240. Search a 2D Matrix II


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 in ascending from left to right.
  • Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix: 

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.


Approach #1 Brute Force [Accepted]

Intuition

As a baseline, we can search the 2D array the same way we might search an unsorted 1D array -- by examining each element.

Algorithm

The algorithm doesn't really do anything more clever than what is explained by the intuition; we loop over the array, checking each element in turn. If we find it, we return true. Otherwise, if we reach the end of the nested for loop without returning, we return false. The algorithm must return the correct answer in all cases because we exhaust the entire search space.

Complexity Analysis

  • Time complexity :

    Becase we perform a constant time operation for each element of an element matrix, the overall time complexity is equal to the size of the matrix.

  • Space complexity :

    The brute force approach does not allocate more additional space than a handful of pointers, so the memory footprint is constant.

Approach #2 Search Space Reduction [Accepted]

Intuition

Because the rows and columns of the matrix are sorted (from left-to-right and top-to-bottom, respectively), we can prune one row or one column when looking at any particular value.

Algorithm

First, we initialize a pointer to the bottom-left of the matrix.1 Then, until we find target and return true (or the pointer points to a that lies outside of the dimensions of the matrix), we do the following: if the currently-pointed-to value is larger than target we can move one row "up". Otherwise, if the currently-pointed-to value is smaller than target, we can move one column "right". It is not too tricky to see why doing this will never prune the correct answer; because the rows are sorted from left-to-right, we know that every value to the right of the current value is larger. Therefore, if the current value is already larger than target, we know that every value to its right will also be too large. A very similar argument can be made for the columns, so this manner of search will always find target in the matrix (if it is present).

Check out some sample runs of the algorithm in the animation below:

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Complexity Analysis

  • Time complexity :

    The key to the time complexity analysis is noticing that, on every iteration (during which we do not return true) either row or col is is decremented/incremented exactly once. Because row can only be decremented times and col can only be incremented times before causing the while loop to terminate, the loop cannot run for more than iterations. Because all other work is constant, the overall time complexity is linear in the sum of the dimensions of the matrix.

  • Space complexity :

    Because this approach only manipulates a few pointers, its memory footprint is constant.

Analysis and solutions written by: @emptyset

Footnotes

  1. This would work equally well with a pointer initialized to the top-right. Neither of the other two corners would work, as pruning a row/column might prevent us from achieving the correct answer. ↩

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