🚀Day 4. Search in a Row-Wise Sorted Matrix 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a row-wise sorted 2D matrix mat[][] of size n x m and an integer x, determine whether the element x is present in the matrix.

Note: A row-wise sorted matrix implies that each row is sorted in non-decreasing order.

🔍 Example Walkthrough:

Input:

mat[][] = [[3, 4, 9], [2, 5, 6], [9, 25, 27]]
x = 9

Output:

true

Explanation: 9 is present in the matrix, so the output is true.

Input:

mat[][] = [[19, 22, 27, 38, 55, 67]]
x = 56

Output:

false

Explanation: 56 is not present in the matrix.

Input:

mat[][] = [[1, 2, 9], [65, 69, 75]]
x = 91

Output:

false

Explanation: 91 is not present in the matrix.

Constraints

• $(1 \leq n, m \leq 1000)$

• $(1 \leq \text{mat}[i][j] \leq 10^5)$

• $(1 \leq x \leq 10^5)$

🎯 My Approach:

1. Iterate Through Rows: Loop through each row of the matrix. Since each row is sorted, we can efficiently determine if the target x exists using binary search.

2. Binary Search for Each Row: For each row:

   • Use the binary_search function (in C++) or equivalent methods in other languages to locate x.

   • The binary search divides the row into halves and compares the middle element with x:

     • If the middle element matches x, return true.

     • If the middle element is greater than x, continue searching in the left half of the row.

     • Otherwise, continue searching in the right half.

3. Final Check: If no row contains x after checking all rows, return false.

🕒 Time and Auxiliary Space Complexity

Expected Time Complexity:

• $(O(n \cdot \log m))$, where $(n)$ is the number of rows and $(m)$ is the number of columns. This is because for each row, a binary search runs in $(O(\log m))$.

Expected Auxiliary Space Complexity:

• $(O(1))$, as the binary search operates in constant space.

📝 Solution Code

Code (Cpp)

class Solution {
public:
    bool searchRowMatrix(vector<vector<int>>& mat, int x) {
        for (auto& row : mat) {
            if (binary_search(row.begin(), row.end(), x)) return true;
        }
        return false;
    }
};

Code (Java)

class Solution {
    public boolean searchRowMatrix(int[][] mat, int x) {
        for (int[] row : mat) {
            if (Arrays.binarySearch(row, x) >= 0) {
                return true;
            }
        }
        return false;
    }
}

Code (Python)

class Solution:
    def searchRowMatrix(self, mat, x):
        for row in mat:
            if x in row:
                return True
        return False

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, please visit my LinkedIn:- Any Questions. Thank you for your input; together, we strive to create a space where learning is a collaborative endeavor.

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