Google Interview Question for SDE-2s


Country: United States




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+ with 3x3 grid, we got 9 spots. Each can be black/white ( 1/0) , making total 2^9 = 512 states
+ Just do BFS with a Queue and keep track of visited states.
+ Increment steps at every BFS iteration and when reaching destination ( all 0s ) return steps.

- Code reviewer April 25, 2019 | Flag Reply
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#include<iostream>

using namespace std;

int getMinFlips(int grid, int flips[9], int i) 
{
    if (grid == 0b111111111) return 0;
    if (i == 9) return -1;

    int ret;
    ret = getMinFlips(grid ^ flips[i], flips, i + 1);
    if (ret > -1) {
        return ret + 1;
    }
    return getMinFlips(grid, flips, i + 1);
}
int getMinFlips(int grid)
{
    int flips[] = {
        0b110100000,
        0b111010000,
        0b011001000,
        0b100110100,
        0b010111010,
        0b001011001,
        0b000100110,
        0b000010111,
    };
    return getMinFlips(grid, flips, 0);
}
int main()
{
    cout << getMinFlips(0b101000101) << endl;
    return 0;
}

- radek April 26, 2019 | Flag Reply
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of 0 vote

I feel like this can be solved with Dynamic programming. If you store all 2^9 combinations in a table, you can start with the one given and start visiting the other potential positions until you hit the all white configeration. This should only take O(2^9). Alternatively this can be thought of as Djkstra's algorithm.

- Anonymous April 30, 2019 | Flag Reply


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