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240. Search a 2D Matrix II

Description

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

 

Example 1:

Input: 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]], target = 5
Output: true

Example 2:

Input: 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]], target = 20
Output: false

 

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= n, m <= 300
  • -109 <= matrix[i][j] <= 109
  • All the integers in each row are sorted in ascending order.
  • All the integers in each column are sorted in ascending order.
  • -109 <= target <= 109

Solutions

Since all elements in each row are sorted in ascending order, we can use binary search to find the first element that is greater than or equal to target for each row, and then check if this element is equal to target. If it equals target, it means the target value has been found, and we directly return true. If it does not equal target, it means all elements in this row are less than target, and we should continue to search the next row.

If all rows have been searched and the target value has not been found, it means the target value does not exist, so we return false.

The time complexity is $O(m \times \log n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively. The space complexity is $O(1)$.

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class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        for row in matrix:
            j = bisect_left(row, target)
            if j < len(matrix[0]) and row[j] == target:
                return True
        return False

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class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for (var row : matrix) {
            int j = Arrays.binarySearch(row, target);
            if (j >= 0) {
                return true;
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        for (auto& row : matrix) {
            int j = lower_bound(row.begin(), row.end(), target) - row.begin();
            if (j < matrix[0].size() && row[j] == target) {
                return true;
            }
        }
        return false;
    }
};

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func searchMatrix(matrix [][]int, target int) bool {
    for _, row := range matrix {
        j := sort.SearchInts(row, target)
        if j < len(matrix[0]) && row[j] == target {
            return true
        }
    }
    return false
}

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function searchMatrix(matrix: number[][], target: number): boolean {
    const n = matrix[0].length;
    for (const row of matrix) {
        let left = 0,
            right = n;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (row[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        if (left != n && row[left] == target) {
            return true;
        }
    }
    return false;
}

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use std::cmp::Ordering;

impl Solution {
    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut i = 0;
        let mut j = n;
        while i < m && j > 0 {
            match target.cmp(&matrix[i][j - 1]) {
                Ordering::Less => {
                    j -= 1;
                }
                Ordering::Greater => {
                    i += 1;
                }
                Ordering::Equal => {
                    return true;