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Leetcode 119. Pascals Triangle II

Problem Statement

Given an integer rowIndex, return the rowIndex-th (0-indexed) row of the Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two directly above it as shown:

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

Clarifying Questions

  1. Constraints:
    • What is the range of rowIndex?
      • Typically, in LeetCode, the constraints are 0 <= rowIndex <= 33.
  2. Output Format:
    • Should the function return the resulting row as a vector of integers?
      • Yes, a vector of integers representing the specified row of Pascal’s triangle.

Strategy

To compute the rowIndex-th row of Pascal’s triangle efficiently:

  1. Initialization:
    • Start with the first row [1].
  2. Iteratively Generate Rows:
    • Use the row from the previous iteration to generate the next row.
  3. Direct Calculations:
    • Without constructing the entire triangle, we can use a single vector and iteratively update it to form each subsequent row.
  4. Update Mechanism:
    • For each row, starting from the end towards the beginning (to avoid overwriting values too early).
  5. Memory Efficiency:
    • By using a single vector and updating it in place, the memory usage remains minimal.

Code

Here’s a C++ function implementing the described strategy:

#include <vector>

std::vector<int> getRow(int rowIndex) {
    // Create a vector initialized with 0s except for the first element.
    std::vector<int> row(rowIndex + 1, 0);
    row[0] = 1;  // The first element is always 1

    // Compute the rowIndex-th row
    for (int i = 1; i <= rowIndex; ++i) {
        // Update the row from the end to the beginning
        for (int j = i; j > 0; --j) {
            row[j] += row[j - 1];
        }
    }

    return row;
}

Explanation

  1. Initialization:
    • A vector row of size rowIndex + 1 is initialized with zeros, and the first element is set to 1.
  2. Iterative Updates:
    • The outer loop runs from 1 to rowIndex, representing the current row being computed.
    • The inner loop updates the vector from the end to the beginning. This ensures that the updates within the same row do not interfere with each other.

Time Complexity

This efficient approach ensures that we generate the desired row of Pascal’s triangle without constructing the entire triangle.

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