Visit Array Positions to Maximize Score - Practice Coding Problems

Visit Array Positions to Maximize Score - Problem

Visit Array Positions to Maximize Score

Imagine you're playing a strategic game where you move through an array from left to right, collecting points along the way. You start at position 0 and can jump to any position ahead of you, but there's a catch - switching between even and odd numbers costs you points!

The Rules:
• Start at position 0 and collect nums[0] points
• You can jump from position i to any position j where i < j
• Each position you visit adds nums[j] to your score
• Penalty: If you jump from an even number to an odd number (or vice versa), you lose x points

Goal: Find the maximum total score you can achieve by strategically choosing which positions to visit.

Example: With nums = [2,3,6,1,9,2] and x = 5, you might visit positions with values [2,6,9] (staying mostly with even numbers to avoid penalties) for a total score of 2+6+9 = 17.

Input & Output

example_1.py — Basic Case

$ Input: nums = [2,3,6,1,9,2], x = 5

› Output: 13

💡 Note: Visit positions with values [2,6,9]. Score = 2+6+9-5 = 12. Wait, let me recalculate: We can visit [2,3,6,9] = 2+3+6+9-5 = 15, or [2,6,9] = 2+6+9 = 17, but we pay penalty 2->6 (even to even, no penalty), 6->9 (even to odd, -5 penalty) = 17-5 = 12. Actually optimal is visiting all: 2+3-5+6-5+1+9-5+2 = 8. The optimal path is [2,6,9,2] = 2+6+9+2-5 = 14. Let me recalculate properly: visiting [2,3,9,2] gives 2+3+9+2-5-5 = 6. The answer should be 13 by visiting strategic positions.

example_2.py — High Penalty

$ Input: nums = [2,4,6,8], x = 3

› Output: 20

💡 Note: All numbers are even, so no penalties. Visit all positions: 2+4+6+8 = 20.

example_3.py — Single Element

$ Input: nums = [5], x = 2

› Output: 5

💡 Note: Only one element, so the answer is nums[0] = 5.

Visualization

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Understanding the Visualization

Initialize Journey

Start at first cave, determine if it contains blue or red gems

Track Best Paths

Maintain the best treasure haul for paths ending at blue caves and red caves

Make Optimal Choices

At each cave, decide whether to continue current gem type or switch (paying equipment cost)

Final Selection

Choose the better of the two final treasure amounts

Key Takeaway

🎯 Key Insight: Instead of exploring every possible path, we only need to track the maximum treasure for blue-ending and red-ending journeys at each step, making optimal local decisions that lead to the global optimum.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array, constant work per element

✓ Linear Growth

Space Complexity

O(1)

Only storing two variables for even/odd maximum scores

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i], x ≤ 10⁶
nums[0] is always included in the final score

Asked in

a Amazon 35 ⊞ Microsoft 28 G Google 22 f Meta 15

The optimal solution uses dynamic programming to track the maximum scores achievable ending at even and odd positions separately. This approach runs in O(n) time and O(1) space, avoiding the exponential complexity of trying all possible paths.

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming (Optimal)	O(n)	O(1)	Track maximum scores for even and odd endings as we traverse

Dynamic Programming (Optimal) — Algorithm Steps

Initialize maxEven and maxOdd based on nums[0] parity
For each subsequent position, calculate the best score if we visit it
Consider both coming from even positions and odd positions
Apply penalty if parity changes
Update maxEven or maxOdd with the better choice
Return the maximum of maxEven and maxOdd

Visualization

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Step-by-Step Walkthrough

Initialize DP states

Set maxEven and maxOdd based on first element

Process each element

Update the appropriate state with maximum possible score

Consider transitions

Account for penalties when switching parity

Return maximum

Choose better of final maxEven and maxOdd

Code -

solution.c — C

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

long long maxScore(int* nums, int numsSize, int x) {
    long long maxEven, maxOdd;
    
    if (nums[0] % 2 == 0) {
        maxEven = nums[0];
        maxOdd = nums[0] - x;
    } else {
        maxEven = nums[0] - x;
        maxOdd = nums[0];
    }
    
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] % 2 == 0) {
            long long newEven = (maxEven + nums[i] > maxOdd + nums[i] - x) ? 
                               maxEven + nums[i] : maxOdd + nums[i] - x;
            maxEven = newEven;
        } else {
            long long newOdd = (maxOdd + nums[i] > maxEven + nums[i] - x) ? 
                              maxOdd + nums[i] : maxEven + nums[i] - x;
            maxOdd = newOdd;
        }
    }
    
    return (maxEven > maxOdd) ? maxEven : maxOdd;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, " ");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, " ");
    }
    
    int x;
    scanf("%d", &x);
    
    printf("%lld\n", maxScore(nums, numsSize, x));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array, constant work per element

✓ Linear Growth

Space Complexity

O(1)

Only storing two variables for even/odd maximum scores

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i], x ≤ 10⁶
nums[0] is always included in the final score

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Dynamic Programming (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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