2219. Maximum Sum Score of Array
Problem Description
In this problem, you're provided with an array of integers named nums
indexed from 0 to n-1
, where n
is the length of the array. You need to calculate what is called the sum score for each index of the array. The sum score at a particular index i
is defined as the maximum between two sums:
- The sum of elements from the start of the array up through index
i
. - The sum of elements from index
i
through the end of the array.
Your task is to determine the maximum sum score that can be obtained at any index i
in the provided array.
Intuition
The key to solving this problem lies in understanding prefix and suffix sums, which are cumulative sums of the elements of the array from the beginning up to a certain element, and from a certain element to the end, respectively.
-
Calculate the prefix sum for the array, which gives you the sum of elements from the start of the array up to each index
i
. This is stored in an arrays
wheres[i]
would represent the sum of elements fromnums[0]
tonums[i-1]
. -
Iterate through the array to calculate the sum score at each index. This can be done in a single pass after the prefix sum array has been built:
- For each index
i
, determine the sum from the start of the array to indexi
. This is given by the prefix sum at indexi + 1
(since we initialized prefix sum with0
at the beginning). - To determine the sum from index
i
to the end of the array, subtract the prefix sum up to indexi
from the total sum of the array, which is the last element of the prefix sum array. - The sum score at index
i
is the maximum of these two sums.
- For each index
-
After calculating the sum scores for each index, the maximum sum score is the answer, which represents the maximum value obtained from any index
i
.
By using a prefix sum array and iterating through the array only once, we can solve this problem efficiently with a time complexity of O(n), where n is the length of the input array.
Learn more about Prefix Sum patterns.
Solution Approach
The given solution implements the intuition behind the problem in Python. It utilizes the accumulate
function from the itertools
module to generate prefix sums efficiently and a simple loop to compare sums at each index.
Here is a step-by-step explanation of the provided code:
Step 1: Calculate Prefix Sums
The line s = [0] + list(accumulate(nums))
is crucial. It creates a new list s
that stores the prefix sums of the nums
array. The accumulate
function takes each element and adds it to the sum of all the previous elements. The [0]
at the start of this list is to simplify calculations for the prefix sum at index 0. After this line, s[i]
will contain the sum of nums
from nums[0]
up to nums[i-1]
.
Step 2: Iterate to Find Maximum Sum Score
The expression (max(s[i + 1], s[-1] - s[i]) for i in range(len(nums)))
uses a generator to calculate sum scores without storing them. For each index i
in nums
, it calculates two sums:
s[i + 1]
gives us the sum of elements from the beginning of the array to indexi
(prefix sum).s[-1] - s[i]
gives us the sum of elements from indexi
to the end of the array (this works becauses[-1]
represents the total sum of the array, which is the last element in the prefix sums list).
Step 3: Find the Maximum Value
The max
function is used again to find the maximum sum score from the generator. This is the maximum value that can be obtained from any of the sum scores calculated in the previous step.
The result of the max
function call is returned as the final answer, representing the overall maximum sum score at any index i
in the input nums
array.
Algorithm Pattern
The algorithm follows a pattern often used for solving cumulative sum problems: it first preprocesses the input array to build a data structure (in this case, a prefix sum array), enabling efficient calculations of subarray sums. It then iterates through the array a single time to find the desired maximum value.
Data Structure
An array (or list in Python) is the primary data structure used here for storing the prefix sums.
Time Complexity
Since each of the steps above runs in O(n) time, where n
is the length of the nums
array, and there are no nested loops, the overall time complexity of the function is O(n).
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Start EvaluatorExample Walkthrough
Let's walk through an example to illustrate the solution approach.
Consider the array nums = [3, 1, -2, 5, 2]
. Our goal is to find the maximum sum score at any index i
after performing the calculations as outlined in the problem description.
Step 1: Calculate Prefix Sums
First, we calculate the prefix sums for the array nums
.
- Initialize an array
s
with an extra0
at the beginning to simplify the prefix sum calculations. - Using the
accumulate
function from theitertools
module, we generate the prefix sums ofnums
. This results in the following array:
In this array,s = [0] + list(accumulate([3, 1, -2, 5, 2])) = [0, 3, 4, 2, 7, 9]
s[i]
represents the sum of elements fromnums[0]
tonums[i-1]
. For example,s[3] = 2
corresponds to the sum of elements3 + 1 - 2
.
Step 2: Iterate to Find Maximum Sum Score
Next, we iterate through the array to calculate the sum score at each index i
.
-
For index
i = 0
, the sum from the start to index0
iss[1] = 3
, and from index0
to the end iss[-1] - s[0] = 9 - 0 = 9
. The sum score for index0
is the maximum of these two, which is9
. -
For index
i = 1
, the sum from the start to index1
iss[2] = 4
, and from index1
to the end iss[-1] - s[1] = 9 - 3 = 6
. The sum score for index1
is the maximum, which is6
. -
Continuing in this way for each index, we compute the sum score:
Index 0: max(3, 9) = 9 Index 1: max(4, 6) = 6 Index 2: max(4, 7) = 7 Index 3: max(2, 7) = 7 Index 4: max(7, 2) = 7
Step 3: Find the Maximum Value
Finally, we find the maximum sum score from all the calculated sum scores, which is 9
(obtained at index 0
).
