Move from top right corner to left down corner with checking if left limit (index 0) is not found.
Elements visited on each itteration for N=5 M=5:
i = 4
J From 4 to 2:
I = 3
J FROM 2 to 0
i = 2
j = 0
I = 1;
J = 0;
i = 0;
J = 0;
public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
int min = Integer.MAX_VALUE;
List<Integer> dimensions = binaryMatrix.dimensions();
int rows = dimensions.get(0);
int columns = dimensions.get(1);
for(int i = rows - 1; i >= 0; --i){
boolean foundOne = false;
for(int j = columns - 1; j >= 0; --j){
int current = binaryMatrix.get(i , j);
if(current == 0 && j != columns - 1 && foundOne){
if(j <= min) {
min = j + 1;
if(min == 0)
return 0;
columns = min;
break;
}
}else if(current == 0)
break;
else if(current == 1 && j == 0)
return 0;
else if(current == 1)
foundOne = true;
}
}
return min == Integer.MAX_VALUE ? -1 : min;
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