Leetcode Search-a-2D-Matrix

2022-06-13 15:33

Algorithms:

#algorithm #binary_search

Data structures:

#DS #vector #vector_2d

Difficulty:

#leetcode #coding_problem #difficulty-easy

tag:#vector OR tag:#vector_2d OR tag:#binary_search

Problem

Write an efficient algorithm that searches for a value target in an m x n integer matrix 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.

Examples

Example 1:

**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
**Output:** true

Example 2:

**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
**Output:** false

Constraints

m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-104 <= matrix[i][j], target <= 104

Thoughts

Binary search algorithm, with simple 2-d to 1-d conversion

[!tip] Binary search while loops In binary search, remember to check for:

l <= r And use:

l = mid + 1

r = mid - 1

Solution

==note how I calculated the mid, i, j, and how I changed r and l==

class Solution {
public:
  bool searchMatrix(vector<vector<int>> &matrix, int target) {
    int m = matrix.size();
    int n = matrix[0].size();
    int l = 0;
    int r = m * n - 1;
    int i; // the key to speed is to precaculate i and j, to save time.
    int j;
    int mid;

    do {
      mid = l + (r - l) / 2;
      i = mid / n;
      j = mid % n;
      if (target == matrix[i][j]) {
        return true;
      } else if (target < matrix[i][j]) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    } while (l <= r);

    return false;
  }
};

2.2 KiB Raw Blame History