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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

Solution 1: Prefix Sum

We use a variable $mx$ to record the maximum value of the first $i$ elements in the array $nums$, and use an array $ans[i]$ to record the score of the first $i$ elements in the array $nums$.

Next, we traverse the array $nums$. For each element $nums[i]$, we update $mx$, i.e., $mx = \max(mx, nums[i])$, and then update $ans[i]$. If $i = 0$, then $ans[i] = nums[i] + mx$, otherwise $ans[i] = nums[i] + mx + ans[i - 1]$.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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