Binary Search - Interview Questions - Question 1

Binary Search - Interview Questions - Question 1

Minimum steps to find a baseball

One baseball has been placed at one spot in an m * n matrix, and we want to quickly find the location of the baseball. For each move (can pick up any locations), the only thing we know is that we are getting closer or further from the baseball comparing to the previous position. A Boolean helper API is provided for this information: helper(x1, y1, x2, y2), where (x1, y1) is the former position and (x2, y2) the current position. When helper(x1, y1, x2, y2) is true, it means closer; false for further. Design an algorithm to find the location.

Analysis:

Suppose the baseball is at (x0, y0), and the matrix size is known as m * n.

If we are on the top side of the matrix above X = x0, then every move down would get us closer to the target; if we are at the down side below X = x0, then every move up gets us closer too.

The horizontal direction can give us the similar results. So we can reduce the 2D problem to 1D.

Let focus on the X directory (or the vertical). If we are at [0, x0-1], then every move down will make the distance smaller; X=x0 will give the minimum; [x0+1, m) will get the distance larger as X increases. The overall behavior of distance as a function of X is like a "V" shape.

So one way to find the baseball is that, we can just make a try first: to move up or down, if one of them gets us closer, choose that direction; if both of them get us further from the baseball, then X is x0 now. Then we apply the same method to the Y direction.

See the code below:

class Solution {
public:
    vector<int, int> findIdx(int m, int n) {
        int x = findX(0, m), y = findY(0, n);
        if(x == -1 || y == -1) {
            cout<<"cannot find"<<endl;
        }
        return {x, y};
    }

private:
    // true for getting closer; false for further
    // can directly call without worrying about the implementation details
    bool helper(int x1, int y1, int x2, int y2){
        
    };

    // return -1 if does not find
    int findX(int x, int range) {
        if(x < 0 || || range < 1 || x >= range) {
            cout<<"wrong range or index"<<endl;
            return -1;
        }
        // some base cases
        if(range == 1) return x;
        if(x == 0 && !helper(x, 0, x+1, 0)) return x;
        if(x + 1 == range && !helper(x, 0, x-1, 0)) return x;
        int up = x - 1, down = x + 1;
        if(up >= 0 || down < range) {
            if(up >= 0 && helper(x, 0, up, 0)) return findX(up, range);
            if(down < range && helper(x, 0, down, 0)) return findX(down, range);
            // find it
            if(up >= 0 && down < range && !helper(x, 0, up, 0) && !helper(x, 0, down, 0)) {
                return x;
            }
        }
        return -1;
    }

    // return -1 if does not find
    int findY(int y, int range) {
        if(y < 0 || range < 1 || y >= range) {
            cout<<"wrong range or index"<<endl;
            return -1;
        }
        // some base cases
        if(range == 1) return y;
        if(y == 0 && !helper(0, y, 0, y+1)) return y;
        if(y + 1 == range && !helper(0, y, 0, y-1)) return y;
        int left = y - 1, right = y + 1;
        if(left >= 0 || right < range) {
            if(left >= 0 && helper(0, y, 0, left)) return findY(left, range);
            if(down < range && helper(0, y, 0, right)) return findY(right, range);
            // find it
            if(left >= 0 && right < range && !helper(0, y, 0, left) && !helper(0, y, 0, right)) {
                return y;
            }
        }
        return -1;
    }
};

The time complexity is O(m + n). Can we do better?

The answer is yes! we can apply binary search to this question, since the search space is kind of sorted, or more specific, it has a "V" shape.

The basic idea is as the following:

Again let focus on the X direction only. The search range is [0, m), the first guess is mid = m/2.  Different from the common binary search, we still need a second guess to use the helper() function. The second guess can be mid+1, then the helper() function can tell us it gets closer or further from mid to mid+1.

If get closer, then the search range becomes [mid+1, m); if not, the search range can be updated as [0, mid+1).

The same procedure can be applied to the Y direction.

The overall time complexity is O(log(m) + log(n)), or O(log(m*n)). 

class Solution {
public:
    vector<int, int> findIdx(int m, int n) {
        int x = findX(0, m-1), y = findY(0, n-1);
        if(x == -1 || y == -1) {
            cout<<"cannot find"<<endl;
        }
        return {x, y};
    }

private:
    // true for getting closer; false for further
    // can directly call without worrying about the implementation details
    bool helper(int x1, int y1, int x2, int y2){
        
    };

    // return -1 if does not find
    int findX(int left, int right) {
        if(left > right) {
            cout<<"wrong range or index"<<endl;
            return -1;
        }
        if(left == right) return left;
        int mid = left + (right - left) / 2;
        if(helper(mid, 0, mid+1, 0)) findX(mid+1, right);
        return findX(left, mid);
    }

    // return -1 if does not find
    int findY(int left, int right) {
        if(left > right) {
            cout<<"wrong range or index"<<endl;
            return -1;
        }
        if(left == right) return left;
        int mid = left + (right - left) / 2;
        if(helper(0, mid, 0, mid+1)) findY(mid+1, right);
        return findY(left, mid);
    }
};

Note:

In general, m*n >> m + n, but log(m*n) usually << m + n

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