### Problem statement

Given an integer **numRows**, return the first numRows of **Pascal's triangle**.

In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:

Problem statement taken from: https://leetcode.com/problems/pascals-triangle

**Example 1:**

```
Input: numRows = 5
Output: [ [1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1] ]
```

**Example 2:**

```
Input: numRows = 1
Output: [[1]]
```

**Constraints:**

```
- 1 <= numRows <= 30
```

### Explanation

#### Brute force approach

A simple method is to run two loops and calculate the value of Binominal Coefficient in the inner loop.

For example, the first line has **1**, the second line has **1 1**, the third line has **1 2 1**,.. and so on. Every entry in a line is the value of a Binomial Coefficient. The value of the ith entry in line number line is C(line, i). The value can be calculated using the following formula.

```
C(line, i) = line! / ( (line-i)! * i! )
```

A small C++ snippet of the above logic is:

```
void printPascal(int n)
{
for (int line = 0; line < n; line++){
for (int i = 0; i <= line; i++)
cout <<" "<< binomialCoefficient(line, i);
cout <<"\n";
}
}
int binomialCoefficient(int n, int k)
{
int result = 1;
if (k > n - k)
k = n - k;
for (int i = 0; i < k; ++i){
result *= (n - i);
result /= (i + 1);
}
return result;
}
```

Since we are generating a coefficient for each iteration the

time complexity of the above problem is **O(N^3)**.

#### Optimized solution (O(N^2) time and O(N^2) extra space)

If we take a look at the Pascal triangle, we can see that every entry is the sum of the two values above it. So we created a 2D array that stores the previously generated

values.

A small C++ snippet of the above logic is:

```
for (int line = 0; line < n; line++) {
for (int i = 0; i <= line; i++) {
if (line == i || i == 0)
arr[line][i] = 1;
else
arr[line][i] = arr[line - 1][i - 1] + arr[line - 1][i];
cout << arr[line][i] << " ";
}
cout << "\n";
}
```

#### Optimized solution (O(N^2) time and O(1) extra space)

This approach is based on the Brute force approach. The binomial coefficient of *ith* entry can be represented as **C(line, i)** and all lines start with value 1. The idea here is to calculate **C(line, i)** using **C(line, i - 1)**. It can be calculated in O(1) time using the following.

```
C(line, i) = line! / ( (line - i)! * i! )
C(line, i - 1) = line! / ( (line - i + 1)! * (i - 1)! )
So using the above approach we can change the formula as below:
C(line, i) = C(line, i - 1) * (line - i + 1) / i
C(line, i) can be calculated from C(line, i - 1) in O(1) time.
```

Let's check the algorithm:

```
- initialize vector<vector<int>> result
- loop for line = 1; line <= n; line++
- initialize vector<int> temp
- set C = 1
- loop for i = 1; i <= line; i++
- temp.push_back(C)
- C = C * (line - i) / i
- result.push_back(temp)
- return result
```

##### C++ Solution

```
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int>> result;
for (int line = 1; line <= numRows; line++){
vector<int> temp;
int C = 1;
for (int i = 1; i <= line; i++){
temp.push_back(C);
C = C * (line - i) / i;
}
result.push_back(temp);
}
return result;
}
};
```

##### Golang Solution

```
func generate(numRows int) [][]int {
var result [][]int
for line := 1; line <= numRows; line++ {
var temp []int
C := 1
for i := 1; i <= line; i++ {
temp = append(temp, C);
C = C * (line - i) / i;
}
result = append(result, temp)
}
return result
}
```

##### Javascript solution

```
var generate = function(numRows) {
var result = [];
for(let line = 1; line <= numRows; line++){
var temp = [];
let C = 1;
for(let i = 1; i <= line; i++){
temp.push(C);
C = C * (line - i) / i;
}
result.push(temp);
}
return result;
};
```

Let's dry-run our algorithm to see how the solution works.

```
Input: numRows = 3
Step 1: initialize vector<vector<int>> result
Step 2: loop for line = 1; line <= numRows
1 <= 3
true
initialize vector<int> temp
C = 1
loop for i = 1; i <= line
1 <= 1
true
temp.push_back(C);
temp = [1]
C = C * (line - i) / i;
C = 1 * (1 - 1) / 1
C = 0
i++
i = 2
loop for i <= line
2 <= 1
false
result.push_back(temp)
result = [[1]]
line++
line = 2
Step 3: loop for line <= numRows
2 <= 3
true
initialize vector<int> temp
C = 1
loop for i = 1; i <= line
1 <= 2
true
temp.push_back(C);
temp = [1]
C = C * (line - i) / i
C = 1 * (2 - 1) / 1
C = 1 * 1 / 1
i++
i = 2
loop for i <= line
2 <= 2
true
loop for i <= line
2 <= 2
true
temp.push_back(C);
temp = [1, 1]
C = C * (line - i) / i
C = 1 * (2 - 2) / 1
C = 1 * 0 / 1
C = 0
i++
i = 3
loop for i <= line
3 <= 2
false
result.push_back(temp)
result = [[1], [1, 1]]
line++
line = 3
Step 4: loop for line <= numRows
3 <= 3
true
initialize vector<int> temp
C = 1
loop for i = 1; i <= line
1 <= 3
true
temp.push_back(C);
temp = [1]
C = C * (line - i) / i
C = 1 * (3 - 1) / 1
C = 1 * 2 / 1
C = 2
i++
i = 2
loop for i <= line
2 <= 3
true
temp.push_back(C);
temp = [1, 2]
C = C * (line - i) / i
C = 2 * (3 - 2) / 2
C = 2 * 1 / 2
C = 1
i++
i = 3
loop for i <= line
3 <= 3
true
temp.push_back(C);
temp = [1, 2, 1]
C = C * (line - i) / i
C = 1 * (3 - 3) / 3
C = 1 * 0 / 3
C = 0
i++
i = 4
loop for i <= line
4 <= 3
false
result.push_back(temp)
result = [[1], [1, 1], [1, 2, 1]]
line++
line = 4
Step 5: loop for line <= numRows
4 <= 3
false
Step 6: return result
So the result is [[1], [1, 1], [1, 2, 1]].
```

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