Weighted Merge Interval - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers intervals.
Each element contains [start, end, weight], meaning that during the inclusive integer interval [start, end] its weight is weight.
Weight is additive so if we have [[1, 5, 6], [2, 7, 3]], then during [2, 5] we have weight of 9.
The input intervals may or may not be overlapping.
Return the list of intervals that have the highest weight, sorted in ascending order.
The returned intervals should be merged, so [[1, 2], [3, 4]] should be merged to [1, 4].
Constraints
0 ≤ n ≤ 100,000 where n is the length of intervals
-2 ** 31 < start ≤ end < 2 ** 31
Example 1
Input
intervals = [
[1, 3, 1],
[2, 6, 1],
[5, 7, 1]
]
Output
[
[2, 3],
[5, 6]
]
Explanation
During [[2, 3], [5, 6]] weight is 2 which is the max possible.
Example 2
Input
intervals = [
[1, 2, 1],
[3, 4, 1],
[5, 6, 1]
]
Output
[
[1, 6]
]
Explanation
All intervals have weight of 1
Solution :
Solution in C++ :
vector<vector<int>> solve(vector<vector<int>>& intervals) {
map<int, int> mp;
for (auto& i : intervals) {
mp[i[0]] += i[2];
mp[i[1] + 1] -= i[2];
}
int sum = 0;
int mx = 0;
for (auto& [k, v] : mp) {
sum += v;
mx = max(sum, mx);
}
sum = 0;
vector<vector<int>> ans;
for (auto it = mp.begin(); it != prev(mp.end()); it++) {
sum += it->second;
int me = it->first;
int him = next(it)->first;
if (sum == mx) {
ans.push_back({me, him - 1});
}
}
vector<vector<int>> res;
for (auto& i : ans) {
if (!res.empty() && res.back()[1] == i[0] - 1) {
res.back()[1] = i[1];
} else {
res.push_back(i);
}
}
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int[][] solve(int[][] n) {
int N = n.length;
E[] events = new E[2 * N];
for (int i = 0; i < N; i++) {
events[i] = new E(n[i][0], 0, n[i][2]);
events[i + N] = new E(n[i][1], 1, n[i][2]);
}
Arrays.sort(events);
// find the maximum weight
int maxw = 0;
int curw = 0;
for (E e : events) {
if (e.type == 0)
curw += e.w;
else
curw -= e.w;
maxw = Math.max(maxw, curw);
}
ArrayList<int[]> ans = new ArrayList<int[]>();
int w = 0;
int last = Integer.MIN_VALUE;
for (E e : events) {
if (e.type == 0) {
w += e.w;
} else {
if (w == maxw)
ans.add(new int[] {last, e.t});
w -= e.w;
}
last = e.t;
}
ArrayList<int[]> merged = new ArrayList<int[]>();
int s = ans.get(0)[0];
int e = ans.get(0)[1];
for (int i = 1; i < ans.size(); i++) {
if (ans.get(i)[0] == e + 1) {
e = ans.get(i)[1];
} else {
merged.add(new int[] {s, e});
s = ans.get(i)[0];
e = ans.get(i)[1];
}
}
merged.add(new int[] {s, e});
return convert(merged);
}
public int[][] convert(ArrayList<int[]> arr) {
int[][] ret = new int[arr.size()][2];
for (int i = 0; i < arr.size(); i++) ret[i] = arr.get(i);
return ret;
}
static class E implements Comparable<E> {
int t;
int type;
int w;
public E(int t, int type, int w) {
this.t = t;
this.type = type;
this.w = w;
}
public int compareTo(E e) {
if (t != e.t)
return t - e.t;
else
return type - e.type;
}
}
}
Solution in Python :
class Solution:
def merge(self, intervals):
intervals.sort(key=lambda x: x[0])
merged = []
for interval in intervals:
if not merged or merged[-1][1] < interval[0] - 1:
merged.append(interval)
else:
merged[-1][1] = max(merged[-1][1], interval[1])
return merged
def solve(self, intervals):
events = []
for start, end, weight in intervals:
events.append([start, weight, 0])
events.append([end + 1, -weight, 1])
events = sorted(events, key=lambda e: (e[0], -e[2]))
highest = float("-inf")
total = 0
for time, weight, event in events:
total += weight
highest = max(highest, total)
res, total = [], 0
during, left = False, -1
for time, weight, event in events:
total += weight
if total == highest:
during, left = True, time
elif during:
res.append([left, time - 1])
during = False
if during:
res.append([left, events[-1][0] - 1])
return self.merge(res)
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