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Minimum block jumps to reach destination

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  • Difficulty Level : Medium
  • Last Updated : 17 Jun, 2022
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Given N lines and one starting point and destination point in 2-dimensional space. These N lines divide the space into some blocks. We need to print the minimum number of jumps to reach destination point from starting point. We can jump from one block to other block only if they share a side. Examples:

Input : Lines = [x = 0, y = 0, x + y – 2 = 0]
        Start point = [1, 1], 
        Dest point  = [-2, -1]
Output : 2        
We need to jump 2 times (B4 -> B3 then B3 -> B5 
or B4 -> B6 then B6 -> B5) to reach destination 
point from starting point shown in below diagram. 
Each block i is given an id Bi in the diagram.

We can solve this problem using a property of lines and points which is stated as, If we put two points in line equation then they will have same sign i.e. positive-positive or negative-negative of evaluated value if both points lies on the same side of line, in case of different sign i.e. positive-negative they will lie on different side of line. Now we can use above property to solve this problem, For each line, we will check if start and destination point lie on the same side or not. If they lie to the different side of a line then that line must be jumped to come closer. As in above diagram start point and the destination point are on the same side of x + y – 2 = 0 line, so that line need not to be jumped, rest two lines need to be jumped because both points lies on opposite side. Finally, we will check the sign of evaluation of points with respect to each line and we will increase our jump count whenever we found opposite signs. Total time complexity of this problem will be linear. 

C++




// C++ program to find minimum jumps to reach
// a given destination from a given source
#include <bits/stdc++.h>
using namespace std;
 
// To represent point in 2D space
struct point
{
    int x, y;
    point(int x, int y) : x(x), y(y)
    {}
};
 
// To represent line of (ax + by + c)format
struct line
{
    int a, b, c;
    line(int a, int b, int c) : a(a), b(b), c(c)
    {}
    line()
    {}
};
 
// Returns 1 if evaluation is greater > 0,
// else returns -1
int evalPointOnLine(point p, line curLine)
{
    int eval = curLine.a* p.x +
               curLine.b * p.y +
               curLine.c;
    if (eval > 0)
        return 1;
    return -1;
}
 
//  Returns minimum jumps to reach
//  dest point from start point
int minJumpToReachDestination(point start,
              point dest, line lines[], int N)
{
    int jumps = 0;
    for (int i = 0; i < N; i++)
    {
        // get sign of evaluation from point
        // co-ordinate and line equation
        int signStart = evalPointOnLine(start, lines[i]);
        int signDest = evalPointOnLine(dest, lines[i]);
 
        // if both evaluation are of opposite sign,
        // increase jump by 1
        if (signStart * signDest < 0)
            jumps++;
    }
 
    return jumps;
}
 
// Driver code to test above methods
int main()
{
    point start(1, 1);
    point dest(-2, -1);
 
    line lines[3];
    lines[0] = line(1, 0, 0);
    lines[1] = line(0, 1, 0);
    lines[2] = line(1, 1, -2);
 
    cout << minJumpToReachDestination(start, dest, lines, 3);
 
    return 0;
}

Java




// Java program to find minimum jumps to reach
// a given destination from a given source
class GFG
{
 
// To represent point in 2D space
static class point
{
    int x, y;
 
    public point(int x, int y)
    {
        super();
        this.x = x;
        this.y = y;
    }
     
};
 
// To represent line of (ax + by + c)format
static class line
{
    public line(int a, int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }
 
    int a, b, c;
     
    line()
    {}
};
 
// Returns 1 if evaluation is greater > 0,
// else returns -1
static int evalPointOnLine(point p, line curLine)
{
    int eval = curLine.a* p.x +
            curLine.b * p.y +
            curLine.c;
    if (eval > 0)
        return 1;
    return -1;
}
 
// Returns minimum jumps to reach
// dest point from start point
static int minJumpToReachDestination(point start,
            point dest, line lines[], int N)
{
    int jumps = 0;
    for (int i = 0; i < N; i++)
    {
        // get sign of evaluation from point
        // co-ordinate and line equation
        int signStart = evalPointOnLine(start, lines[i]);
        int signDest = evalPointOnLine(dest, lines[i]);
 
        // if both evaluation are of opposite sign,
        // increase jump by 1
        if (signStart * signDest < 0)
            jumps++;
    }
 
    return jumps;
}
 
// Driver code
public static void main(String[] args)
{
    point start = new point(1, 1);
    point dest = new point(-2, -1);
 
    line []lines = new line[3];
    lines[0] = new line(1, 0, 0);
    lines[1] = new line(0, 1, 0);
    lines[2] = new line(1, 1, -2);
 
