LeetCode #3767 — MEDIUM

Maximize Points After Choosing K Tasks

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two integer arrays, technique1 and technique2, each of length n, where n represents the number of tasks to complete.

If the i^th task is completed using technique 1, you earn technique1[i] points.
If it is completed using technique 2, you earn technique2[i] points.

You are also given an integer k, representing the minimum number of tasks that must be completed using technique 1.

You must complete at least k tasks using technique 1 (they do not need to be the first k tasks).

The remaining tasks may be completed using either technique.

Return an integer denoting the maximum total points you can earn.

Example 1:

Input: technique1 = [5,2,10], technique2 = [10,3,8], k = 2

Output: 22

Explanation:

We must complete at least k = 2 tasks using technique1.

Choosing technique1[1] and technique1[2] (completed using technique 1), and technique2[0] (completed using technique 2), yields the maximum points: 2 + 10 + 10 = 22.

Example 2:

Input: technique1 = [10,20,30], technique2 = [5,15,25], k = 2

Output: 60

Explanation:

We must complete at least k = 2 tasks using technique1.

Choosing all tasks using technique 1 yields the maximum points: 10 + 20 + 30 = 60.

Example 3:

Input: technique1 = [1,2,3], technique2 = [4,5,6], k = 0

Output: 15

Explanation:

Since k = 0, we are not required to choose any task using technique1.

Choosing all tasks using technique 2 yields the maximum points: 4 + 5 + 6 = 15.

Constraints:

1 <= n == technique1.length == technique2.length <= 10⁵
1 <= technique1[i], technique2[i] <= 10⁵
0 <= k <= n

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays, technique1 and technique2, each of length n, where n represents the number of tasks to complete. If the ith task is completed using technique 1, you earn technique1[i] points. If it is completed using technique 2, you earn technique2[i] points. You are also given an integer k, representing the minimum number of tasks that must be completed using technique 1. You must complete at least k tasks using technique 1 (they do not need to be the first k tasks). The remaining tasks may be completed using either technique. Return an integer denoting the maximum total points you can earn.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[5,2,10]
[10,3,8]
2

Example 2

[10,20,30]
[5,15,25]
2

Example 3

[1,2,3]
[4,5,6]
0

Step 02

Core Insight

What unlocks the optimal approach

Initially complete all tasks using way 1; let the initial total be <code>sum(reward1)</code>.
The delta when switching task <code>i</code> from way 1 to way 2 is <code>reward2[i] - reward1[i]</code>.
Sort deltas descending and apply the largest <code>n - k</code>; after each switch update the maximum.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
class Solution {
    public long maxPoints(int[] technique1, int[] technique2, int k) {
        int n = technique1.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, (i, j) -> technique1[j] - technique2[j] - (technique1[i] - technique2[i]));
        long ans = 0;
        for (int x : technique2) {
            ans += x;
        }
        for (int i = 0; i < k; i++) {
            int index = idx[i];
            ans -= technique2[index];
            ans += technique1[index];
        }
        for (int i = k; i < n; i++) {
            int index = idx[i];
            if (technique1[index] >= technique2[index]) {
                ans -= technique2[index];
                ans += technique1[index];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
func maxPoints(technique1 []int, technique2 []int, k int) int64 {
	n := len(technique1)
	idx := make([]int, n)
	for i := 0; i < n; i++ {
		idx[i] = i
	}

	sort.Slice(idx, func(i, j int) bool {
		return technique1[idx[j]]-technique2[idx[j]] < technique1[idx[i]]-technique2[idx[i]]
	})

	var ans int64
	for _, x := range technique2 {
		ans += int64(x)
	}

	for i := 0; i < k; i++ {
		index := idx[i]
		ans -= int64(technique2[index])
		ans += int64(technique1[index])
	}

	for i := k; i < n; i++ {
		index := idx[i]
		if technique1[index] >= technique2[index] {
			ans -= int64(technique2[index])
			ans += int64(technique1[index])
		}
	}

	return ans
}

# Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
class Solution:
    def maxPoints(self, technique1: List[int], technique2: List[int], k: int) -> int:
        n = len(technique1)
        idx = sorted(range(n), key=lambda i: -(technique1[i] - technique2[i]))
        ans = sum(technique2)
        for i in idx[:k]:
            ans -= technique2[i]
            ans += technique1[i]
        for i in idx[k:]:
            if technique1[i] >= technique2[i]:
                ans -= technique2[i]
                ans += technique1[i]
        return ans

// Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
// class Solution {
//     public long maxPoints(int[] technique1, int[] technique2, int k) {
//         int n = technique1.length;
//         Integer[] idx = new Integer[n];
//         Arrays.setAll(idx, i -> i);
//         Arrays.sort(idx, (i, j) -> technique1[j] - technique2[j] - (technique1[i] - technique2[i]));
//         long ans = 0;
//         for (int x : technique2) {
//             ans += x;
//         }
//         for (int i = 0; i < k; i++) {
//             int index = idx[i];
//             ans -= technique2[index];
//             ans += technique1[index];
//         }
//         for (int i = k; i < n; i++) {
//             int index = idx[i];
//             if (technique1[index] >= technique2[index]) {
//                 ans -= technique2[index];
//                 ans += technique1[index];
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3767: Maximize Points After Choosing K Tasks
function maxPoints(technique1: number[], technique2: number[], k: number): number {
    const n = technique1.length;
    const idx = Array.from({ length: n }, (_, i) => i);

    idx.sort((i, j) => technique1[j] - technique2[j] - (technique1[i] - technique2[i]));

    let ans = technique2.reduce((sum, x) => sum + x, 0);

    for (let i = 0; i < k; i++) {
        const index = idx[i];
        ans -= technique2[index];
        ans += technique1[index];
    }

    for (let i = k; i < n; i++) {
        const index = idx[i];
        if (technique1[index] >= technique2[index]) {
            ans -= technique2[index];
            ans += technique1[index];
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.