Leftmost Column with at Least a One | LeetCode 30 day Challenge | Day 21 | (C++, Java, Python)

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LeetCode 1428,
LeetCode 30 day Challenge | Problem 21 | Leftmost Column with at Least a One | 21 April,
Facebook Coding Interview question,
google coding interview question,
leetcode,
Leftmost Column with at Least a One,
lefmost column with 1,
leftmost non-zero column,

#Amazon #Facebook #CodingInterview #LeetCode #30DayChallenge #Google #LeftMostColumn

[KnowledgeCenter]

Don't forget to suggest other approaches in the comments section below. Thanks.

[Anand-wiyb]

My C++ Solution:
int &binaryMatrix) {
vector<int> dim = binaryMatrix.dimensions();
int rows = dim[0];
int cols = dim[1];
for(int i = 0; i < rows; i++){
for(int j = cols-1; j>=0; j--){
if(binaryMatrix.get(i, j) == 1){
cols = j;
}
else break;
}
}
return cols==dim[1]?-1:cols;
}

[mayureshrao]

Solution in swift

class Solution {
func leftMostColumnWithOne(_ binaryMatrix: BinaryMatrix) -> Int {
var dim : [Int] = binaryMatrix.dimensions()
var rows = dim[0]
var cols = dim[1]

if (rows == 0 || cols == 0){
return -1
}

var result = -1
var r = 0
var c = cols - 1
while(r < rows && c >= 0){
if (binaryMatrix.get(r, c) == 1){
result = c;
c -= 1 ;
} else {
r += 1
}
}
return result
}
}

[triveninaik]

Similar Bottom-Up Approach
Accepted Solution

class Solution
{
public int binaryMatrix)
{
List<Integer> list = binaryMatrix.dimensions();

int rows = list.get(0) - 1;
int cols = list.get(1) - 1;

int result = -1;

if(rows == 0 || cols == 0)
{
return -1;
}

while(rows >= 0 && cols >= 0)
{
if(binaryMatrix.get(rows, cols) == 1)
{
result = cols;
cols--;
}
else
{
rows--;
}
}

return result;
}
}

[williamwambua]

Always liking the way you have an open mind to such problems and simplify then into an understandable way. Nice work as usual.

[indranilthakur]

Awesome solution. Subscribed to your channel.

[shubhamsagar]

# """
# This is BinaryMatrix's API interface.
# You should not implement it, or speculate about its implementation
# """
#class BinaryMatrix(object):
# def get(self, x: int, y: int) -> int:
# def dimensions(self) -> list[]:

class Solution:
def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int:
if not binaryMatrix:
return -1
m, n=binaryMatrix.dimensions()
res=-1
i, j=0, n-1
while(i<m and j>=0):
if (binaryMatrix.get(i, j)==1):
res=j
j-=1
else:
i+=1
return (res)

[officialspock]

Thank you very much for this clear explanation, I was having a hard time understanding it, liked and subscribed to you.

[adarshmv]

Even I did in 2nd Hint approach... I haven't seen the binary search approach. But will the usual binary search work with duplicates present?

[vincent-uhuo]

would it be faster if you can do an extra binary search for each row rather then just "c--"?