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Solution: Running Sum of 1d Array (easy)

Problem Statement

Given a one-dimensional array of integers, create a new array that represents the running sum of the original array.

The running sum at position i in the new array is calculated as the sum of all the numbers in the original array from the 0th index up to the i-th index (inclusive). Formally, the resulting array should be computed as follows: result[i] = sum(nums[0] + nums[1] + ... + nums[i]) for each i from 0 to the length of the array minus one.

Examples

Example 1

  • Input: [2, 3, 5, 1, 6]
  • Expected Output: [2, 5, 10, 11, 17]

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Aravind Badiger

Aravind Badiger

· 2 years ago

Whether we consider a returning result variable or replace elements in the original array the space complexity is O(1)

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E

ethanedge

· 19 days ago

The solution states:

  • Check for Edge Cases:
    • If the input array is null or has no elements, return an empty array since there's nothing to process.

However, the question states:

Constraints:

  • 1 <= nums.length <= 1000

Which could be confusing to some as it's contradictory.

Show 1 reply
Rafael Scarduelli

Rafael Scarduelli

· 2 years ago

This happens on line 4 when trying to get nums[0].

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rcreddyn

rcreddyn

· 7 days ago

Is the check for empty array or null necessary as contraints state the minimum size of the array is 1?