    System.out.print(minJumpToReachDestination(start, dest, lines, 3));
}
}
 
// This code is contributed by Rajput-Ji

C#




// C# program to find minimum jumps to reach
// a given destination from a given source
using System;
 
class GFG
{
 
// To represent point in 2D space
class point
{
    public int x, y;
 
    public point(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
     
};
 
// To represent line of (ax + by + c)format
class line
{
    public int a, b, c;
    line()
    {}
    public line(int a, int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }
};
 
// Returns 1 if evaluation is greater > 0,
// else returns -1
static int evalPointOnLine(point p, line curLine)
{
    int eval = curLine.a* p.x +
            curLine.b * p.y +
            curLine.c;
    if (eval > 0)
        return 1;
    return -1;
}
 
// Returns minimum jumps to reach
// dest point from start point
static int minJumpToReachDestination(point start,
            point dest, line []lines, int N)
{
    int jumps = 0;
    for (int i = 0; i < N; i++)
    {
        // get sign of evaluation from point
        // co-ordinate and line equation
        int signStart = evalPointOnLine(start, lines[i]);
        int signDest = evalPointOnLine(dest, lines[i]);
 
        // if both evaluation are of opposite sign,
        // increase jump by 1
        if (signStart * signDest < 0)
            jumps++;
    }
 
    return jumps;
}
 
// Driver code
public static void Main(String[] args)
{
    point start = new point(1, 1);
    point dest = new point(-2, -1);
 
    line []lines = new line[3];
    lines[0] = new line(1, 0, 0);
    lines[1] = new line(0, 1, 0);
    lines[2] = new line(1, 1, -2);
 
    Console.Write(minJumpToReachDestination(start, dest, lines, 3));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program to find minimum jumps to reach
# a given destination from a given source
 
# To represent point in 2D space
 
 
class point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
 
# To represent line of (ax + by + c)format
 
 
class line:
    def __init__(self, a, b, c):
        self.a = a
        self.b = b
        self.c = c
 
# Returns 1 if evaluation is greater > 0,
# else returns -1
 
 
def evalPointOnLine(p, curLine):
    eval = curLine.a * p.x + curLine.b * p.y + curLine.c
    if (eval > 0):
        return 1
    return -1
 
# Returns minimum jumps to reach
# dest point from start point
 
 
def minJumpToReachDestination(start, dest, lines, N):
    jumps = 0
    for i in range(N):
        # get sign of evaluation from point
        # co-ordinate and line equation
        signStart = evalPointOnLine(start, lines[i])
        signDest = evalPointOnLine(dest, lines[i])
 
        # if both evaluation are of opposite sign,
        # increase jump by 1
        if (signStart * signDest < 0):
            jumps = jumps + 1
 
    return jumps
 
 
# Driver code to test above methods
start = point(1, 1)
dest = point(-2, -1)
 
lines = []
lines.append(line(1, 0, 0))
lines.append(line(0, 1, 0))
lines.append(line(1, 1, -2))
 
print(minJumpToReachDestination(start, dest, lines, 3))
 
# The code is contributed by Gautam goel (gautamgoel962)

Javascript




<script>
     
    // Javascript program to find minimum jumps to reach
    // a given destination from a given source
     
    // To represent point in 2D space
    class Point
    {
        constructor(x, y)
        {
            this.x = x;
                this.y = y;
        }
    }
     
    // To represent line of (ax + by + c)format
    class line
    {
        constructor(a,b,c)
        {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }
     
     
    // Returns 1 if evaluation is greater > 0,
    // else returns -1
    function evalPointOnLine(p, curLine){
        var eval = curLine.a* p.x +
                curLine.b * p.y +
                curLine.c;
        if (eval > 0)
            return 1;
        return -1;
    }
     
    // Returns minimum jumps to reach
    // dest point from start point
    function minJumpToReachDestination(start, dest, lines, N){
        var jumps = 0;
        for(var i = 0; i < N; i++){
            // get sign of evaluation from point
            // co-ordinate and line equation
            var signStart = evalPointOnLine(start, lines[i]);
            var signDest = evalPointOnLine(dest, lines[i]);
       
            // if both evaluation are of opposite sign,
            // increase jump by 1
            if (signStart * signDest < 0)
                jumps++;
        }
        return jumps;
    }
     
    // Driver code
    let start = new Point(1, 1);
    let dest = new Point(-2, -1);
     
    let lines = new Array(3);
    lines[0] = new line(1, 0, 0);
    lines[1] = new line(0, 1, 0);
    lines[2] = new line(1, 1, -2);
     
    document.write(minJumpToReachDestination(start, dest, lines, 3));
    // This code is contributed by shruti456rawal
</script>

Output:

2

Time Complexity: O(1) 
Auxiliary Space: O(1)

This article is contributed by Aarti_Rathi and Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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