Search a 2D Matrix LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Search a 2D Matrix in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemSearch a 2D Matrix– LeetCode Problem

Search a 2D Matrix– LeetCode Problem


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 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


  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 100
  • -104 <= matrix[i][j], target <= 104
Search a 2D Matrix– LeetCode Solutions
Search a 2D Matrix in C++:
class Solution {
  bool searchMatrix(vector<vector<int>>& matrix, int target) {
    if (matrix.empty())
      return false;

    const int m = matrix.size();
    const int n = matrix[0].size();

    int l = 0;
    int r = m * n;

    while (l < r) {
      const int mid = l + (r - l) / 2;
      const int i = mid / n;
      const int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
        r = mid;

    return false;
Search a 2D Matrix in Java:
class Solution {
  public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix.length == 0)
      return false;

    final int m = matrix.length;
    final int n = matrix[0].length;

    int l = 0;
    int r = m * n;

    while (l < r) {
      final int mid = l + (r - l) / 2;
      final int i = mid / n;
      final int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
        r = mid;

    return false;
Search a 2D Matrix in Python:
class Solution:
  def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
    if not matrix:
      return False

    m = len(matrix)
    n = len(matrix[0])

    l = 0
    r = m * n

    while l < r:
      mid = l + (r - l) // 2
      i = mid // n
      j = mid % n
      if matrix[i][j] == target:
        return True
      if matrix[i][j] < target:
        l = mid + 1
        r = mid

    return False

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