Solution For Pascals Triangle
Problem Statement:
Given a non-negative integer numRows, generate the first numRows of Pascal’s triangle.
Example:
Input: 5
Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
Solution:
The problem requires us to generate the first numRows of Pascal’s triangle. For this, we can iterate through each row from 0 to numRows and generate each row of the Pascal’s triangle.
The first and the last element of each row is always 1. For all other elements, we can calculate them based on the elements in the previous row. Specifically, the value of element i in row j is equal to the sum of element i-1 and i in row j-1.
The solution approach can be summarized in the following steps:
- Create an empty 2D list.
- Iterate through each row from 0 to numRows.
- If the current row is the first row, then add [1] to the 2D list.
- If the current row is not the first row:
a. Create an empty list.
b. Append 1 to the list.
c. Iterate through each element in the previous row (except the first and last elements):
i. Calculate the value of the current element using the formula mentioned above.
ii. Append the value to the list.
d. Append 1 to the list.
e. Add the list to the 2D list. - Return the 2D list.
Time Complexity:
The time complexity is O(numRows^2) because we are iterating through numRows for generating each row of the Pascal’s triangle.
Space Complexity:
The space complexity is also O(numRows^2) because we are storing all the elements of the Pascal’s triangle in a 2D list.
Python Code:
def generate(numRows: int) -> List[List[int]]:
triangle = []
for i in range(numRows):
row = []
if i == 0:
row = [1]
else:
prev_row = triangle[i-1]
row.append(1)
for j in range(1, i):
row.append(prev_row[j-1] + prev_row[j])
row.append(1)
triangle.append(row)
return triangle
Note:
Here, we are using type hints for function parameters and return values. Also, we are importing the List type from the typing module which is used for type hints.
Step by Step Implementation For Pascals Triangle
/* Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] */ class Solution { public List> generate(int numRows) { List
> triangle = new ArrayList
>(); // First row is always [1] triangle.add(new ArrayList<>()); triangle.get(0).add(1); for (int rowNum = 1; rowNum < numRows; rowNum++) { List
row = new ArrayList<>(); List prevRow = triangle.get(rowNum-1); // The first row element is always 1 row.add(1); // Each triangle element (other than the first and last of each row) // is equal to the sum of the elements above-and-to-the-left and // above-and-to-the-right. for (int j = 1; j < rowNum; j++) { row.add(prevRow.get(j-1) + prevRow.get(j)); } // The last row element is always 1 row.add(1); triangle.add(row); } return triangle; } }
# Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. # In Pascal's triangle, each number is the sum of the two numbers directly above it. # Example: # Input: 5 # Output: # [ # [1], # [1,1], # [1,2,1], # [1,3,3,1], # [1,4,6,4,1] # ] def generate(numRows): """ :type numRows: int :rtype: List[List[int]] """ # Base case: If numRows is 0 or 1, return empty or single-element list if numRows == 0: return [] elif numRows == 1: return [[1]] # Otherwise, initialize triangle with two rows triangle = [[1], [1, 1]] # Iteratively generate remaining rows for i in range(2, numRows): # Start with the list corresponding to the row above row = triangle[i-1] # Initialize current row with 1 in the first position current_row = [1] # Generate elements of current row by summing adjacent elements of the row above for j in range(len(row)-1): current_row.append(row[j] + row[j+1]) # Append 1 to the end of the current row current_row.append(1) # Append current row to triangle triangle.append(current_row) return triangle
var generate = function(numRows) { // Create an empty array of arrays to hold the pascal triangle let pascalTriangle = []; // For loop to iterate through each row of the triangle for(let i = 0; i < numRows; i++) { // Create an empty array to store the values in the current row let row = []; // For loop to iterate through each column of the current row for(let j = 0; j <= i; j++) { // If we are on the first or last column of the row, the value will be 1 if(j === 0 || j === i) { row.push(1); } else { // Otherwise, the value will be the sum of the values in the previous row at the current column and the column before it row.push(pascalTriangle[i-1][j] + pascalTriangle[i-1][j-1]); } } // Push the row array into the pascalTriangle array pascalTriangle.push(row); } // Return the pascalTriangle array return pascalTriangle; };
vector> generate(int numRows) { vector > result; if (numRows == 0) { return result; } result.push_back(vector (1, 1)); for (int i = 1; i < numRows; i++) { vector row(1, 1); for (int j = 1; j < i; j++) { row.push_back(result[i-1][j-1] + result[i-1][j]); } row.push_back(1); result.push_back(row); } return result; }
public IList> Generate(int numRows) { //Create an empty list of lists IList > triangle = new List >(); //For each row in the triangle for (int i = 0; i < numRows; i++) { //Create a new list IList row = new List (); //For each element in the row for (int j = 0; j <= i; j++) { //If the element is the first or last in the row if (j == 0 || j == i) { //Set the value to 1 row.Add(1); } else { //Set the value to the sum of the elements in the previous row row.Add(triangle[i-1][j-1] + triangle[i-1][j]); } } //Add the row to the triangle triangle.Add(row); } //Return the triangle return triangle; }