Job Scheduling to Maximize Profit - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers intervals where each list contains three values [start, finish, profit].
Given you can only perform one task at a time, return the most amount of profit you can gain.
Constraints
n ≤ 10,000 where n is the length of intervals.
Example 1
Input
intervals = [
[1, 2, 50],
[3, 5, 20],
[6, 19, 100],
[2, 100, 200]
]
Output
250
Explanation
We can take intervals [1, 2, 50] and [2, 100, 200]
Example 2
Input
intervals = [
[10, 12, 250],
[3, 5, 20],
[6, 19, 100],
[2, 100, 200]
]
Output
270
Explanation
We can take the intervals [3, 5, 20] and [10, 12, 250]
Solution :
Solution in C++ :
int solve(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end());
int n = intervals.size();
vector<int> dp(n);
dp[n - 1] = intervals[n - 1][2];
for (int i = n - 2; i >= 0; --i) {
auto it = lower_bound(intervals.begin() + i + 1, intervals.end(),
vector<int>{intervals[i][1], 0, 0});
int idx = it - intervals.begin();
dp[i] = intervals[i][2];
if (it != intervals.end()) dp[i] += dp[idx];
dp[i] = max(dp[i], dp[i + 1]);
}
return dp[0];
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
int n = intervals.length;
int[] dp = new int[n];
dp[0] = intervals[0][2];
for (int i = 1; i < n; i++) {
dp[i] = dp[i - 1];
for (int j = i - 1; j >= 0; j--) {
if (intervals[i][0] >= intervals[j][1]) {
dp[i] = Math.max(dp[i], dp[j] + intervals[i][2]);
}
}
dp[i] = Math.max(dp[i], intervals[i][2]);
}
return dp[n - 1];
}
}
Solution in Python :
class Solution:
def solve(self, intervals):
intervals.sort(key=lambda x: (x[0], x[1]))
def nxt_index(
i,
): # given the current interval index, binary search for the next valid interval index
if i == len(intervals) - 1:
return i + 1
_, prev_end, _ = intervals[i]
l = i + 1
r = len(intervals) - 1
res = len(intervals)
while l <= r:
m = l + (r - l) // 2
s, e, _ = intervals[m]
if s >= prev_end:
res = min(m, res)
r = m - 1
else:
l = m + 1
return res
@lru_cache(None)
def f(idx):
if idx >= len(intervals):
return 0
_, _, c = intervals[idx]
nxt = nxt_index(idx)
res = max(c + f(nxt), f(idx + 1)) # use this index or dont
return res
return f(0)
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