Intuition

The problem is to generate the first n rows of Pascals triangle, where each element is the sum of the two elements above it. One possible way to solve this problem is to use a list of lists (or in other words, a 2D array) to store the rows, and iterate from the first row to the nth row, adding new elements based on the previous row.

Approach

• Initialize an empty list of lists to store the rows.
• Add the first row, which is a list containing only 1, to the list of rows.
• For each row from 1 to n-1, do the following:
  • Initialize an empty list to store the current row.
  • Add 1 to the beginning and end of the current row.
  • For each element from 1 to i-1, where i is the index of the current row, do the following:
    • Get the previous row from the list of rows.
    • Add the element at index j-1 and j from the previous row, and append it to the current row.
  • Add the current row to the list of rows.
• Return the list of rows.

Complexity

• Time complexity: O(n^2)

We need to iterate over n rows, and for each row i, we need to iterate over i elements. The total number of elements is n(n+1) / 2, which is O(n^2) in asymptotic notation.

• Space complexity: O(n^2)

We need to store n rows, and each row i has i elements. The total space required is n(n+1) / 2, which is O(n^2) in asymptotic notation.

Code

public class Solution {    public static List<List<Integer>> generate(int numRows) {        List<List<Integer>> allRows = new ArrayList<>(numRows);        List<Integer> firstRow = new ArrayList<>();        firstRow.add(1);        allRows.add(firstRow);        for (int i = 1; i < numRows; i++) {            List<Integer> row = new ArrayList<>(i + 1);            row.add(1);            List<Integer> prevRow = allRows.get(i - 1);            for (int j = 1; j < i; j++) {                int sum = 0;                if (j - 1 >= 0 && j - 1 < prevRow.size()) {                    sum += prevRow.get(j - 1);                }                if (j >= 0 && j < prevRow.size()) {                    sum += prevRow.get(j);                }                row.add(sum);            }            row.add(1);            allRows.add(row);        }        return allRows;    }}

For a more in-depth explanation, please view my LeetCode post: https://leetcode.com/problems/pascals-triangle/solutions/3563758/pascal-s-triangle-extremely-simple-logic-solution-beats-96-of-java-solutions-in-memory-runtime/