119. 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

Recursive Solution

1. What we know?
   • We Know that, if rowIndex==0 (first row) then we only have {1} in the row.
   • We also know that first and last element of our row will be 1 always.
   • We also Know that we can find currentRow[i] by adding previousRow[i] and previousRow[i+1].
2. Data for recursion,
   • from above discussion, now we can get 2 important things for recursion
     1. Base Case: if(rowIndex==0) return {1};
     2. Recurrence Relation: currentRow[i] = (previousRow[i-1]+previousRow[i]);

        Before reading out please give it a try now.

3. Algorithm
   • We can start with currentRow[0] = 1
   • Now believe that, we get the previous row with recursive call previousRow = getRow(rowIndex-1);
   • Then fill the remaining elements of current row with recurrence relation currentRow[i] = (previousRow[i-1]+previousRow[i]);
   • As we know, last element of row is 1, push 1 in the vector.
   • Return the current row i.e. currentRow.

Code

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        if(rowIndex==0) return {1}; // Base Case

        vector<int> currentRow = {1}; // current row with 1 value in it
        vector<int> previousRow = getRow(rowIndex-1); // get the previous row

        // Now fill the current row based on previous row
        for(int i=1;i<rowIndex;i++){
            currentRow.push_back(previousRow[i-1]+previousRow[i]);
        }

        currentRow.push_back(1); // fill the last element of current row
        return currentRow;
    }
};