Solving LeetCode Problem 2574: Left and Right Sum Differences

 In this blog post, we will explore an easy LeetCode problem, number 2574, titled "Left and Right Sum Differences." The problem provides us with a 0-indexed integer array called "nums." Our task is to find another 0-indexed integer array called "answer," where each element "answer[i]" is the absolute difference between "leftSum[i]" and "rightSum[i]."

To clarify, "leftSum[i]" represents the sum of elements to the left of the index "i" in the "nums" array, and if there is no element to the left, then "leftSum[i]" is considered to be 0. Similarly, "rightSum[i]" represents the sum of elements to the right of the index "i" in the "nums" array, and if there is no element to the right, then "rightSum[i]" is considered to be 0.

CODE : 

class Solution {

public:

    vector<int> leftRightDifference(vector<int>& nums) {

        int n = nums.size();

        vector<int> dk1;

        vector<int> dk2(n, 0);

        dk1.push_back(0);

        vector<int> ans(n, 0);

        

        int x =0;

        x+=nums[0];

        for (int i = 1; i < n; i++) {

            dk1.push_back(x);

            x += nums[i];

        }

        x=0;

        dk2[n-1]=0;

        x+=nums[n-1];

        for (int j = n - 2; j >= 0; j--) {

            dk2[j]=x;

            x += nums[j];

        }

        for (int d = 0; d < n; d++) {

            ans[d] = abs(dk1[d] - dk2[d]);

        }

        return ans;

    }

};

The problem requires us to efficiently compute the "answer" array based on the given "nums" array. We will discuss possible approaches to solve the problem, including algorithmic insights and step-by-step implementations. Additionally, we will analyze the time and space complexity of the solutions to identify the most optimal approach.

Whether you are new to LeetCode problems or seeking to improve your problem-solving skills, this blog post will guide you through a comprehensive solution for Problem 2574. So, join us on this coding journey and learn how to tackle the challenge of finding left and right sum differences efficiently!