2640

2640. Find the Score of All Prefixes of an Array (Medium)

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

Companies: TikTok

Related Topics:
Array, Prefix Sum

Similar Questions:

• Most Beautiful Item for Each Query (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/find-the-score-of-all-prefixes-of-an-array
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) extra space
class Solution {
public:
    vector<long long> findPrefixScore(vector<int>& A) {
        vector<long long> ans;
        long long N = A.size(), sum = 0, mx = 0;
        for (int i = 0; i < N; ++i) {
            mx = max(mx, (long long)A[i]);
            sum += A[i] + mx;
            ans.push_back(sum);
        }
        return ans;
    }
};