Find if there exists a direction for ranges such that no two range intersect

Given N ranges [L, R] with velocities vel[]. The task is to assign each range a direction i.e. either left or right. All the ranges will start moving in the assigned direction at time t = 0. Find, if there is an assignment of directions possible given that no two ranges overlap at infinite time.

Examples:

Input: range[][] = {{1, 2}, {2, 5}, {3, 10}, {4, 4}, {5, 7}},
vel[] = {3, 1, 1, 1, 10}
Output: No
Intervals {2, 5}, {3, 10} and {4, 4} share a common point 4
and have the same velocity.

Input: range[][] = {{1, 2}, {2, 5}, {3, 10}, {4, 4}, {5, 7}},
vel[] = {3, 1, 11, 1, 10}
Output: Yes

Approach:



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define MAX 100001
using namespace std;
  
// Structure to hold details of
// each interval
typedef struct
{
    int l, r, v;
} interval;
  
// Comparator to sort intervals
// based on velocity
bool cmp(interval a, interval b)
{
    return a.v < b.v;
}
  
// Function that returns true if the
// assignment of directions is possible
bool isPossible(int range[][3], int N)
{
    interval test[N];
    for (int i = 0; i < N; i++) {
        test[i].l = range[i][0];
        test[i].r = range[i][1];
        test[i].v = range[i][2];
    }
  
    // Sort the intervals based on velocity
    sort(test, test + N, cmp);
  
    for (int i = 0; i < N; i++) {
        int count[MAX] = { 0 };
        int current_velocity = test[i].v;
  
        int j = i;
  
        // Test the condition for all intervals
        // with same velocity
        while (j < N && test[j].v == current_velocity) {
            for (int k = test[j].l; k <= test[j].r; k++) {
                count[k]++;
  
                // If for any velocity, 3 or more intervals
                // share a common point return false
                if (count[k] >= 3)
                    return false;
            }
            j++;
        }
  
        i = j - 1;
    }
  
    return true;
}
  
// Driver code
int main()
{
    int range[][3] = { { 1, 2, 3 },
                       { 2, 5, 1 },
                       { 3, 10, 1 },
                       { 4, 4, 1 },
                       { 5, 7, 10 } };
    int n = sizeof(range) / sizeof(range[0]);
  
    if (isPossible(range, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}
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// Java implementation of the approach
import java.util.*;
  
class GFG
{
static int MAX = 100001;
  
// Structure to hold details of
// each interval
static class interval
{
    int l, r, v;
  
static class Sort implements Comparator<interval> 
{
    // Comparator to sort intervals
    // based on velocity
    public int compare(interval a, interval b)
    {
        return (a.v < b.v ? 1 : 0);
    }
}
  
// Function that returns true if the
// assignment of directions is possible
static boolean isPossible(int range[][], int N)
{
    interval test[] = new interval[N];
    for (int i = 0; i < N; i++) 
    {
        test[i] = new interval();
        test[i].l = range[i][0];
        test[i].r = range[i][1];
        test[i].v = range[i][2];
    }
  
    // Sort the intervals based on velocity
    Arrays.sort(test, new Sort());
  
    for (int i = 0; i < N; i++)
    {
        int count[] = new int[MAX];
        int current_velocity = test[i].v;
  
        int j = i;
  
        // Test the condition for all intervals
        // with same velocity
        while (j < N && test[j].v == current_velocity)
        {
            for (int k = test[j].l; k <= test[j].r; k++) 
            {
                count[k]++;
  
                // If for any velocity, 3 or more intervals
                // share a common point return false
                if (count[k] >= 3)
                    return false;
            }
            j++;
        }
        i = j - 1;
    }
    return true;
}
  
// Driver code
public static void main(String args[])
{
    int range[][] = {{ 1, 2, 3 },
                     { 2, 5, 1 },
                     { 3, 10, 1 },
                     { 4, 4, 1 },
                     { 5, 7, 10 }};
    int n = range.length;
  
    if (isPossible(range, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Arnab Kundu
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# Python implementation of the approach 
MAX = 100001
  
# Function that returns true if the 
# assignment of directions is possible 
def isPossible(rangee, N):
      
    # Structure to hold details of 
    # each interval 
    test = [[0 for x in range(3)] for x in range(N)]
    for i in range(N):
        test[i][0] = rangee[i][0
        test[i][1] = rangee[i][1
        test[i][2] = rangee[i][2
          
    # Sort the intervals based on velocity 
    test.sort(key = lambda x: x[2]) 
    for i in range(N):
        count = [0] * MAX
        current_velocity = test[i][2
        j =
          
        # Test the condition for all intervals 
        # with same velocity 
        while (j < N and test[j][2] == current_velocity):
            for k in range(test[j][0], test[j][1] + 1):
                count[k] += 1
                  
                # If for any velocity, 3 or more intervals 
                # share a common poreturn false 
                if (count[k] >= 3):
                    return False
            j += 1
        i = j - 1
      
    return True
      
# Driver code 
rangee = [[1, 2, 3] ,[2, 5, 1] ,[3, 10, 1],
        [4, 4, 1 ],[5, 7, 10 ]]
n = len(rangee)
if (isPossible(rangee, n)):
    print("Yes")
else:
    print("No")
      
# This code is contributed by SHUBHAMSINGH10
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Output:
No

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