## Interview Question

Country: United States

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1
of 1 vote

``````Thanks for the question. We can solve it as the following:
Assume the following matrix:
1, 1, D, D, D, D
1, 1, D, D, D, D
1, 1, 1, D, D, D
1, 1, 1, 1, 1, D
1, 1, 1, 1, 1, D
1, 1, 1, 1, 1, 1

If you look at this matrix, you will see the tric here, for each row, we just need to check from the end position of '1' in the previous row.
Here in this example,
the end position of each row is 1, 1, 2, 4, 4, 5.
the end position is a non-decreasing sequence. So we end up checking from 1 - n. Then the runtime is O(n).``````

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0

Thank you. Can you write the pseudocode if possible.

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1
of 1 vote

in the example you have taken number of 1's in row i is less than the number in row i+1, while the question says otherwise.
just pointing this out..it will not have much have on the answer though ( just reversing the loop condition)

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1
of 1 vote

``````count_ones(A, n):
b = 0
for (i = 0; i < n; ++i)
if (A[n - 1][i] != '1') break
++b

c = b
for (j = n - 2; j >= 0; --j)
c += b
for (i = b; i < n; ++i)
if (A[j][i] != '1') break
++b
++c
return c``````

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0

This will give O(n^2) i want it in O(n)

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1
of 1 vote

This is O(n). The top for-loop will run n times to count the number of 1's in the n-1 row. The bottom for-loops will run from rows n-2 to 0 counting the 1's in those rows. However, the inner-loop will only count from b to n, where b is the number of 1's in the row below it. The worst case is when the number of 1's from n-1 to 0 increase by just one per row, forcing two comparisons per row (one for the new '1', and another for the 'D' next to it). The number of comparisons made can be represented by the relation, T(n) = T(n -1) + 2. The 2 is from two comparison.
T(n) = T(n - 1) + 2
T(n - 1) = T(n - 2) + 2
T(n - 2) = T(n - 3) + 2
-------------------------------
T(n - 1) = [T(n - 3) + 2] + 2 = T(n - 3) + 4
T(n) = [T(n - 3) + 4] + 2 = T(n - 3) + 6
T(n) = T(n - i) + 2i
T(1) = 1, since one element will only require one comparison
for i = n - 1,
T(n) = T(n - (n - 1)) + 2(n - 1) = T(1) + 2n - 2 = 1 + 2n -2 = 2n - 1 = O(n)
So, O(n) + O(n) = O(n)

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0

You should observe that the number of comparisons (in the stated worst case) will be two (the new '1' in the current row plus the following 'D') per row. That is O(2 * n), where n is the number of rows.

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0
of 0 vote

what will happened in this situation ?

1, 1, D, D, D, D
1, 1, 1, 1, 1, D
1, 1, 1, D, D, D
1, 1, 1, 1, 1, D
1, 1, 1, 1, 1, D
1, 1, 1, 1, 1, 1

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1
of 1 vote

That can't happen due to "number of 1's in row i is at least the number in row i+ 1"

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0

Thanks for clearing my doubt...

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0
of 0 vote

``````Or, more succinctly:

count_ones(A, n):
b = 0
c = 0
for (j = n - 1; j >= 0; --j)
c += b
for (i = b; i < n; ++i)
if (A[j][i] != '1') break
++b
++c
return c``````

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0
of 0 vote

``````int countones(int arr[][], int n) {
int prev = 0;
int count = 0;
for (int i = n-1; i > =0 ; i++) {
num+=prev;
for (int j = prev - 1; j < n; j++) {
if (arr[i][j]) {
num++;
prev++;
} else break;
}
}``````

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0

This will give O(n^2) i want it in O(n)

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0
of 2 vote

``````{int count_ones(A, n)
{
int num = 0, j=0;
for(i=n-1;i>=0;i++)
{
while(A[i][j] != 'D')
j++;
j--;
num += j;
}
return num;``````

}

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0
of 0 vote

int count = 0;
int column = 0;

for (int row = arr.length - 1; row > 0; row--) {
while (column <= arr.length - 1 && arr[row][column] == 1) {
count = count + row + 1;
column = column + 1;
}
}

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0

This is o(n).

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0
of 0 vote

``````public int count1s(char[][] arr){
int tot = 0;
int pos = 0;
for(char[] row : arr){
while(pos < row.length && row[pos] == '1'){
pos++;
}
tot += pos;
}
}``````

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0
of 0 vote

``````public int countingOnes(int[][] ones){
int size = ones.length();
int counter = 0;
for(int i=0;i<size;i++){
//check for the last row because that will have max number of ones.
if(ones[size-1][i]!='1') break;

counter++;
}
int t = counter;
int j=n-2;
while(j>=0){
if(ones[j][b]!='1') break;
b++;
j++;
}
}

}``````

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0

Sorry there are 2 typos:
1. There is an extra } at the end. 2. it should be j--. It works perfectly then, giving O(n)

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0
of 0 vote

GIVEN A MULTIDIMENSIONAL ARRAY WITH VALUES SHOWN
40 1 3 270 21
6 73 44 10 5
72 127 99 82 72
WRITE A PROGRAM THAT OUTPUTS THE SMALLEST NUMBER IN EACH ROW

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0
of 0 vote

count_ones(A) { |MAX Times|
[row,col] = size(A); | 1 time | // Get rows and columns of Array A
result = 0; | 1 time |
i = 1 ; j = 1; | 1 time | // First element A(1,1)
for(i=1; i<=row; i++){ | N times |
while( A(i,j) != 0) |col times|
j++; |col times|
result += j; | N times |
j = 0; | N times |
}
return result; | 1 time |
}

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0
of 0 vote

``````count_ones(A){			   |MAX Times|
[row,col] = size(A);             	   | 1 time  |       // Get rows and columns of Array A
result = 0;			   | 1 time  |
i = 1 ; j = 1;		         	   | 1 time  |	  // First element A(1,1)
for(i=1; i<=row; i++){	       	   | N times |
while( A(i,j) != 0)      	   |col times|
j++;                   	   |col times|
result += j;		   | N times |
j = 0;			   | N times |
}
return result;		             | 1 time  |
}``````

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