☑️BEST C++ SOLUTION JUST CHECK ✅
PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT\n# CONNECT WITH ME\n### https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n#### https://www.instagram.com/pratay_nand
# **PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT**\n# **CONNECT WITH ME**\n### **[https://www.linkedin.com/in/pratay-nandy-9ba57b229/]()**\n#### **[https://www.instagram.com/pratay_nandy/]()**\n\n# Approach\nThe provided C++ code defines a function `getSumAbsoluteDifferences` that takes a vector of integers `nums` as input and returns a vector of integers `ans`. The goal is to calculate the sum of absolute differences between each element in the array and the rest of the elements.\n\nHere\'s a breakdown of the approach:\n\n1. **Calculate the Total Sum:**\n   - Initialize a variable `sum` to zero. Use a loop to iterate through the elements of the `nums` array and calculate the total sum of all elements.\n\n   ```cpp\n   int sum = 0;\n   for (int i = 0; i < nums.size(); i++) {\n       sum += nums[i];\n   }\n   ```\n\n2. **Calculate Absolute Differences:**\n   - Initialize a vector `ans` of the same size as `nums` to store the results.\n   - Initialize a variable `leftsum` to zero to keep track of the cumulative sum of elements to the left of the current element during the iteration.\n\n   ```cpp\n   int n = nums.size();\n   vector<int> ans(n);\n   int leftsum = 0;\n   ```\n\n3. **Iterate Through the Elements:**\n   - Use a loop to iterate through each element in the `nums` array.\n   - For each element at index `i`, calculate the absolute difference sum using the formula: `(nums[i] * i - leftsum) + ((sum - leftsum - nums[i]) - (nums[i] * (n - i - 1)))`.\n   - Update the `leftsum` by adding the current element.\n\n   ```cpp\n   for (int i = 0; i < n; i++) {\n       ans[i] = (nums[i] * i - leftsum) + ((sum - leftsum - nums[i]) - (nums[i] * (n - i - 1)));\n       leftsum += nums[i];\n   }\n   ```\n\n4. **Return the Result:**\n   - Return the vector `ans` containing the sum of absolute differences for each element.\n\n   ```cpp\n   return ans;\n   ```\n\nThe approach leverages cumulative sums and the properties of absolute differences to efficiently calculate the desired result without explicitly iterating through all pairs of elements.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:***0(N)***\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:**0(1)**\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n![upvote2.png](https://assets.leetcode.com/users/images/0bf65a30-962d-4855-b7f3-05c795c2f3ca_1700919344.2965546.png)\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> getSumAbsoluteDifferences(vector<int>& nums) {\n    int sum = 0 ; \n    for(int i =0 ;i<nums.size();i++)\n    {\n        sum+=nums[i];\n    }   \n    int n = nums.size();\n    vector<int>ans(n);\n    int leftsum = 0;\n    for(int i = 0; i < n;i++)\n    {\n        ans[i] = (nums[i] * i - leftsum  )+( (sum - leftsum - nums[i]) - (nums[i] * (n-i-1)));\n        leftsum+=nums[i];\n    }\n    return ans;\n}\n};\n```
['Array', 'Math', 'Prefix Sum', 'C++']