# Pascal’s Triangle – Extremely Simple Logic Solution Beats 96% of Java Solutions in Memory & Runtime

May 26, 2023

The problem is to generate the first n rows of Pascal’s 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.

• 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.

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.

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.

public class Solution {
public static List<List<Integer>> generate(int numRows) {
List<List<Integer>> allRows = new ArrayList<>(numRows);
List<Integer> firstRow = new ArrayList<>();
for (int i = 1; i < numRows; i++) {
List<Integer> row = new ArrayList<>(i + 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);
}