DevExCode

2640. Find the Score of All Prefixes of an Array

ArrayPrefix Sum

Explanation

To solve this problem, we can iterate over the given array nums and calculate the score for each prefix as described in the problem statement. We will use a variable maxVal to keep track of the maximum value encountered so far while iterating. For each element nums[i], the score for that prefix will be nums[i] + maxVal. We update maxVal whenever we encounter a larger element. The score for the next prefix will include the updated maxVal.

Algorithm

  1. Initialize an array ans of length n to store the scores for each prefix.
  2. Initialize maxVal to 0.
  3. Iterate over the elements of nums:
    • Update maxVal to be the maximum of the current element and the previous maxVal.
    • Calculate the score for the current prefix as nums[i] + maxVal and store it in ans[i].
  4. Return the ans array.

Time Complexity

The time complexity of this algorithm is O(n), where n is the number of elements in the given array nums.

Space Complexity

The space complexity is O(n) for the ans array to store the scores.

View Leetcode Problem
class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        int maxVal = 0;

        for (int i = 0; i < n; i++) {
            maxVal = Math.max(maxVal, nums[i]);
            ans[i] = nums[i] + maxVal;
        }

        return ans;
    }
}

Code Editor (Testing phase)

Improve Your Solution

Use the editor below to refine the provided solution. Select a programming language and try the following:

  • Add import statement if required.
  • Optimize the code for better time or space complexity.
  • Add test cases to validate edge cases and common scenarios.
  • Handle error conditions or invalid inputs gracefully.
  • Experiment with alternative approaches to deepen your understanding.

Click "Run Code" to execute your solution and view the output. If errors occur, check the line numbers and debug accordingly. Resize the editor by dragging its bottom edge.

Previous QuestionNext Question