Related Articles
Find minimum value of y for the given x values in Q queries from all the given set of lines
• Last Updated : 21 Jan, 2021

Given a 2 dimensional array arr[][] consisting of slope(m) and intercept(c) for large number of lines of the form y = mx + c and Q queries such that each query contains a value x. The task is to find the minimum value of y for the given x values from all the given set of lines.

Examples:

Input: arr[][] ={ {1, 1}, {0, 0}, {-3, 3} }, Q = {-2, 2, 0}
Output: -1, -3, 0
Explanation:
For query x = -2, y values from the equations are -1, 0, 9. So the minimum value is -1
Similarly, for x = 2, y values are 3, 0, -3. So the minimum value is -3
And for x = 0, values of y = 1, 0, 3 so min value is 0.

Input: arr[][] ={ {5, 6}, {3, 2}, {7, 3} }, Q = { 1, 2, 30 }
Output: 5, 8, 92

Naive Approach: The naive approach is to substitute the values of x in every line and compute the minimum of all the lines. For each query, it will take O(N) time and so the complexity of the solution becomes O(Q * N) where N is the number of lines and Q is the number of queries.

Efficient approach: The idea is to use convex hull trick:

• From the given set of lines, the lines which carry no significance (for any value of x they never give the minimal value y) can be found and deleted thereby reducing the set.
• Now, if the ranges (l, r) can be found where each line gives the minimum value, then each query can be answered using binary search.
• Therefore, a sorted vector of lines with decreasing order of slopes is created and the lines are inserted in decreasing order of the slopes.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `struct` `Line {``    ``int` `m, c;` `public``:``    ``// Sort the line in decreasing``    ``// order of their slopes``    ``bool` `operator<(Line l)``    ``{` `        ``// If slopes arent equal``        ``if` `(m != l.m)``            ``return` `m > l.m;` `        ``// If the slopes are equal``        ``else``            ``return` `c > l.c;``    ``}` `    ``// Checks if line L3 or L1 is better than L2``    ``// Intersection of Line 1 and``    ``// Line 2 has x-coordinate (b1-b2)/(m2-m1)``    ``// Similarly for Line 1 and``    ``// Line 3 has x-coordinate (b1-b3)/(m3-m1)``    ``// Cross multiplication will give the below result``    ``bool` `check(Line L1, Line L2, Line L3)``    ``{``        ``return` `(L3.c - L1.c) * (L1.m - L2.m)``               ``< (L2.c - L1.c) * (L1.m - L3.m);``    ``}``};` `struct` `Convex_HULL_Trick {` `    ``// To store the lines``    ``vector l;` `    ``// Add the line to the set of lines``    ``void` `add(Line newLine)``    ``{` `        ``int` `n = l.size();` `        ``// To check if after adding the new line``        ``// whether old lines are``        ``// losing significance or not``        ``while` `(n >= 2``               ``&& newLine.check(l[n - 2],``                                ``l[n - 1],``                                ``newLine)) {``            ``n--;``        ``}` `        ``l.resize(n);` `        ``// Add the present line``        ``l.push_back(newLine);``    ``}` `    ``// Function to return the y coordinate``    ``// of the specified line for the given coordinate``    ``int` `value(``int` `in, ``int` `x)``    ``{``        ``return` `l[in].m * x + l[in].c;``    ``}` `    ``// Function to Return the minimum value``    ``// of y for the given x coordinate``    ``int` `minQuery(``int` `x)``    ``{``        ``// if there is no lines``        ``if` `(l.empty())``            ``return` `INT_MAX;` `        ``int` `low = 0, high = (``int``)l.size() - 2;` `        ``// Binary search``        ``while` `(low <= high) {``            ``int` `mid = (low + high) / 2;` `            ``if` `(value(mid, x) > value(mid + 1, x))``                ``low = mid + 1;``            ``else``                ``high = mid - 1;``        ``}` `        ``return` `value(low, x);``    ``}``};` `// Driver code``int` `main()``{``    ``Line lines[] = { { 1, 1 },``                     ``{ 0, 0 },``                     ``{ -3, 3 } };``    ``int` `Q[] = { -2, 2, 0 };``    ``int` `n = 3, q = 3;``    ``Convex_HULL_Trick cht;` `    ``// Sort the lines``    ``sort(lines, lines + n);` `    ``// Add the lines``    ``for` `(``int` `i = 0; i < n; i++)``        ``cht.add(lines[i]);` `    ``// For each query in Q``    ``for` `(``int` `i = 0; i < q; i++) {``        ``int` `x = Q[i];``        ``cout << cht.minQuery(x) << endl;``    ``}` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.ArrayList;``import` `java.util.Arrays;` `class` `GFG{` `static` `class` `Line ``implements` `Comparable``{``    ``int` `m, c;` `    ``public` `Line(``int` `m, ``int` `c)``    ``{``        ``this``.m = m;``        ``this``.c = c;``    ``}` `    ``// Sort the line in decreasing``    ``// order of their slopes``    ``@Override``    ``public` `int` `compareTo(Line l)``    ``{``        ` `        ``// If slopes arent equal``        ``if` `(m != l.m)``            ``return` `l.m - ``this``.m;` `        ``// If the slopes are equal``        ``else``            ``return` `l.c - ``this``.c;``    ``}` `    ``// Checks if line L3 or L1 is better than L2``    ``// Intersection of Line 1 and``    ``// Line 2 has x-coordinate (b1-b2)/(m2-m1)``    ``// Similarly for Line 1 and``    ``// Line 3 has x-coordinate (b1-b3)/(m3-m1)``    ``// Cross multiplication will give the below result``    ``boolean` `check(Line L1, Line L2, Line L3)``    ``{``        ``return` `(L3.c - L1.c) * (L1.m - L2.m) <``               ``(L2.c - L1.c) * (L1.m - L3.m);``    ``}``}` `static` `class` `Convex_HULL_Trick``{``    ` `    ``// To store the lines``    ``ArrayList l = ``new` `ArrayList<>();` `    ``// Add the line to the set of lines``    ``void` `add(Line newLine)``    ``{``        ``int` `n = l.size();` `        ``// To check if after adding the new``        ``// line whether old lines are``        ``// losing significance or not``        ``while` `(n >= ``2` `&&``               ``newLine.check(l.get(n - ``2``),``                             ``l.get(n - ``1``), newLine))``        ``{``            ``n--;``        ``}` `        ``// l = new Line[n];` `        ``// Add the present line``        ``l.add(newLine);``    ``}` `    ``// Function to return the y coordinate``    ``// of the specified line for the given``    ``// coordinate``    ``int` `value(``int` `in, ``int` `x)``    ``{``        ``return` `l.get(in).m * x + l.get(in).c;``    ``}` `    ``// Function to Return the minimum value``    ``// of y for the given x coordinate``    ``int` `minQuery(``int` `x)``    ``{``        ` `        ``// If there is no lines``        ``if` `(l.isEmpty())``            ``return` `Integer.MAX_VALUE;` `        ``int` `low = ``0``, high = (``int``)l.size() - ``2``;` `        ``// Binary search``        ``while` `(low <= high)``        ``{``            ``int` `mid = (low + high) / ``2``;` `            ``if` `(value(mid, x) > value(mid + ``1``, x))``                ``low = mid + ``1``;``            ``else``                ``high = mid - ``1``;``        ``}``        ``return` `value(low, x);``    ``}``};` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Line[] lines = { ``new` `Line(``1``, ``1``),``                     ``new` `Line(``0``, ``0``),``                     ``new` `Line(-``3``, ``3``) };``    ``int` `Q[] = { -``2``, ``2``, ``0` `};``    ``int` `n = ``3``, q = ``3``;``    ` `    ``Convex_HULL_Trick cht = ``new` `Convex_HULL_Trick();` `    ``// Sort the lines``    ``Arrays.sort(lines);` `    ``// Add the lines``    ``for``(``int` `i = ``0``; i < n; i++)``        ``cht.add(lines[i]);` `    ``// For each query in Q``    ``for``(``int` `i = ``0``; i < q; i++)``    ``{``        ``int` `x = Q[i];``        ``System.out.println(cht.minQuery(x));``    ``}``}``}` `// This code is contributed by sanjeev2552`
Output:
```-1
-3
0```

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up