# 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 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 possible.

**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 increasing left values, if left values are equal, sort it on the basis of right values first keeping an order of the original indices of the segments. Then, iterate over the segments and check if any of the previous segments 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++

`// C++ Program to divide N segments` `// into two non empty groups such that` `// given condition is satisfied` `#include<bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to print the answer if` `// it exists using the concept of` `// merge overlapping segments` `void` `printAnswer(vector<pair<` `int` `,` `int` `>> v,` `int` `n)` `{` ` ` `// vec[i].first -> left value` ` ` `// vec[i].second.first -> right value` ` ` `// vec[i].second.second -> position` ` ` `vector<pair<` `int` `,pair<` `int` `,` `int` `>>> vec;` ` ` `for` `(` `int` `i=0;i<n;i++)` ` ` `{` ` ` `vec.push_back({v[i].first,{v[i].second,i}});` ` ` `}` ` ` `sort(vec.begin(),vec.end());` ` ` ` ` `// Resultant array` ` ` `// Initialise all the values in resultant array with '2'` ` ` `// except the position at the first index of sorted vec which is initialised as '1'` ` ` `// Initialise maxR to store the maximum of all right values encountered so far` ` ` `vector<` `int` `> res(n,2);` ` ` `int` `maxR = vec[0].second.first;` ` ` `res[vec[0].second.second] = 1;` `// If the i-th index has any point in common with the (i-1)th index classify it as '1'` `// in resultant array and update maxR if necessary` `// else we have found the breakpoint and we can exit the loop` ` ` ` ` `bool` `ok=` `false` `;` ` ` `for` `(` `int` `i=1;i<n;i++)` ` ` `{` ` ` `if` `(maxR>=vec[i].first)` ` ` `{` ` ` `res[vec[i].second.second] = res[vec[i-1].second.second];` ` ` `maxR=max(maxR,vec[i].second.first);` ` ` `}` ` ` `else` ` ` `{` ` ` `ok=` `true` `;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `if` `(ok)` ` ` `{` ` ` `for` `(` `auto` `x:res)` ` ` `cout << x << ` `" "` `;` ` ` `cout << ` `'\n'` `;` ` ` `}` ` ` `else` ` ` `cout << ` `"Not possible\n"` `;` ` ` `}` `int` `main()` `{` ` ` `vector<pair<` `int` `,` `int` `>> v = {{2,8}, {3,4} , {5,8}, {9,10}};` ` ` `int` `n = (` `int` `)v.size();` ` ` ` ` `printAnswer(v,n);` `}` |

## Python3

`# Python3 Program to divide N segments` `# into two non empty groups such that` `# given condition is satisfied` ` ` `# Function to print the answer if` `# it exists using the concept of` `# merge overlapping segments` `def` `printAnswer(v, n):` ` ` `# Sort the indices based on their corresponding value in V ` ` ` `indices ` `=` `list` `(` `range` `(n))` ` ` `indices.sort(key` `=` `lambda` `i: v[i])` ` ` ` ` `# Resultant array` ` ` `# Initialise all the values in resultant array with '2'` ` ` `# except the first index of 'indices' which is initialised as '1'` ` ` `# Initialise maxR to store the maximum of all right values encountered so far` ` ` `res ` `=` `[` `2` `] ` `*` `n` ` ` `res[indices[` `0` `]] ` `=` `1` ` ` `maxR` `=` `v[indices[` `0` `]][` `1` `]` ` ` ` ` `#If the i-th index has any point in common with the (i-1)th index classify it as '1' in resultant array and update maxR if necessary` ` ` `#else we have found the breakpoint and we can exit the loop ` ` ` `for` `i ` `in` `range` `(` `1` `, n):` ` ` `if` `maxR>` `=` `v[indices[i]][` `0` `]:` ` ` `res[indices[i]]` `=` `res[indices[i` `-` `1` `]]` ` ` `maxR` `=` `max` `(maxR,v[indices[i]][` `1` `])` ` ` `else` `:` ` ` `break` ` ` `else` `:` ` ` `print` `(` `"Not possible"` `)` ` ` `return` ` ` `print` `(` `" "` `.join(` `map` `(` `str` `,res)))` ` ` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `v ` `=` `[[` `2` `, ` `8` `], [` `3` `, ` `4` `], [` `5` `, ` `8` `], [` `9` `, ` `10` `]]` ` ` `n ` `=` `len` `(v)` ` ` `printAnswer(v, n)` |

**Output**

1 1 1 2

**Complexity Analysis:**

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