# 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. 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.

# Solution

Approach 1: O(mlogn).

Approach 2: O(m+n).

Approach 3: O(logm+logn).

Approach 4: O(log4_(m*n))?

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

// O(mlogn)
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        for (int i = 0; i < m; ++i)
        {
            int lo = 0;
            int hi = matrix[i].size() - 1;
            while (lo <= hi)
            {
                int mid = lo + (hi - lo) / 2;
                if (target < matrix[i][mid])
                    hi = mid - 1;
                else if (target > matrix[i][mid])
                    lo = mid + 1;
                else
                    return true;
            }
        }
        return false;
    }
};

Approach 2:

// O(m+n)
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if (m == 0) return false;
        int n = matrix[0].size();
        if (n == 0) return false;
        int i = 0;
        int j = n - 1;
        while (i < m && j >= 0)
        {
            if (target < matrix[i][j])
                j--;
            else if (target > matrix[i][j])
                i++;
            else
                return true;
        }
        return false;
    }
};

Approach 3:

// O(logm+logn)
class Solution {
private:
    int binarySearchCol(vector<vector<int>>& matrix, int i, int j, int target) {
        int lo = 0;
        int hi = j;
        while (lo <= hi)
        {
            int mid = lo + (hi - lo) / 2;
            if (target < matrix[i][mid])
                hi = mid - 1;
            else if (target > matrix[i][mid])
                lo = mid + 1;
            else
                return mid;
        }
        return lo - 1;
    }
    int binarySearchRow(vector<vector<int>>& matrix, int i, int j, int target) {
        int lo = 0;
        int hi = i;
        while (lo <= hi)
        {
            int mid = lo + (hi - lo) / 2;
            if (target < matrix[mid][j])
                hi = mid - 1;
            else if (target > matrix[mid][j])
                lo = mid + 1;
            else
                return mid;
        }
        return lo;
    }
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if (m == 0) return false;
        int n = matrix[0].size();
        if (n == 0) return false;
        int i = 0;
        int j = n - 1;
        while (i < m && j >= 0)
        {
            if (target < matrix[i][j])
                j = binarySearchCol(matrix, i, j, target);
            else if (target > matrix[i][j])
                i = binarySearchRow(matrix, i, j, target);
            else
                return true;
        }
        return false;
    }
};

Approach 4:

// O(log4_(m*n)?)
class Solution {
private:
    bool doSearchMatrix(vector<vector<int>>& matrix, int i1, int j1, int i2, int j2, int target) {
        if (i1 < 0 || i1 >= matrix.size() ||
            i2 < 0 || i2 >= matrix.size() ||
            j1 < 0 || j1 >= matrix[0].size() ||
            j2 < 0 || j2 >= matrix[0].size() ||
            target < matrix[i1][j1] || target > matrix[i2][j2])
        {
            return false;
        }
        if (target == matrix[i1][j1] || target == matrix[i2][j2])
            return true;
        int midI = (i1 + i2) / 2;
        int midJ = (j1 + j2) / 2;
        return
            doSearchMatrix(matrix, i1, j1, midI, midJ, target) ||
            doSearchMatrix(matrix, midI+1, j1, i2, midJ, target) ||
            doSearchMatrix(matrix, i1, midJ+1, midI, j2, target) ||
            doSearchMatrix(matrix, midI+1, midJ+1, i2, j2, target);
    }
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if (m == 0) return false;
        int n = matrix[0].size();
        if (n == 0) return false;
        return doSearchMatrix(matrix, 0, 0, m-1, n-1, target);
    }
};