LeetCode #2640 — MEDIUM

Find the Score of All Prefixes of an Array

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

The Problem

Problem Statement

We define the conversion array conver of an array arr as follows:

conver[i] = arr[i] + max(arr[0..i]) where max(arr[0..i]) is the maximum value of arr[j] over 0 <= j <= i.

We also define the score of an array arr as the sum of the values of the conversion array of arr.

Given a 0-indexed integer array nums of length n, return an array ans of length n where ans[i] is the score of the prefix nums[0..i].

Example 1:

Input: nums = [2,3,7,5,10]
Output: [4,10,24,36,56]
Explanation: 
For the prefix [2], the conversion array is [4] hence the score is 4
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56

Example 2:

Input: nums = [1,1,2,4,8,16]
Output: [2,4,8,16,32,64]
Explanation: 
For the prefix [1], the conversion array is [2] hence the score is 2
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: We define the conversion array conver of an array arr as follows: conver[i] = arr[i] + max(arr[0..i]) where max(arr[0..i]) is the maximum value of arr[j] over 0 <= j <= i. We also define the score of an array arr as the sum of the values of the conversion array of arr. Given a 0-indexed integer array nums of length n, return an array ans of length n where ans[i] is the score of the prefix nums[0..i].

Baseline thinking

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

Pattern signal: Array

Example 1

[2,3,7,5,10]

Example 2

[1,1,2,4,8,16]

Core Insight

What unlocks the optimal approach

Keep track of the prefix maximum of the array
Establish a relationship between ans[i] and ans[i-1]
for 0 < i < n, ans[i] = ans[i-1]+conver[i]. In other words, array ans is the prefix sum array of the conversion array

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 #2640: Find the Score of All Prefixes of an Array
class Solution {
    public long[] findPrefixScore(int[] nums) {
        int n = nums.length;
        long[] ans = new long[n];
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            mx = Math.max(mx, nums[i]);
            ans[i] = nums[i] + mx + (i == 0 ? 0 : ans[i - 1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2640: Find the Score of All Prefixes of an Array
func findPrefixScore(nums []int) []int64 {
	n := len(nums)
	ans := make([]int64, n)
	mx := 0
	for i, x := range nums {
		mx = max(mx, x)
		ans[i] = int64(x + mx)
		if i > 0 {
			ans[i] += ans[i-1]
		}
	}
	return ans
}

# Accepted solution for LeetCode #2640: Find the Score of All Prefixes of an Array
class Solution:
    def findPrefixScore(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * n
        mx = 0
        for i, x in enumerate(nums):
            mx = max(mx, x)
            ans[i] = x + mx + (0 if i == 0 else ans[i - 1])
        return ans

// Accepted solution for LeetCode #2640: Find the Score of All Prefixes of an Array
fn find_prefix_score(nums: Vec<i32>) -> Vec<i64> {
    let mut max = 0i64;
    let mut acc = 0i64;

    let mut ret: Vec<i64> = vec![];

    for num in nums {
        let num = num as i64;

        max = std::cmp::max(max, num);
        acc += max + num;

        ret.push(acc);
    }

    ret
}

fn main() {
    let nums = vec![2, 3, 7, 5, 10];
    let ret = find_prefix_score(nums);
    println!("ret={ret:?}");
}

#[test]
fn test_find_prefix_score() {
    {
        let nums = vec![2, 3, 7, 5, 10];
        let expected = vec![4i64, 10, 24, 36, 56];
        let ret = find_prefix_score(nums);
        assert_eq!(ret, expected);
    }
    {
        let nums = vec![1, 1, 2, 4, 8, 16];
        let expected = vec![2, 4, 8, 16, 32, 64i64];
        let ret = find_prefix_score(nums);
        assert_eq!(ret, expected);
    }
}

// Accepted solution for LeetCode #2640: Find the Score of All Prefixes of an Array
function findPrefixScore(nums: number[]): number[] {
    const n = nums.length;
    const ans: number[] = new Array(n);
    let mx: number = 0;
    for (let i = 0; i < n; ++i) {
        mx = Math.max(mx, nums[i]);
        ans[i] = nums[i] + mx + (i === 0 ? 0 : ans[i - 1]);
    }
    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.