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:

  • At infinite time, if there are two ranges with different velocities then they can never overlap irrespective of their directions since range with greater velocity will always be ahead of that with slower velocity.
  • Now, consider ranges with same velocity. If there exist a point lying in at least 3 ranges with same velocity then the assignment of directions is not possible because even if two of these ranges are assigned a different direction but the third range will definitely intersect with the any one of the other ranges with the same direction.
  • So, for all the velocities, find whether there exists at least 3 ranges such that they share a common point, if yes then assignment is not possible. Otherwise it is possible.

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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Output:

No


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