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74. Search a 2D Matrix

DifficultyMedium

Description

You are given an m x n integer matrix matrix with the following two properties:

  • Each row is sorted in non-decreasing order.
  • The first integer of each row is greater than the last integer of the previous row.

Given an integer target, return true if target is in matrix or false otherwise.

You must write a solution in O(log(m * n)) time complexity.

 

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

Solutions

Thinking

A full scan is \(O(mn)\). \(m,n \le 100\) would pass, but the problem asks for \(O(\log(mn))\).

Two dimensions hide a single binary search. Each row increases, and the last of row \(i\) is less than the first of row \(i+1\), so concatenating rows is a sorted array of length \(mn\). Logical index \(\textit{mid}\) maps to \(\textit{matrix}[\lfloor \textit{mid}/n\rfloor][\textit{mid} \bmod n]\), so we binary-search \([0,mn-1]\) without materializing the flatten.

We can logically unfold the two-dimensional matrix and then perform binary search.

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

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class Solution:
    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n - 1
        while left < right:
            mid = (left + right) >> 1
            x, y = divmod(mid, n)
            if matrix[x][y] >= target:
                right = mid
            else:
                left = mid + 1
        return matrix[left // n][left % n] == target

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class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
}

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class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = left + right >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
};

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func searchMatrix(matrix [][]int, target int) bool {
    m, n := len(matrix), len(matrix[0])
    left, right := 0, m*n-1
    for left < right {
        mid := (left + right) >> 1
        x, y := mid/n, mid%n
        if matrix[x][y] >= target {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return matrix[left/n][left%n] == target
}

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function searchMatrix(matrix: number[][], target: number): boolean {
    const m = matrix.length;
    const n = matrix[0].length;
    let left = 0;
    let right = m * n;
    while (left < right) {
        const mid = (left + right) >>> 1;
        const i = Math.floor(mid / n);
        const j = mid % n;
        if (matrix[i][j] === target) {
            return true;
        }

        if (matrix[i][j] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return false;
}

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impl Solution {
    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut left = 0;
        let mut right = m * n - 1;
        while left < right {
            let mid = (left + right) >> 1;
            let x = mid / n;
            let y = mid % n;
            if matrix[x][y] >= target {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        matrix[left / n][left % n] == target
    }
}

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/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
    const m = matrix.length,
        n = matrix[0].length;
    let left = 0,
        right = m * n - 1;
    while (left < right) {
        const mid = (left + right + 1) >> 1;
        const x = Math.floor(mid / n);
        const y = mid % n;
        if (matrix[x][y] <= target) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return matrix[Math.floor(left / n)][left % n] == target;
};

Solution 2: Search from the Bottom Left or Top Right

Thinking

Method 1 flattens the matrix and binary-searches, which is \(O(\log(mn))\), but it maps \(\textit{mid}\) to coordinates and needs the last of one row to be less than the first of the next. Without that index map, row/column monotonicity lets a corner walk drop a whole row or column per comparison. The code is simpler; the cost is \(O(m+n)\) time.

Here, we start searching from the bottom left corner and move towards the top right direction. We compare the current element \(\textit{matrix}[i][j]\) with \(\textit{target}\):

  • If \(\textit{matrix}[i][j] = \textit{target}\), we have found the target value and return true.
  • If \(\textit{matrix}[i][j] > \textit{target}\), all elements to the right of the current position in this row are greater than target, so we should move the pointer \(i\) upwards, i.e., \(i = i - 1\).
  • If \(\textit{matrix}[i][j] < \textit{target}\), all elements above the current position in this column are less than target, so we should move the pointer \(j\) to the right, i.e., \(j = j + 1\).

If we still can't find \(\textit{target}\) after the search, return false.

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

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class Solution:
    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        i, j = m - 1, 0
        while i >= 0 and j < n:
            if matrix[i][j] == target:
                return True
            if matrix[i][j] > target:
                i -= 1
            else:
                j += 1
        return False

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class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (matrix[i][j] == target) {
                return true;
            }
            if (matrix[i][j] > target) {
                --i;
            } else {
                ++j;
            }
        }
        return false;
    }
}

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

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func searchMatrix(matrix [][]int, target int) bool {
    m, n := len(matrix), len(matrix[0])
    for i, j := m-1, 0; i >= 0 && j < n; {
        if matrix[i][j] == target {
            return true
        }
        if matrix[i][j] > target {
            i--
        } else {
            j++
        }
    }
    return false
}

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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 = m as i32 - 1;
        let mut j: i32 = 0;
        while i >= 0 && j < n as i32 {
            let cur = matrix[i as usize][j as usize];
            if cur == target {
                return true;
            }
            if cur > target {
                i -= 1;
            } else {
                j += 1;
            }
        }
        false
    }
}

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

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