119. Pascal’s Triangle II LeetCode Solution

In this guide, you will get 119. Pascal’s Triangle II LeetCode Solution with the best time and space complexity. The solution to Pascal’s Triangle II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Pascal’s Triangle II solution in C++
Pascal’s Triangle II solution in Java
Pascal’s Triangle II solution in Python
Additional Resources

119. Pascal's Triangle II LeetCode Solution image

Problem Statement of Pascal’s Triangle II

Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal’s triangle.
In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:
Input: rowIndex = 3
Output: [1,3,3,1]
Example 2:
Input: rowIndex = 0
Output: [1]
Example 3:
Input: rowIndex = 1
Output: [1,1]

Constraints:

0 <= rowIndex <= 33

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n)

119. Pascal’s Triangle II LeetCode Solution in C++

class Solution {
 public:
  vector<int> getRow(int rowIndex) {
    vector<int> ans(rowIndex + 1, 1);

    for (int i = 2; i < rowIndex + 1; ++i)
      for (int j = 1; j < i; ++j)
        ans[i - j] += ans[i - j - 1];

    return ans;
  }
};
/* code provided by PROGIEZ */

119. Pascal’s Triangle II LeetCode Solution in Java

class Solution {
  public List<Integer> getRow(int rowIndex) {
    Integer[] ans = new Integer[rowIndex + 1];
    Arrays.fill(ans, 1);

    for (int i = 2; i < rowIndex + 1; ++i)
      for (int j = 1; j < i; ++j)
        ans[i - j] += ans[i - j - 1];

    return Arrays.asList(ans);
  }
}
// code provided by PROGIEZ

119. Pascal’s Triangle II LeetCode Solution in Python

class Solution:
  def getRow(self, rowIndex: int) -> list[int]:
    ans = [1] * (rowIndex + 1)

    for i in range(2, rowIndex + 1):
      for j in range(1, i):
        ans[i - j] += ans[i - j - 1]

    return ans
 # code by PROGIEZ