Maximal Score After Applying K Operations

Maximal Score After Applying K Operations - Problem

Maximize Your Score Through Strategic Operations!

You're given a 0-indexed integer array nums and an integer k. Starting with a score of 0, you need to perform exactly k operations to maximize your total score.

Each operation allows you to:
• Choose any index i where 0 ≤ i < nums.length
• Add nums[i] to your current score
• Replace nums[i] with ⌈nums[i] / 3⌉ (ceiling division by 3)

The challenge is to strategically choose which elements to pick to achieve the maximum possible score after exactly k operations.

Goal: Return the maximum score you can achieve.
Key Insight: Always pick the largest available number for maximum immediate gain!

Input & Output

example_1.py — Basic case

$ Input: nums = [10,10,10,10,10], k = 5

› Output: 50

💡 Note: Pick 10 five times. First operation: score=10, array becomes [4,10,10,10,10] (since ceil(10/3)=4). We continue picking 10s until we get score = 10+10+10+10+10 = 50.

example_2.py — Different values

$ Input: nums = [1,10,3,3,3], k = 3

› Output: 17

💡 Note: Operation 1: Pick 10, score=10, array becomes [1,4,3,3,3]. Operation 2: Pick 4, score=14, array becomes [1,2,3,3,3]. Operation 3: Pick 3, score=17, array becomes [1,2,1,3,3]. Total score = 17.

example_3.py — Single element

$ Input: nums = [7], k = 3

› Output: 12

💡 Note: Operation 1: Pick 7, score=7, array becomes [3] (ceil(7/3)=3). Operation 2: Pick 3, score=10, array becomes [1] (ceil(3/3)=1). Operation 3: Pick 1, score=11, array stays [1]. Total = 7+3+1 = 11.

Visualization

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Understanding the Visualization

Setup Orchard

Arrange all apple trees by size (max heap)

Pick Largest Apple

Always harvest from the tree with biggest apples

Tree Updates

After picking, apples on that tree shrink to 1/3 size

Re-organize

Re-arrange trees by new apple sizes

Repeat

Continue until you've made k picks

Key Takeaway

🎯 Key Insight: The greedy approach works perfectly here because elements shrink independently - picking the largest element at each step maximizes immediate gain without affecting the relative order of other elements' future potential.

Time & Space Complexity

Time Complexity

⏱️

O(n^k)

At each of k operations, we try all n indices, leading to n^k total combinations

✓ Linear Growth

Space Complexity

O(k)

Recursion depth is k, each frame uses constant space

✓ Linear Space

Constraints

1 ≤ nums.length, k ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
Note: Total score can exceed 32-bit integer range

Asked in

G Google 42 a Amazon 38 f Meta 25 ⊞ Microsoft 31

The optimal approach uses a greedy strategy with max heap to always pick the largest available element. This works because maximizing immediate gain at each step leads to the globally optimal solution. The algorithm runs in O(n + k log n) time by maintaining a priority queue of elements and performing k extract-max and insert operations.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(n^k)	O(k)	Try every possible sequence of k operations using recursion
Greedy with Max Heap (Optimal)	O(n + k log n)	O(n)	Always pick the largest element using a priority queue for O(k log n) efficiency

Brute Force (Try All Combinations) — Algorithm Steps

Use recursive function with current array state, operations left, and current score
Base case: if k operations done, return current score
For each index in array, try picking it: add to score, divide by 3 (ceiling)
Recursively solve for remaining k-1 operations
Return maximum score among all choices

Visualization

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Step-by-Step Walkthrough

Initial State

Start with original array and k operations

Branch Decision

Try picking each index for current operation

Update & Recurse

Add score, update element, recurse for remaining operations

Backtrack

Restore original value and try next option

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

long long backtrack(int* arr, int n, int operationsLeft) {
    if (operationsLeft == 0) {
        return 0;
    }
    
    long long maxScore = 0;
    for (int i = 0; i < n; i++) {
        long long score = arr[i];
        int original = arr[i];
        arr[i] = (int)ceil((double)arr[i] / 3);
        
        long long remainingScore = backtrack(arr, n, operationsLeft - 1);
        if (score + remainingScore > maxScore) {
            maxScore = score + remainingScore;
        }
        
        arr[i] = original;
    }
    
    return maxScore;
}

long long maxKelements(int* nums, int numsSize, int k) {
    int* arr = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        arr[i] = nums[i];
    }
    
    long long result = backtrack(arr, numsSize, k);
    free(arr);
    return result;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n^k)

At each of k operations, we try all n indices, leading to n^k total combinations

✓ Linear Growth

Space Complexity

O(k)

Recursion depth is k, each frame uses constant space

✓ Linear Space

Constraints

1 ≤ nums.length, k ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
Note: Total score can exceed 32-bit integer range

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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