# 2640. Find the Score of All Prefixes of an Array

## Description

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

## Solutions

• 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;
}
}

• class Solution {
public:
vector<long long> findPrefixScore(vector<int>& nums) {
int n = nums.size();
vector<long long> ans(n);
int mx = 0;
for (int i = 0; i < n; ++i) {
mx = max(mx, nums[i]);
ans[i] = nums[i] + mx + (i == 0 ? 0 : ans[i - 1]);
}
return ans;
}
};

• 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


• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• 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;
}