


We will use mathematical methods to find the maximum value of the sum of the index times the value of the elements in the array. By rotating the array, we can maximize this sum by placing the maximum value of the array at the index with the largest product. The algorithm we will use involves finding the sum of the products of the index times the element values, and then adding to that sum the difference between that sum and the product of the array length times the sum of the index values.
In the future, we will keep applying this algorithm to different arrays to find the maximum value of the sum of the index and the product of the element values that only allow rotation. This solution is very efficient as it only requires one pass through the array and has a time complexity of O(n). By using this algorithm we can quickly and easily find the maximum sum of the products of the indexes and values of the elements in the array.
method
The sum of all rotations can be obtained by multiplying each element in the array by its corresponding index and adding the results.
The maximum value can be obtained by finding the index of the maximum value and rotating the array so that the maximum value is the first element.
The maximum value can be found by summing the value of each element multiplied by its index and comparing it to the current maximum value.
The sum of all spins can be found by adding the sum of all spins to the current sum and dividing by the number of spins.
Can return the maximum value as the result.
Example
The way to solve this problem is to first sum up all the elements in the array, then iterate over the rotated array and update the sum by adding the difference of the current rotation to the previous sum. The maximum sum is the answer. Here is a complete JavaScript example -
function maxSum(arr) { let n = arr.length; let arrSum = 0; let currVal = 0; for (let i = 0; i < n; i++) { arrSum += arr[i]; currVal += i * arr[i]; } let maxVal = currVal; for (let j = 1; j < n; j++) { currVal = currVal + arrSum - n * arr[n - j]; maxVal = Math.max(maxVal, currVal); } return maxVal; } let arr = [1, 20, 2, 10]; console.log(maxSum(arr)); // Output: 72
illustrate
Function maxSum takes an array as input and returns the maximum sum that can be obtained by rotating the array and taking the sum of i * arr[i] b> for each rotation .
VariablenStores the length of the array.
VariablearrSum stores the sum of all elements in the array and is initialized to 0.
Variable currVal stores the sum of i * arr[i] of the current rotation and is initialized to 0.
The first loop calculates the sum of all elements in the array and the sum of i * arr[i] for the first rotation.
Variable maxVal stores the maximum sum and is initialized to currVal.
The second loop iteratively rotates the array and updates the sum of i * arr[i] for each rotation. The i * arr[i] sum for the current rotation is updated by adding the difference of the current rotation to the previous sum.
The value of currVal is updated by adding the difference between the i * arr[i] sum of the current rotation and the sum >i * arr[ i] is used for the last rotation. The difference is calculated by subtracting n * arr[n - j] from arrSum.
-
The currVal maximum value for each spin is stored in maxVal using the Math.max function.
Finally returns the value of maxVal as the answer.
The above is the detailed content of JavaScript program to find the maximum value of Sum( i*arr), allowing only rotation of the given array. For more information, please follow other related articles on the PHP Chinese website!

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