Divide N segments into two non-empty groups such that given condition is satisfied

Given N segments (or ranges) represented by two non-negative integers L and R. Divide these segments into two non-empty groups such that there are no two segments from different groups that share a common point. If if it is possible to do so, assign each segment a number from the set {1, 2} otherwise print Not Possible.

Examples:

Input: arr[][] = {{5, 5}, {2, 3}, {3, 4}}
Output: 2 1 1
Since 2nd and 3rd segment have one point common i.e. 3, they should be contained in same group.

Input: arr[][] = {{3, 5}, {2, 3}, {1, 4}}
Output: Not Possible
All segments should be contained in the same group since every pair has a common point with each other. Since the other group is empty, answer is not possbile.

Prerequisites: Merge Overlapping Intervals



Approach: Using the concept of merging overlapping intervals, we can assign the same group to all those segments that are overlapping and alternatively changing the group number.
To merge overlapping segments, sort all the segments with respect to their right indexes keeping in order of the original indexes of the segments. Then, iterate over the segments and check if the previous segment is overlapping with the current segment. If it does then merge it making it one segment and if it doesn’t create a new one.
At last, check if one of the group is empty of not. If one of them is empty, answer is not possible, otherwise print all the assigned values of segments.

Below is the implementation of the above approach:

C++

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// CPP Program to divide N segments into
// two non empty groups such that given
// condition is satisfied
#include <bits/stdc++.h>
using namespace std;
  
// Structure to hold the elements of
// a segment
struct segment {
  
    // left index
    int l;
  
    // right index
    int r;
  
    // index of the segment
    int idx;
};
  
// Comparator function to sort the segments
// according to the right indexes
bool comp(const segment& a, const segment& b)
{
    if (a.r == b.r)
        return a.idx < b.idx;
    return a.r < b.r;
}
  
// Function to print the answer if it exists
// using the concept of merge overlapping segments
void printAnswer(vector<pair<int, int> > v, int n)
{
    segment seg[n];
  
    // Assigning values from the vector v
    for (int i = 0; i < n; i++) {
        seg[i].l = v[i].first;
        seg[i].r = v[i].second;
        seg[i].idx = i;
    }
  
    // Sort the array of segments
    // according to right indexes
    sort(seg, seg + n, comp);
  
    // Resultant array
    int res[n];
    memset(res, 0, sizeof(res));
  
    // Let's denote first group with 0 and second
    // group with 1
    // Current segment
    int prev = 0;
  
    // Assigning group 1 to first segment
    res[seg[prev].idx] = 0;
    for (int i = 1; i < n; i++) {
  
        // If the current segment overlaps
        // with the previous segment, merge it
        if (seg[i].l <= seg[prev].r) {
  
            // Assigning same group value
            res[seg[i].idx] = res[seg[prev].idx];
            seg[prev].r = max(seg[prev].r, seg[i].r);
        }
        else {
  
            // Change group number and create
            // new segment
            res[seg[i].idx] = res[seg[prev].idx] ^ 1;
            ++prev;
            seg[prev] = seg[i];
        }
    }
  
    // Check if one of the groups is empty or not
    int one = 0, two = 0;
    for (int i = 0; i < n; i++) {
        if (!res[i])
            one++;
        else
            two++;
    }
  
    // If both groups are non-empty
    if (one && two) {
        for (int i = 0; i < n; i++)
            cout << res[i] + 1 << " ";
        cout << endl;
    }
    else
        cout << "Not Possible" << endl;
}
  
// Driver Code
int main()
{
    vector<pair<int, int> > v = { { 1, 2 }, { 3, 4 }, { 5, 6 } };
    int n = v.size();
    printAnswer(v, n);
    return 0;
}

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Python3

# Python3 Program to divide N segments
# into two non empty groups such that
# given condition is satisfied

# Structure to hold the elements of
# a segment
class segment:

def __init__(self):
self.l = None # left index
self.r = None # right index
self.idx = None # index of the segment

# Function to print the answer if
# it exists using the concept of
# merge overlapping segments
def printAnswer(v, n):

seg = [segment() for i in range(n)]

# Assigning values from the vector v
for i in range(0, n):
seg[i].l = v[i][0]
seg[i].r = v[i][1]
seg[i].idx = i

# Sort the array of segments
# according to right indexes
seg.sort(key = lambda x: (x.r, x.idx))

# Resultant array
res = [0] * (n)

# Let’s denote first group with 0 and
# second group with 1 Current segment
prev = 0

# Assigning group 1 to first segment
res[seg[prev].idx] = 0
for i in range(1, n):

# If the current segment overlaps
# with the previous segment, merge it
if seg[i].l <= seg[prev].r: # Assigning same group value res[seg[i].idx] = res[seg[prev].idx] seg[prev].r = max(seg[prev].r, seg[i].r) else: # Change group number and create # new segment res[seg[i].idx] = res[seg[prev].idx] ^ 1 prev += 1 seg[prev] = seg[i] # Check if one of the groups is # empty or not one, two = 0, 0 for i in range(0, n): if not res[i]: one += 1 else: two += 1 # If both groups are non-empty if one and two: for i in range(0, n): print(res[i] + 1, end = " ") print() else: print("Not Possible") # Driver Code if __name__ == "__main__": v = [[1, 2], [3, 4], [5, 6]] n = len(v) printAnswer(v, n) # This code is contributed # by Rituraj Jain [tabbyending]

Output:

1 2 1

Time Complexity: O(n * log n), where n is the number of segments



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Improved By : rituraj_jain