In this guide, you will get 1480. Running Sum of 1d Array LeetCode Solution with the best time and space complexity. The solution to Running Sum of d Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).
Return the running sum of nums.
Example 1:
Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
Example 2:
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]
Constraints:
1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
1480. Running Sum of 1d Array LeetCode Solution in C++
class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
vector<int> ans;
int sum = 0;
for (const int num : nums)
ans.push_back(sum += num);
return ans;
}
}; /* code provided by PROGIEZ */
1480. Running Sum of 1d Array LeetCode Solution in Java
class Solution {
public int[] runningSum(int[] nums) {
int[] ans = new int[nums.length];
int sum = 0;
for (int i = 0; i < nums.length; ++i)
ans[i] = sum += nums[i];
return ans;
}
} // code provided by PROGIEZ
1480. Running Sum of 1d Array LeetCode Solution in Python
class Solution:
def runningSum(self, nums: list[int]) -> list[int]:
return itertools.accumulate(nums) # code by PROGIEZ
