You have been given a 2-D array 'mat' of size 'M x N' where 'M' and 'N' denote the number of rows and columns, respectively. The elements of each row are sorted in non-decreasing order.
Moreover, the first element of a row is greater than the last element of the previous row (if it exists).
You are given an integer ‘target’, and your task is to find if it exists in the given 'mat' or not.
Input: ‘M’ = 3, 'N' = 4, ‘mat’ = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]], ‘target’ = 8
Output: true
Explanation: The output should be true as '8' exists in the matrix.
The first line contains three space-separated integers 'M', 'N', and ‘target’ where 'M' and 'N' denote the number of rows and columns of the 'mat', respectively, and ‘target’ is the integer to be found.
From the second line, the next 'M' lines represent the rows of the 'mat'. Every row contains 'N' single space-separated integers.
Output Format :
Return 'true' if ‘target’ is present in the 'mat'; else, return 'false'.
Note :
You do not need to print anything. It has already been taken care of. Just implement the function.
Follow-Up:
Can you solve this problem in less than O(M*N) time complexity?
3 4 8
1 2 3 4
5 6 7 8
9 10 11 12
true
The ‘target’ = 8 exists in the 'mat' at index (1, 3).
3 3 78
1 2 4
6 7 8
9 10 34
false
The ‘target' = 78 does not exist in the 'mat'. Therefore in the output, we see 'false'.
1 <= N <= 50
1 <= M <= 50
-10^5 <= mat[i], target <= 10^5
Time Limit: 1 sec
Naively search the MAT to find the TARGET.
We have a brute force solution to this problem. We can simply traverse the matrix and check if ‘TARGET’ exists or not.
O(M*N) where ‘M’ and ‘N’ denote the number of rows and columns in ‘MAT’, respectively.
In the worst case, we will be traversing the entire matrix of M*N elements.
O(1)
Since, we are not using any extra space to find the number of pairs. Therefore, the overall space complexity will be O(1).