: https://leetcode-solution-leetcode-pp.gitbook.io/leetcode-solution/medium/238.product-of-array-except-self
+Reference 2: 
+Author:
+  - Xinyang YU
+Author Profile:
+  - https://linkedin.com/in/xinyang-yu
+Creation Date: 2024-01-08, 12:52
+Last Date: 2024-06-01T15:10:38+08:00
+tags:
+  - cp
+draft: 
+description: Leetcode 238. Product of Array Except Self, detailed solution with hand-crafted visual
+---
+
+## Abstract
+---
+- [Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/) gives us an [[Array]], and we need to produce a new array with the same length, the elements inside the new array is the product of all the elements in the given array, excluding the element at the same index
+- The solution must run in `O(n)`
+- Can't use division operation
+- The new array does not count as extra space for [[Algorithm Complexity Analysis#Worst Space Complexity]]
+- We can make use of the idea presented in [[Prefix Sum (前缀和)]], **storing** the **intermediate product** to **avoid duplicated computation**. This allows us to obtain the answer in $O(n)$ time 
+
+## Solution
+---
+
+![[product_of_array_except_self.svg|600]]
+
+- We can make use of the idea presented in [[Prefix Sum (前缀和)]], **storing** the **intermediate product** to **avoid duplicated computation** as shown in the diagram above. This allows us to obtain the answer in $O(n)$ time 
+- The element at a **particular index of the new array** is the **product** of **all elements at the left side** of that particular index and **all element at the right side** of that particular index. So we have **prefix product array** that keeps track of the **product of all elements at the left side** for all indexes, and **suffix product array** that keeps track of the **product of all elements at the right side** for all indexes. Then the answer is just the product of the **intermediate product** from **prefix product array** and **suffix product array**
+- Below is the code implementation with Java based idea described above. From `prefix[i] = prefix[i - 1] * nums[i - 1]` and `suffix[i] = suffix[i + 1] * nums[i + 1]`, we can see we are making use of [[Dynamic Programming#Optimal Substructure (最优子结构)]] to avoid [[Dynamic Programming#Overlapping Subproblems (重复子问题)]]. We are able to use [[Dynamic Programming#Top-down DP Approach]] to create the **prefix product array** and **suffix product array**. `prefix[0] = 1` and `suffix[nums.length - 1] = 1` are the base case
+
+```java
+class Solution {
+  public int[] productExceptSelf(int[] nums) {
+    int[] prefix = new int[nums.length];
+    int[] suffix = new int[nums.length];
+    int[] res = new int[nums.length];
+
+    prefix[0] = 1;
+    suffix[nums.length - 1] = 1;
+
+    for (int i = 1; i < nums.length; i++) {
+      prefix[i] = prefix[i - 1] * nums[i - 1];
+    }
+
+    for (int i = nums.length - 2; i >= 0; i--) {
+      suffix[i] = suffix[i + 1] * nums[i + 1];
+    }
+
+    for (int i = 0; i < nums.length; i++) {
+      res[i] = prefix[i] * suffix[i];
+    }
+
+    return res;
+  }
+}
+```
+
+- We can further optimise the solution by replacing the **suffix product array** with **a variable** `suffix` as shown below. We are using `res` to produce the **prefix product array**, then compute the final answer array with the `suffix` variable, the **final product** at a **particular index** just needs the **suffix at that particular index**, **optimal substructure**
+
+```java
+class Solution {
+  public int[] productExceptSelf(int[] nums) {
+    int[] res = new int[nums.length];
+    res[0] = 1;
+
+    for (int i = 1; i < nums.length; i++) {
+      res[i] = res[i - 1] * nums[i - 1];
+    }
+
+    int suffix = 1;
+    for (int i = nums.length - 2; i >= 0; i--) {
+      suffix *= nums[i + 1];
+      res[i] = res[i] * suffix;
+    }
+
+    return res;
+  }
+}
+```
+
+
+
+## Space & Time Analysis
+---
+The analysis method we are using is [[Algorithm Complexity Analysis]]
+### Space - O(1)
+- *Ignore input size & language dependent space*
+- The output array isn't counted as stated by the question 
+### Time - O(n)
+- We need to loop through the elements of the given array 2 times
+- 1 time to obtain the **prefix product array**, another time to calculate the suffix to obtain the final answer
+ 
+
+## Personal Reflection
+---
+- **Why it takes so long to solve:** Didn't read the problem carefully and implementing [[Prefix Sum (前缀和)]] on products of elements
+- **What you could have done better:** Sketch out the calculation process clearly on the paper
+- **What you missed:** *NIL*
+- **Ideas you've seen before:** Prefix Sum (前缀和)
+- **Ideas you found here that could help you later:** Prefix Sum (前缀和)'s ability to provide the product of a range of elements in O(1)
+- **Ideas that didn't work and why:** *NIL*