22_row-with-minimum-number-of-1s.md

Row with minimum number of 1's

Difficulty: Easy

Points: 2

Given a 2D binary matrix (1-based indexed) a of dimensions nxm , determine the row that contains the minimum number of 1's .
Note: The matrix contains only 1's and 0's . Also, if two or more rows contain the minimum number of 1's , the answer is the lowest of those indices .

Example 1:

Input:
n = 4,m = 4
a = [[1, 1, 1, 1],
[1, 1, 0, 0],
[0, 0, 1, 1],
[1, 1, 1, 1]]
Output:
2
Explanation:
Rows 2 and 3 contain the minimum number
of 1's(2 each).Since,row 2 is less than row 3.
Thus, the answer is 2.

Example 2:

Input:
n = 3,m = 3
a = [[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
Output:
1
Explanation:
All the rows contain the same number
of 1's(0 each).Among them, index 1
is the smallest, so the answer is 1.

Your Task:
You don't need to read input or print anything. Your task is to complete the function minRow() which takes the two integers n , m as well as the 2D binary matrix a as input parameters and returns the minimum index of the row which contains the least number of 1's .

Expected Time Complexity: O(n*m)
Expected Auxillary Space: O(1)

Constraints:
1 <= n,m <= 1000

0 <= a[i][j] <= 1

Company Tags

Amazon

Topic Tags

Matrix Data Structures