Using this approach, we can efficiently determine that the maximum sum score for the array [3, 1, -2, 5, 2]
is 9
. The solution only iterates through the array once (after computing the prefix sums), thus ensuring a time complexity of O(n).
Solution Implementation
1from typing import List
2from itertools import accumulate
3
4class Solution:
5 def maximum_sum_score(self, nums: List[int]) -> int:
6 # Prefix sum array initialization with an extra zero at the beginning
7 # to handle the case where we accumulate from the start of the array.
8 prefix_sums = [0] + list(accumulate(nums))
9
10 # Calculate the maximum sum score by iterating through each element
11 # in the nums array while taking into account both the sum of elements
12 # before the current element (prefix sum) and the sum of elements
13 # after the current element (suffix sum).
14 max_score = max(
15 max(prefix_sums[i + 1], prefix_sums[-1] - prefix_sums[i])
16 for i in range(len(nums))
17 )
18
19 # Return the maximum sum score found.
20 return max_score
21
1class Solution {
2 public long maximumSumScore(int[] nums) {
3 int length = nums.length;
4 // Create an array to store the prefix sums
5 long[] prefixSums = new long[length + 1];
6
7 // Calculate the prefix sums
8 for (int i = 0; i < length; ++i) {
9 prefixSums[i + 1] = prefixSums[i] + nums[i];
10 }
11
12 // Initialize the maximum sum as the smallest possible value
13 long maxSum = Long.MIN_VALUE;
14
15 // Find the maximum sum score by choosing the larger between
16 // the sum from the start to the current element or
17 // the sum from the current element to the end
18 for (int i = 0; i < length; ++i) {
19 long sumFromStart = prefixSums[i + 1];
20 long sumFromEnd = prefixSums[length] - prefixSums[i];
21 maxSum = Math.max(maxSum, Math.max(sumFromStart, sumFromEnd));
22 }
23
24 // Return the maximum score found
25 return maxSum;
26 }
27}
28
1#include <vector>
2#include <algorithm> // for std::max
3#include <climits> // For INT_MIN constant
4
5class Solution {
6public:
7 long long maximumSumScore(vector<int>& nums) {
8 int n = nums.size(); // Get the size of the input vector
9 vector<long long> prefixSums(n + 1, 0); // Initialize prefix sums vector with an extra element
10
11 // Calculate prefix sums for the entire array
12 for (int i = 0; i < n; ++i) {
13 prefixSums[i + 1] = prefixSums[i] + nums[i];
14 }
15
16 long long maxScore = LLONG_MIN; // Initialize the maxScore with the smallest possible value for long long
17
18 // Iterate through the array to find the maximum sum score
19 for (int i = 0; i < n; ++i) {
20 // For each element, we consider two cases:
21 // 1. The sum of elements from start up to i (prefixSums[i + 1])
22 // 2. The sum of elements from i to the end of the array (prefixSums[n] - prefixSums[i])
23 // The maximum of these two values is compared with the maxScore to update it
24 maxScore = max(maxScore, max(prefixSums[i + 1], prefixSums[n] - prefixSums[i]));
25 }
26
27 // Return the maximum sum score
28 return maxScore;
29 }
30};
31
1function maximumSumScore(nums: number[]): number {
2 const numElements = nums.length; // Total number of elements in the array
3
4 // Create and initialize a prefix sum array with an additional
5 // first element set to 0 for ease of calculation
6 let prefixSums = new Array(numElements + 1).fill(0);
7
8 // Populate the prefix sum array with cumulative sums,
9 // where prefixSums[i] will hold the sum of nums[0] to nums[i - 1]
10 for (let i = 0; i < numElements; ++i) {
11 prefixSums[i + 1] = prefixSums[i] + nums[i];
12 }
13
14 // Initialize the answer variable to the minimum possible value
15 // since we are looking for the maximum
16 let maxSumScore = -Infinity;
17
18 // Iterate through each element and calculate the sum score
19 // by taking the higher value between:
20 // 1. The sum of elements from the start to the current index (inclusive)
21 // 2. The sum of elements from the current index (excluding) to the end
22 for (let i = 0; i < numElements; ++i) {
23 // Compare the sum score at each index with the maxSumScore found so far
24 maxSumScore = Math.max(
25 maxSumScore,
26 Math.max(prefixSums[i + 1], prefixSums[numElements] - prefixSums[i])
27 );
28 }
29
30 // Return the maximum sum score found
31 return maxSumScore;
32}
33
Time and Space Complexity
The time complexity of the given code is O(n)
, where n
is the number of elements in the input list nums
. This is due to the following reasons:
- We first precompute the prefix sums with
accumulate(nums)
, which takesO(n)
time. - Then, we perform a single pass over the
nums
list to calculate the maximum sum score. During each iteration, we calculate the maximum betweens[i + 1]
ands[-1] - s[i]
. There aren
such calculations, each taking constant time.
The space complexity of the code is O(n)
. This is because we are creating a new list called s
that contains the prefix sums of the nums
array, which is of size n + 1
.
Learn more about how to find time and space complexity quickly using problem constraints.
Depth first search is equivalent to which of the tree traversal order?
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