Most Frequent Number in Intervals - Facebook Top Interview Questions
Problem Statement :
You are given a list of list of integers intervals where each element contains the inclusive interval [start, end].
Return the most frequently occurring number in the intervals. If there are ties, return the smallest number.
Constraints
n ≤ 100,000 where n is the length of intervals
Example 1
Input
intervals = [
[1, 4],
[3, 5],
[6, 9],
[7, 9]
]
Output
3
Explanation
The most frequent numbers are [3, 4, 7, 8, 9] and they all occur twice but 3 is the smallest one.
Solution :
Solution in C++ :
int solve(vector<vector<int>>& intervals) {
map<int, int> counts;
for (const auto& i : intervals) {
counts[i[0]]++;
counts[i[1] + 1]--;
}
int level = 0;
int max_level = 0;
int ans = 0;
for (const auto& p : counts) {
level += p.second;
if (level > max_level) {
max_level = level;
ans = p.first;
}
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
private class Node {
int val;
int state;
public Node(int v, int state) {
this.val = v;
this.state = state;
}
}
private int res;
private int ans = -1;
private int count = 0;
List<Node> list = new ArrayList();
public int solve(int[][] intervals) {
for (int[] interval : intervals) {
list.add(new Node(interval[0], 0));
list.add(new Node(interval[1], 1));
}
Collections.sort(list, (n1, n2) -> {
if (n1.val == n2.val)
return n1.state - n2.state;
return n1.val - n2.val;
});
for (int i = 0; i < list.size(); i++) {
Node node = list.get(i);
int diff = (node.state == 0) ? 1 : -1;
count += diff;
if (count > ans) {
ans = count;
res = node.val;
}
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, intervals):
OPEN, CLOSE = 0, 1
events = []
for s, e in intervals:
events.append([s, OPEN])
events.append([e, CLOSE])
events.sort()
active = max_active = 0
for x, cmd in events:
active += 1 if cmd == OPEN else -1
max_active = max(max_active, active)
for x, cmd in events:
active += 1 if cmd == OPEN else -1
if active == max_active:
return x
return 0
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