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Posts tagged as “matrix”

花花酱 LeetCode 1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:

Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0

Example 3:

Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3

Example 4:

Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

Constraints:

  • 1 <= m, n <= 300
  • m == mat.length
  • n == mat[i].length
  • 0 <= mat[i][j] <= 10000
  • 0 <= threshold <= 10^5

Solution: DP + Brute Force

Precompute the sums of sub-matrixes whose left-top corner is at (0,0).

Try all possible left-top corner and sizes.

Time complexity: O(m*n*min(m,n))
Space complexity: O(m*n)

C++

Solution 2: Binary Search

Search for the smallest size k that is greater than the threshold, ans = k – 1.

C++

Solution 3: Bounded Search

Time complexity: O(m*n + min(m,n))

C++

花花酱 LeetCode 48. Rotate Image

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

Solution: 2 Passes

First pass: mirror around diagonal
Second pass: mirror around y axis

Time complexity: O(n^2)
Space complexity: O(1)

C++

花花酱 LeetCode 59. Spiral Matrix II

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

Example:

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

Solution: Simulation

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

C++

花花酱 LeetCode 54. Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

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

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Solution: Simulation

Keep track of the current bounds (left, right, top, bottom).

Init: left = 0, right = n – 1, top = 0, bottom = m – 1

Each time we move in one direction and shrink the bounds and turn 90 degrees:
1. go right => –top
2. go down => –right
3. go left => ++bottom
4. go up => ++left

C++

花花酱 LeetCode 74. Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

  • Integers in each row are sorted from left to right.
  • The first integer of each row is greater than the last integer of the previous row.

Example 1:

Example 2:

Solution: Binary Search

Treat the 2D array as a 1D array. matrix[index / cols][index % cols]

Time complexity: O(log(m*n))
Space complexity: O(1)

C++