LeetCode 2640. Find the Score of All Prefixes of an Array Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)