How to solve Coding Challenge 1480. Running Sum of 1d Array

How to solve Coding Challenge 1480. Running Sum of 1d Array

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Nhut Nguyen
Jun 1, 2022·

2 min read

Problem statement

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.

Solution 1: Unchange nums

Code

#include <vector>
#include <iostream>
using namespace std;
vector<int> runningSum(vector<int>& nums) {
    vector<int> rs;
    int s = 0;
    for (int n : nums) {
        s += n;
        rs.push_back(s);
    }
    return rs;
}
void printResult(vector<int>& sums) {
    cout << "[";
    for (int s: sums) {
        cout << s << ",";
    }
    cout << "]\n";
}
int main() {
    vector<int> nums{1,2,3,4};
    auto rs = runningSum(nums);
    printResult(rs);
    nums = {1,1,1,1,1};
    rs = runningSum(nums);
    printResult(rs);
    nums = {3,1,2,10,1};
    rs = runningSum(nums);
    printResult(rs);
}

Output:
[1,3,6,10,]
[1,2,3,4,5,]
[3,4,6,16,17,]

Complexity

  • Runtime: O(N), where N = nums.length.

  • Extra space: O(1).

Solution 2: Change nums

If nums is allowed to be changed, you could use it to store the result directly.

Code

#include <vector>
#include <iostream>
using namespace std;
vector<int> runningSum(vector<int>& nums) {
    for (int i = 1; i < nums.size(); i++) {
        nums[i] += nums[i - 1];
    }
    return nums;
}
void printResult(vector<int>& sums) {
    cout << "[";
    for (int s: sums) {
        cout << s << ",";
    }
    cout << "]\n";
}
int main() {
    vector<int> nums{1,2,3,4};
    auto rs = runningSum(nums);
    printResult(rs);
    nums = {1,1,1,1,1};
    rs = runningSum(nums);
    printResult(rs);
    nums = {3,1,2,10,1};
    rs = runningSum(nums);
    printResult(rs);
}

Output:
[1,3,6,10,]
[1,2,3,4,5,]
[3,4,6,16,17,]

Complexity

  • Runtime: O(N), where N = nums.length.

  • Extra space: O(1).

References