59. Spiral Matrix II LeetCode Solution

In this guide, you will get 59. Spiral Matrix II LeetCode Solution with the best time and space complexity. The solution to Spiral Matrix II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Spiral Matrix II solution in C++
Spiral Matrix II solution in Java
Spiral Matrix II solution in Python
Additional Resources

59. Spiral Matrix II LeetCode Solution image

Problem Statement of Spiral Matrix II

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

Example 1:

Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

1 <= n <= 20

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n^2)

59. Spiral Matrix II LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> generateMatrix(int n) {
    vector<vector<int>> ans(n, vector<int>(n));
    int count = 1;

    for (int mn = 0; mn < n / 2; ++mn) {
      const int mx = n - mn - 1;
      for (int i = mn; i < mx; ++i)
        ans[mn][i] = count++;
      for (int i = mn; i < mx; ++i)
        ans[i][mx] = count++;
      for (int i = mx; i > mn; --i)
        ans[mx][i] = count++;
      for (int i = mx; i > mn; --i)
        ans[i][mn] = count++;
    }

    if (n % 2 == 1)
      ans[n / 2][n / 2] = count;

    return ans;
  }
};
/* code provided by PROGIEZ */

59. Spiral Matrix II LeetCode Solution in Java

class Solution {
  public int[][] generateMatrix(int n) {
    int[][] ans = new int[n][n];
    int count = 1;

    for (int mn = 0; mn < n / 2; ++mn) {
      final int mx = n - mn - 1;
      for (int i = mn; i < mx; ++i)
        ans[mn][i] = count++;
      for (int i = mn; i < mx; ++i)
        ans[i][mx] = count++;
      for (int i = mx; i > mn; --i)
        ans[mx][i] = count++;
      for (int i = mx; i > mn; --i)
        ans[i][mn] = count++;
    }

    if (n % 2 == 1)
      ans[n / 2][n / 2] = count;

    return ans;
  }
}
// code provided by PROGIEZ

59. Spiral Matrix II LeetCode Solution in Python

class Solution:
  def generateMatrix(self, n: int) -> list[list[int]]:
    ans = [[0] * n for _ in range(n)]
    count = 1

    for mn in range(n // 2):
      mx = n - mn - 1
      for i in range(mn, mx):
        ans[mn][i] = count
        count += 1
      for i in range(mn, mx):
        ans[i][mx] = count
        count += 1
      for i in range(mx, mn, -1):
        ans[mx][i] = count
        count += 1
      for i in range(mx, mn, -1):
        ans[i][mn] = count
        count += 1

    if n % 2 == 1:
      ans[n // 2][n // 2] = count

    return ans
 # code by PROGIEZ