Pascal’s Triangle II LeetCode Solution | Easy Approach

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

Pascal’s Triangle II Solutions

✅Time: O(n)
✅Space: O(n)

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;
  }
};

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);
  }
}

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

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