} // Initialize an array to store the running sum int[] result = new int[nums.length]; result[0] = nums[0]; // Loop through the array starting from index 1, adding the previous sum to the current element for(int i = 1; i < nums.length; i++){ result[i] = result[i-1] + nums[i]; } // Return the array containing the running sum return result; } public static void main(String[] args) { Solution solution = new Solution(); // Test cases int[][] testInputs = { {2, 3, 5, 1, 6}, {1, 1, 1, 1, 1}, {-1, 2, -3, 4, -5} }; for (int[] input : testInputs) { int[] output = solution.runningSum(input); // Print the output array for (int val : output) { System.out.print(val + " "); } System.out.println(); } } } 2c:T4b8,#include #include using namespace std; class Solution { public: vector runningSum(vector& nums) { // Check if the vector is empty and return an empty vector if true if(nums.empty()){ return {}; } // Initialize a vector to store the running sum vector result(nums.size()); result[0] = nums[0]; // Loop through the vector starting from index 1, adding the previous sum to the current element for(int i = 1; i < nums.size(); i++){ result[i] = result[i-1] + nums[i]; } // Return the vector containing the running sum return result; } }; int main() { Solution solution; // Test cases vector> testInputs = { {2, 3, 5, 1, 6}, {1, 1, 1, 1, 1}, {-1, 2, -3, 4, -5} }; for (auto& input : testInputs) { vector output = so

Space complexity is O(1) not O(N)

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c a year ago