Search a 2D Matrix II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Search a 2D Matrix II LeetCode Solution

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

Java Search a 2D Matrix II LeetCode Solution

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length < 1 || matrix[0].length <1) {
            return false;
        }
        int col = matrix[0].length-1;
        int row = 0;
        while(col >= 0 && row <= matrix.length-1) {
            if(target == matrix[row][col]) {
                return true;
            } else if(target < matrix[row][col]) {
                col--;
            } else if(target > matrix[row][col]) {
                row++;
            }
        }
        return false;
    }
}
 

Python 3 Search a 2D Matrix II LeetCode Solution

def searchMatrix(matrix, target):
	m, n = len(matrix), len(matrix) and len(matrix[0])
	r, c = 0, n-1
	while r < m and c >= 0:
		if target > matrix[r][c]:
			r += 1
		elif target < matrix[r][c]:
			c -= 1
		else: return True
	return False

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

Leave a comment below

Related posts:

Search a 2D Matrix – LeetCode Solution Search in Rotated Sorted Array – LeetCode Solution Search in Rotated Sorted Array II – LeetCode Solution Word Search – LeetCode Solution Convert Sorted Array to Binary Search Tree – LeetCode Solution Word Search II – LeetCode Solution