Intersecting Lines - Facebook Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers lines and integers lo and hi. 

Each element in lines contains [m, b] which represents the line y = mx + b. 

Return the number of lines that intersect with at least one other line between x = lo and x = hi, inclusive.

Constraints

0 ≤ n ≤ 100,000 where n is the length of lines

Example 1


Input

lines = [

    [2, 3],

    [-3, 5],

    [4, 6]

]

lo = 0

hi = 1

Output

2

Explanation

Lines [2, 3] and [-3, 5] intersect at x = 0.4, which is between x = 0 and x = 1.

Example 2

Input


lines = [

    [-1, 0],

    [-1, 1],

    [-1, 2],

    [-1, 3]

]

lo = 0

hi = 1

Output

0

Explanation

None of the lines intersect anywhere.



Solution :



title-img




                        Solution in C++ :

void eval(vector<pair<long long, long long>>& ret, vector<pair<long long, long long>>& lines,
          int x) {
    ret.clear();
    for (int i = 0; i < lines.size(); i++) {
        ret.emplace_back(lines[i].first * x + lines[i].second, i);
    }
    sort(ret.begin(), ret.end());
}

int solve(vector<vector<int>>& olines, int lo, int hi) {
    vector<pair<long long, long long>> lines;
    for (auto& out : olines) lines.emplace_back(out[0], out[1]);
    vector<pair<long long, long long>> lhs, rhs;
    eval(lhs, lines, lo);
    eval(rhs, lines, hi);
    int ret = 0;
    map<int, int> dp;
    for (int i = 0; i < lhs.size();) {
        int j = i + 1;
        while (j < lhs.size() && lhs[i].first == lhs[j].first) j++;
        for (int k = i; k < j; k++) {
            dp[lhs[k].second] = j - 1;
        }
        i = j;
    }
    multiset<int> seen;
    int reallhs = 0;
    for (int i = 0; i < rhs.size();) {
        int j = i + 1;
        while (j < rhs.size() && rhs[j].first == rhs[i].first) j++;
        for (int k = i; k < j; k++) {
            seen.insert(dp[rhs[k].second]);
        }
        if (reallhs + seen.size() - 1 == *seen.rbegin()) {
            if (seen.size() > 1) ret += seen.size();
            reallhs += seen.size();
            seen.clear();
        }
        i = j;
    }
    assert(seen.size() == 0);
    return ret;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] lines, int lo, int hi) {
        long[][] p = new long[lines.length][2];
        for (int i = 0; i < lines.length; i++) {
            p[i][0] = lines[i][0] * lo + lines[i][1];
            p[i][1] = lines[i][0] * hi + lines[i][1];
        }
        Arrays.sort(p, new Comparator<long[]>() {
            public int compare(long[] a, long[] b) {
                if (a[0] == b[0])
                    return Long.compare(b[1], a[1]);
                return Long.compare(a[0], b[0]);
            }
        });
        long max = p[0][1], min = p[p.length - 1][1];
        HashSet<Integer> set = new HashSet();
        for (int i = 1; i < p.length; i++) {
            if (p[i][1] <= max)
                set.add(i);
            max = Math.max(max, p[i][1]);
        }
        for (int i = p.length - 2; i > -1; i--) {
            if (p[i][1] >= min)
                set.add(i);
            min = Math.min(min, p[i][1]);
        }
        return set.size();
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, lines, lo, hi):
        segments = [(m * lo + b, m * hi + b, i) for i, (m, b) in enumerate(lines)]
        segments.sort()
        res = [0 for _ in lines]

        cstarting = Counter([a for a, b, i in segments])
        for (x, y, i) in segments:
            if cstarting[x] > 1:
                res[i] = 1

        curmax = -(10 ** 10)
        prevx = -(10 ** 10)
        for (x, y, i) in segments:
            if x == prevx:
                res[i] = 1
            if y <= curmax:
                res[i] = 1
            curmax = max(curmax, y)
            prevx = x

        curmin = 10 ** 10
        prevx = 10 ** 10

        for (x, y, i) in segments[::-1]:
            if x == prevx:
                res[i] = 1
            if y >= curmin:
                res[i] = 1
            curmin = min(curmin, y)
            prevx = x

        return sum(res)
                    


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