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 :



title-img




                        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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