Algorithm
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.
Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
5
1 10
2 9
3 9
2 3
2 9
2 1
3
1 5
2 6
6 20
-1 -1
In the first example the following pairs are considered correct:
- (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
- (3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
- (5, 2), (2, 5) — match exactly.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int const N = 2e5 + 1;
int t, n, m;
pair < pair<int, int>, int> seg[N];
bool check(int i, int j) {
if(seg[i].first.first <= seg[j].first.first && seg[i].first.second >= seg[j].first.second)
return true;
return false;
}
int main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i) {
scanf("%d %d", &seg[i].first.first, &seg[i].first.second);
seg[i].second = i + 1;
}
if(n == 1) {
puts("-1 -1");
return 0;
}
sort(seg, seg + n);
for(int i = 1; i < n; ++i) {
if(check(i, i - 1)) {
printf("%d %d\n", seg[i - 1].second, seg[i].second);
return 0;
} else if(check(i - 1, i)) {
printf("%d %d\n", seg[i].second, seg[i - 1].second);
return 0;
}
}
puts("-1 -1");
return 0;
}Input
1 10
2 9
3 9
2 3
2 9
Output
Demonstration
Codeforces Solution-C. Nested Segments-Solution in C, C++, Java, Python ,Nested Segments,Codeforces Solution