Leetcode 2574: Left and Right Sum Differences
You are given a 0-indexed integer array nums. Find a 0-indexed integer array answer where: answer[i] = |leftSum[i] - rightSum[i]|, where leftSum[i] is the sum of elements to the left of the index i, and rightSum[i] is the sum of elements to the right of the index i.
Problem
Approach
Steps
Complexity
Input: The input consists of a single integer array nums.
Example: For example, nums = [5, 3, 7, 2].
Constraints:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10^5
Output: Return an array answer where answer[i] = |leftSum[i] - rightSum[i]|, with leftSum[i] being the sum of elements to the left of index i and rightSum[i] being the sum of elements to the right of index i.
Example: For nums = [5, 3, 7, 2], the output is [12, 2, 4, 12].
Constraints:
• The output array must have the same length as the input array.
Goal: The goal is to compute the leftSum and rightSum arrays and use them to calculate the absolute difference at each index.
Steps:
• 1. Calculate the leftSum array where leftSum[i] is the sum of all elements before index i.
• 2. Calculate the rightSum array where rightSum[i] is the sum of all elements after index i.
• 3. For each index i, compute answer[i] = |leftSum[i] - rightSum[i]|.
Goal: The input size is at most 1000 and each element can go up to 10^5.
Steps:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10^5
Assumptions:
• The input will always have at least one element.
• Input: For nums = [5, 3, 7, 2], the output is [12, 2, 4, 12].
• Explanation: We calculate the leftSum and rightSum for each index, and then take the absolute difference of the sums.
Approach: The approach involves calculating the cumulative sum to the left and right for each index, and then taking their absolute difference.
Observations:
• We need to calculate the sum of the elements on the left and right of each index efficiently.
• By using cumulative sums, we can compute both leftSum and rightSum in linear time.
Steps:
• 1. Initialize the leftSum and rightSum arrays.
• 2. Compute the cumulative sum from the left and right.
• 3. Calculate the answer array by finding the absolute difference between leftSum and rightSum for each index.
Empty Inputs:
• nums will always have at least one element.
Large Inputs:
• Ensure the solution handles large arrays efficiently.
Special Values:
• Handle cases where all elements are the same.
Constraints:
• The output should handle large integers resulting from the sums of large elements.
vector<int> leftRightDifference(vector<int>& nums) {
int n = nums.size();
vector<int> sum(n + 1, 0), ans(n, 0);
for(int i = 0; i < n; i++)
sum[i + 1] = nums[i] + sum[i];
int net = sum[n];
for(int i = 0; i < n; i++) {
ans[i] = abs(net - sum[i + 1] - sum[i]);
}
return ans;
}
1 : Function Definition
vector<int> leftRightDifference(vector<int>& nums) {
Define the function 'leftRightDifference' which takes a vector of integers as input and returns a vector of integers containing the left-right differences.
2 : Variable Initialization
int n = nums.size();
Declare an integer 'n' to store the size of the input vector 'nums'.
3 : Variable Initialization
vector<int> sum(n + 1, 0), ans(n, 0);
Declare two vectors: 'sum' to store the prefix sums of the input array, and 'ans' to store the resulting differences. Initialize both vectors with zeros.
4 : Loop
for(int i = 0; i < n; i++)
Start a loop to iterate through the input array and compute the prefix sums.
5 : Prefix Sum Calculation
sum[i + 1] = nums[i] + sum[i];
Compute the prefix sum for each element by adding the current element of 'nums' to the previous value in the 'sum' array.
6 : Blank Line
An empty line used for readability and to separate sections of code.
7 : Variable Initialization
int net = sum[n];
Store the total sum of the input array in the variable 'net', which is the last element of the 'sum' array.
8 : Loop
for(int i = 0; i < n; i++) {
Start a loop to calculate the left-right differences for each element.
9 : Difference Calculation
ans[i] = abs(net - sum[i + 1] - sum[i]);
For each element, compute the absolute difference between the total sum 'net' and the sum of the elements to the left and right of the current element.
10 : Return Statement
return ans;
Return the result vector 'ans', which contains the left-right differences for each element in the input array.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in the size of the input array, as we iterate over the array a few times.
Best Case: O(n)
Worst Case: O(n)
Description: We use extra space to store the leftSum and rightSum arrays, which are both of size n.
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