# 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 2^{nd}and 3^{rd}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++

`// 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; ` `} ` |

*chevron_right*

*filter_none*

## 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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