Algorithm
You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers li and ri (1≤li≤ri≤m) — coordinates of the left and of the right endpoints.
Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l;r] if and only if l≤x≤r.
The first line of the input contains two integers n and m (1≤n,m≤100) — the number of segments and the upper bound for coordinates.
The next n lines contain two integers each li and ri (1≤li≤ri≤m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that li=ri, i.e. a segment can degenerate to a point.
In the first line print one integer k — the number of points that don't belong to any segment.
In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.
If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.
3 5
2 2
1 2
5 5
2
3 4
1 7
1 7
0
In the first example the point 1 belongs to the second segment, the point 2 belongs to the first and the second segments and the point 5 belongs to the third segment. The points 3 and 4 do not belong to any segment.
In the second example all the points from 1 to 7 belong to the first segment.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int const N = 1e2 + 1;
int n, m, a[N], b[N];
vector<int> sol;
int main() {
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%d %d", a + i, b + i);
for(int i = 1; i <= m; ++i) {
bool ok = true;
for(int j = 0; j < n; ++j) {
if(i >= a[j] && i <= b[j]) {
ok = false;
break;
}
}
if(ok)
sol.push_back(i);
}
printf("%d\n", int(sol.size()));
for(int i = 0; i < sol.size(); ++i)
printf("%d ", sol[i]);
puts("");
return 0;
}
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Input
2 2
1 2
5 5
Output
3 4
Demonstration
Codeforces Solution-A. Points in Segments-Solution in C, C++, Java, Python,Points in Segments,Codeforces Solution