Array sum question

#1

my code-https://ide.geeksforgeeks.org/Pa50ARrsfY
Problem Statement
Given a N*M binary matrix that is, it contains 0s and 1s only, we need to find and return sum of coverage of all zeros of the input matrix. Coverage for a particular 0 is defined as, total number of ones around a zero (i.e. in left, right, up and bottom directions).
Input Format :
Line 1: N and M (space-separated positive integers)
Next N lines: M elements of each row (separated by space).
Output Format :
Line 1: Print Sum Of Coverage Of All Rows
Constraints :
1 <= N <= 10^3
1 <= M <= 10^3

Time Limit : 1 sec
Sample Input 1 :
2 2
1 0
0 1
Sample Output 1 :
4
Sample Input 1 Explanation :
The 0 in first row have 2 1’s i.e. left and down, whereas 0 at second row also have 2 1’s i.e. top and right. So the total coverage value is 2 + 2 = 4.

#2

why failing some testcases

#3


you have miss this line
The ones can be anywhere till corner point in a direction.
it is not necessary to have 1 in just left or just right
1 can be anywhere in right and anywhere in left
same for up and down

Input :

mat[][] ={         0 0 0 0
                   1 0 0 1
                   0 1 1 0
                   0 1 0 0}

Output : 20
First four zeros are surrounded by only
one 1. So coverage for zeros in first
row is 1 + 1 + 1 + 1
Zeros in second row are surrounded by
three 1’s. Note that there is no 1 above.
There are 1’s in all other three directions.
Coverage of zeros in second row = 3 + 3.
Similarly counting for others also, we get
overall count as below.
1 + 1 + 1 + 1 + 3 + 3 + 2 + 2 + 2 + 2 + 2 = 20

#4

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