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Running Sum of 1d Array (easy)
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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] - Justification:
- For i=0: 2
- For i=1: 2 + 3 = 5
- For i=2: 2 + 3 + 5 = 10
- For i=3: 2 + 3 + 5 + 1 = 11
- For i=4: 2 + 3 + 5 + 1 + 6 = 17
Example 2
- Input:
[1, 1, 1, 1, 1] - Expected Output:
[1, 2, 3, 4, 5] - Justification: Each element is simply the sum of all preceding elements plus the current element.
Example 3
- Input:
[-1, 2, -3, 4, -5] - Expected Output:
[-1, 1, -2, 2, -3] - Justification: Negative numbers are also summed up in the same manner as positive ones.
Constraints:
1 <= nums.length <= 1000- -10^6 <= nums[i] <= 10^6
Try it yourself
Try solving this question here:
Please check the next lesson for a complete solution.
rcreddyn
· 2 months ago
Is the check for empty array or null necessary as contraints state the minimum size of the array is 1?
ethanedge
· 2 months ago
The solution states:
- Check for Edge Cases:
- If the input array is
nullor has no elements, return an empty array since there's nothing to process.
- If the input array is
However, the question states:
Constraints:
1 <= nums.length <= 1000
Which could be confusing to some as it's contradictory.
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)
Rafael Scarduelli
· 3 years ago
This happens on line 4 when trying to get nums[0].
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