0074 - Search a 2D Matrix (Medium)

Problem Link

https://leetcode.com/problems/search-a-2d-matrix/

Problem Statement

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.

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
• -10^4 <= matrix[i][j], target <= 10^4

Approach 1: Searching Row & Column

C++

Written by @wingkwong

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int n = matrix.size(), m = matrix[0].size();
        int targetRow = 0;
        // search for the target row
        for (int row = 0; row < n; row++) {
            // target should be within [matrix[row][0] .. matrix[row][m - 1]]
            if (matrix[row][0] <= target && target <= matrix[row][m - 1]) {
                // target row is found
                targetRow = row;
                break;
            }
        }
        // then search for the target col
        for (int col = 0; col < m; col++) {
            if (matrix[targetRow][col] == target) {
                return true;
            }
        }
        return false;
    }
};

Python

Written by @radojicic23

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        rows, cols = len(matrix), len(matrix[0])
        r = 0
        for i in range(rows):
            if target >= matrix[i][0] and target <= matrix[i][-1]:
                r = i
                break
        for i in range(cols):
            if (matrix[r][i] == target):
                return True
        return False

JavaScript

Written by @radojicic23

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function(matrix, target) {
    let rows = matrix.length, cols = matrix[0].length;
    let r = 0;
    for (let i = 0; i < rows; i++) {
        if (target >= matrix[i][0] && target <= matrix[i][cols - 1]) {
            r = i;
            break;
        }
    }
    for (let i = 0; i < cols; i++) {
        if (matrix[r][i] == target) {
            return true;
        }
    }
    return false;
};

Approach 2: Binary Search

Find which row and coloumn the element belongs to by using Binary Search

Java

Written by @ganajayant

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for (int i = 0; i < matrix.length; i++) {
            int low = 0;
            int high = matrix[i].length - 1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (matrix[i][mid] == target) {
                    return true;
                } else if (matrix[i][mid] > target) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
        }
        return false;
    }
}

Python

Written by @radojicic23

# Time Complexity: O(log n)
# Space Complexity: O(1)
class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        # initialize rows and cols 
        rows, cols = len(matrix), len(matrix[0]) 
        # top row and bottom row
        top, bot = 0, rows - 1
        
        # binary search
        while top <= bot: 
            # compute the middle row
            mid = (top + bot) // 2
            # if this target value is greater then 
            # the largest value in the middle row
            if target > matrix[mid][-1]:
                # look at rows with larger value
                top = mid + 1
            # if this target value is smaller then 
            # the smallest value in this row
            elif target < matrix[mid][0]:
                # look at rows with smaller value
                bot = mid - 1
            else:
                break 
            
        if not (top <= bot):
            return False
        
        # second binary search portion
        # run binary search on the current (middle row)
        mid = (top + bot) // 2
        # leftmost value and rightmost value
        l, r = 0, cols - 1
        while l <= r:
            # compute the middle value
            m = (l + r) // 2
            if target > matrix[mid][m]:
                l = m + 1
            elif target < matrix[mid][m]:
                r = m - 1
            else:
                return True
        
        return False

JavaScript

Written by @radojicic23

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function(matrix, target) {
    let row = matrix.length;
    let col = matrix[0].length;
    let top = 0;
    let bottom = row - 1;
    while (top <= bottom) {
        let mid = Math.floor((top + bottom) / 2);
        if (target > matrix[mid][col - 1]) {
            top = mid + 1;
        } else if (target < matrix[mid][0]) {
            bottom = mid - 1;
        } else {
            break;
        }
    }
    if (!(top <= bottom)) {
        return false;
    }
    let l = 0;
    let r = col - 1;
    let mid = Math.floor((top + bottom) / 2);
    while (l <= r) {
        let m = Math.floor((l + r) / 2);
        if (target > matrix[mid][m]) {
            l = m + 1;
        } else if (target < matrix[mid][m]) {
            r = m - 1;
        } else {
            return true;
        }
    }
    return false;
};

C++

Written by @radojicic23

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int rows = matrix.size(), cols = matrix[0].size();
        int top = 0, bottom = rows - 1;
        while (top <= bottom) {
            int mid = top + (bottom - top) / 2;
            if (target > matrix[mid][cols - 1]) {
                top = mid + 1;
            } else if (target < matrix[mid][0]) {
                bottom = mid - 1;
            } else {
                break;
            }
        }
        if (!(top <= bottom)) {
            return false;
        }
        int mid = top + (bottom - top) / 2;
        int left = 0, right = cols - 1;
        while (left <= right) {
            int m = left + (right - left) / 2;
            if (target > matrix[mid][m]) {
                left = m + 1;
            } else if (target < matrix[mid][m]) {
                right = m - 1;
            } else {
                return true;
            }
        }
        return false;
    }
};