CareerCup

I don't understand this question! Given any matrix, what is the output?

it should use stack to solve

public int solve(int[] array){
        if(array.length<=1){
            return 0;
        }

        int mod = (int)(1e9+7);
        int sum = 0;
        int pre_sum = array[0];
        for(int i=1;i<array.length;i++){
              sum += (pre_sum%array[i]);
              pre_sum+=array[i];
              sum %= mod;
              pre_sum %= mod;
        }

        int suffix_sum = array[array.length-1];
        for(int i=array.length-2;i>=0;i--){
            sum += (suffix_sum%array[i]);
            suffix_sum += array[i];
            sum %= mod;
            suffix_sum %= mod;
        }

        return sum;
    }

what do you mean set all of values to be true? Does it mean that we need to set all of values which have been appeared to be true?

I think create two maps, one is hashmap which is name -> rank, another is treemap, rank->list<String>(the names), then we can use treemap's properties to extract all of people who has same rank. Correct me if I am wrong! Any other idea?

are all of names different with each other in this list? Or, there are duplicate names in the list?

I feel this question can be solved by stack. once we found previous element in the stack is larger than current element, than pop out

I feel like it is like a bfs problem, but I am not sure whether it satisfies interviewer? start from the origin point, and bfs search its surrounding area. Any other idea?

correct me if my ideas are incorrect

general idea: combine two sensors if these two sensors intersect with each other, and store their y range{(min_y, max_y)} which can reach in the room. If some union sensors' y_range satisfies min_y<=0 and max_y<=room_width, then thief cannot reach the right side, otherwise the thief can reach right side

I think this question it should be using union find to solve it

I think we can use KMP algorithm to solve it, and it takes linear time to solve it.

bfs would be ok!