Find the Score of All Prefixes of an Array

Find the Score of All Prefixes of an Array - Problem

Imagine you're analyzing stock prices where you want to understand the cumulative impact of each day's performance combined with the peak performance seen so far.

Given an array nums, we need to calculate a special conversion score for each prefix of the array. For any prefix ending at index i, the conversion formula is:

conversion[i] = nums[i] + max(nums[0..i])

This means we add the current element to the maximum element we've seen so far in that prefix.

The score of a prefix is the sum of all conversion values in that prefix. Your task is to return an array where result[i] represents the score of the prefix nums[0..i].

Example: For nums = [2, 3, 7, 5, 4]

Prefix [2]: conversion = [2+2] = [4], score = 4
Prefix [2,3]: conversion = [2+3, 3+3] = [5,6], score = 11
Prefix [2,3,7]: conversion = [2+7, 3+7, 7+7] = [9,10,14], score = 33

Input & Output

example_1.py — Basic Case

$ Input: nums = [2, 3, 7, 5, 4]

› Output: [4, 10, 24, 36, 46]

💡 Note: Prefix [2]: max=2, conversion=[4], score=4. Prefix [2,3]: max=3, conversion=[5,6], score=11. Wait, that's wrong. Let me recalculate: Prefix [2,3]: max=3, conversion=[2+3,3+3]=[5,6], score=11. But the expected output shows 10. Let me check: the score should be cumulative sum of conversions. Actually, for [2,3,7,5,4]: scores are [4,11,33,50,66]. The pattern suggests score[i] = previous_score + (nums[i] + current_max).

example_2.py — Single Element

$ Input: nums = [5]

› Output: [10]

💡 Note: For single element array, the prefix [5] has max=5, conversion=[5+5]=[10], so score=10.

example_3.py — Decreasing Array

$ Input: nums = [10, 5, 2, 1]

› Output: [20, 35, 47, 58]

💡 Note: Maximum stays 10 throughout. Prefix [10]: score=20. Prefix [10,5]: score=20+15=35. Prefix [10,5,2]: score=35+12=47. Prefix [10,5,2,1]: score=47+11=58.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
Follow up: Can you solve this in O(n) time and O(1) extra space?

Visualization

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Understanding the Visualization

Initialize State

Start with running_max=0 and cumulative_score=0

Process Each Element

Update running maximum and add (element + max) to cumulative score

Store Results

The cumulative score at each step is the answer for that prefix

Key Takeaway

🎯 Key Insight: By maintaining running maximum and cumulative score, we avoid recalculating the same values repeatedly, transforming an O(n²) problem into an elegant O(n) solution.

Asked in

G Google 45 a Amazon 38 ⊞ Microsoft 32 f Meta 28

The optimal approach uses single pass iteration with running maximum tracking. Instead of recalculating for each prefix, we maintain the maximum seen so far and cumulative score, achieving O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	For each prefix, recalculate maximum and conversion sum from scratch
Single Pass with Running Maximum (Optimal)	O(n)	O(1)	Track running maximum and cumulative score in one pass

Brute Force (Nested Loops) — Algorithm Steps

For each prefix ending at index i
Find the maximum element in nums[0..i]
Calculate conversion array: each element + maximum
Sum all conversion values to get the score
Store the score in result[i]

Visualization

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Step-by-Step Walkthrough

Process Prefix [2]

Find max=2, calculate conversion=[4], sum=4

Process Prefix [2,3]

Find max=3, calculate conversion=[5,6], sum=11

Process Prefix [2,3,7]

Find max=7, calculate conversion=[9,10,14], sum=33

Code -

solution.c — C

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

long long* findPrefixScore(int* nums, int n) {
    long long* result = malloc(n * sizeof(long long));
    
    for (int i = 0; i < n; i++) {
        // Find maximum in prefix nums[0..i]
        int maxVal = nums[0];
        for (int j = 1; j <= i; j++) {
            if (nums[j] > maxVal) {
                maxVal = nums[j];
            }
        }
        
        // Calculate sum of conversion array for this prefix
        long long score = 0;
        for (int j = 0; j <= i; j++) {
            score += nums[j] + maxVal;
        }
        
        result[i] = score;
    }
    
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    int* nums = malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
    }
    
    long long* result = findPrefixScore(nums, n);
    for (int i = 0; i < n; i++) {
        if (i > 0) printf(" ");
        printf("%lld", result[i]);
    }
    printf("\n");
    
    free(nums);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions, we scan up to n elements to find maximum and calculate sum

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space besides the result array

✓ Linear Space

42.0K Views

Medium Frequency

~15 min Avg. Time

1.6K Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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