
Given an integer array, nums, which is sorted in non-decreasing order, your task is to construct a new integer array result of the same size. Each element in result, denoted as result[i], must represent the summation of the absolute differences between the element nums[i] and each of the other elements in the nums array. More specifically, for each i, compute result[i] as the sum of |nums[i] - nums[j]| for all j such that 0 <= j < nums.length and j != i.
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Explanation:
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2 <= nums.length <= 1051 <= nums[i] <= nums[i + 1] <= 104The key to solving this problem efficiently lies in understanding the terms involved in absolute differences in a sorted array. Here is a step-by-step approach to derive the solution:
Understanding Absolute Differences:
i, all elements before i are less than or equal to nums[i] and all elements after i are greater than or equal to nums[i].nums[i], the absolute difference with any prior element nums[j] where j < i can be simplified to nums[i] - nums[j] because nums[i] >= nums[j]. nums[k] where k > i, the absolute value of nums[i] - nums[k] simply translates to nums[k] - nums[i] since nums[k] >= nums[i].Constructing the Result Using Prefix and Suffix Sums:
i holds the sum of elements from the start of 'nums' array to the i-th element. This allows for quick calculation of the sum of any subarray from the start to a given position.prefix and suffix sums, compute for each i the contribution of all elements before i and all elements after i, in O(1) time using these summations.Final Calculation:
i, calculate the sum of absolute differences with elements before i using the prefix sum and with elements after i using the suffix sum. Summing these two gives result[i].By using this method, each result element is efficiently computed in constant time after the initial prefix and suffix arrays are built, resulting in a linear overall time complexity relative to the length of nums, which is vital given the constraints.
Understanding and applying these optimizations efficiently tackle the potential computational overhead, especially for larger arrays, within the prescribed limits.
This solution involves computing the sum of absolute differences in a sorted array. Here's a breakdown of how the provided C++ code achieves this:
Initialize necessary variables:
count to store the number of elements in the input array.sumAll to calculate the sum of all elements in the array using std::accumulate.sumLeft to keep a running sum of the elements processed so far from the left.result vector to store the results.Iterate through each element of the array:
sumRight, which is the sum of elements to the right of the current element.countLeft and countRight, representing the number of elements on the left and right of the current element, respectively.totalLeft as the difference between countLeft times the current element and sumLeft.totalRight as the difference between sumRight and countRight times the current element.totalLeft and totalRight to the result vector.sumLeft by adding the current element to it.Return the result vector containing the calculated values.
The core logic revolves around leveraging the sorted nature of the array to efficiently compute the sums of absolute differences for each element relative to all other elements. By keeping track of the sums and counts of elements to the left and right of the current element, the solution ensures optimal performance.
The provided solution details a method for computing the sum of absolute differences for each element in a sorted array. Here's a breakdown of how the method functions:
Initialize elementCount to ascertain the number of elements in the array.
Calculate sumAllElements which is the total sum of all the elements in the array.
Initialize sumPreviousElements to keep track of the sum of elements before the current index as the iterations proceed.
Create an array result to store the results for each element.
Use a loop to iterate through the array. For each element at the specified index:
sumFollowingElements which is the sum of the elements after the current index.previousCount and followingCount respectively count the number of elements before and after the current index.sumLeftDifference, which is the difference multiplied by the count of previous elements, minus sumPreviousElements.sumRightDifference, which is sumFollowingElements less the product of followingCount times the current element.result array.sumPreviousElements by adding the current element.Return the result array containing the desired sums of absolute differences for each element.
This algorithm efficiently leverages cumulative sums to calculate the required differences, ensuring each element’s result depends on previously computed totals, thus optimizing the performance by avoiding redundant recalculations.
The given Python solution calculates the sum of absolute differences for each element in a sorted array. The function calcSumOfAbsDifferences takes a list of integers (elements) and returns a new list with the computed result. Below, understand each step that this function executes:
elements and calculate the total sum of all elements.sum_to_left to zero and prepare an empty list result for storing the final output.sum_to_left and the current element from the total_sum.count_to_left as the current index and count_to_right as the number of elements to the right of the current element.sum_left, the difference between the product of count_to_left and the current element and the sum_to_left.sum_right, the difference between sum_to_right and the product of count_to_right and the current element.sum_left and sum_right to the result list.sum_to_left by adding the current element to it.result list containing the sum of absolute differences for each element.This implementation efficiently uses a single loop to manage the calculations, leveraging the properties of sorted arrays to optimize the computation of differences.
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