95 lines
2 KiB
Markdown
95 lines
2 KiB
Markdown
# Leetcode Pascal's-Triangle
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#### 2022-06-12
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---
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##### Data structures:
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#DS #array
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##### Algorithms:
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#algorithm #recursion #iteration
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##### Difficulty:
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#leetcode #coding_problem #difficulty-easy
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##### Related topics:
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```expander
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tag:#recursion OR tag:#iteration OR tag:#array -tag:#Kadane_s_algorithm
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```
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- [[Binary Search Algorithm]]
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- [[cpp_Range_based_for_loop]]
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- [[Leetcode Binary-Search]]
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- [[Leetcode First-Bad-Version]]
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- [[Leetcode First-Unique-Character-In-a-String]]
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- [[Leetcode Insert-Into-a-Binary-Search-Tree]]
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- [[Leetcode Invert-Binary-Tree]]
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- [[Leetcode Path-Sum]]
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- [[Leetcode Ransom-Note]]
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- [[Leetcode Reshape-The-Matrix]]
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- [[Leetcode Reverse-Linked-List]]
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- [[Leetcode Search-Insert-Position]]
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- [[Leetcode Symmetric-Tree]]
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- [[Leetcode Two-Sum]]
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- [[Leetcode Valid-Anagram]]
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- [[Two pointers approach]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/pascals-triangle/)
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- [Additional resources]()
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___
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### Problem
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Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
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In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
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#### Examples
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**Example 1:**
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```markdown
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Input: numRows = 5
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Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
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```
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**Example 2:**
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```markdown
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Input: numRows = 1
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Output: [[1]]
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```
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#### Constraints
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- 1 <= numRows <= 30
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### Thoughts
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Very simple and straightforward problem.
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> [!summary]
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> answer[i][j] = answer[i - 1][j - 1] + answer[i - 1][j]
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I worked around special cases using the for loop in j: j = 1 and j < i.
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### Solution
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```cpp
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class Solution {
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public:
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vector<vector<int>> generate(int numRows) {
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vector<vector<int>> answer(numRows);
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for (int i = 0; i < numRows; i++) {
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// initialize sub-array
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answer[i] = vector<int>(i + 1);
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answer[i][0] = 1;
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for (int j = 1; j < i; j++) {
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answer[i][j] = answer[i - 1][j] + answer[i - 1][j - 1];
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}
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answer[i][i] = 1;
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}
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return answer;
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}
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};
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``` |