只允许给定数组旋转的情况下，寻找 Sum( i*arr[i])最大值的 Php 程序
原文:https://www . geesforgeks . org/PHP-for-program-for-find-sum-value-iarri-仅允许给定数组上的旋转/
给定一个数组，只允许对数组进行旋转操作。我们可以随意旋转阵列多次。返回 i*arr[i]的最大可能总和。
示例:
Input: arr[] = {1, 20, 2, 10}
Output: 72
We can get 72 by rotating array twice.
{2, 10, 1, 20}
20*3 + 1*2 + 10*1 + 2*0 = 72
Input: arr[] = {10, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Output: 330
We can get 330 by rotating array 9 times.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
0*1 + 1*2 + 2*3 ... 9*10 = 330
我们强烈建议你尽量减少浏览器，先自己试试这个。 A 简单解就是逐个找到所有的旋转，检查每个旋转的和，返回最大和。这个解的时间复杂度为 O(n 2 )。
我们可以使用高效解决方案在 O(n)时间内解决这个问题。 设 R j 为 i*arr[i]的值，j 旋转。其思路是由前一次旋转计算下一次旋转值，即由 R j-1 计算出 R j 。我们可以将结果的初始值计算为 R 0 ，然后继续计算下一个旋转值。
如何从 R j-1 高效计算 R j ？ 这可以在 O(1)时间内完成。以下是细节。
Let us calculate initial value of i*arr[i] with no rotation
R0 = 0*arr[0] + 1*arr[1] +...+ (n-1)*arr[n-1]
After 1 rotation arr[n-1], becomes first element of array,
arr[0] becomes second element, arr[1] becomes third element
and so on.
R1 = 0*arr[n-1] + 1*arr[0] +...+ (n-1)*arr[n-2]
R1 - R0 = arr[0] + arr[1] + ... + arr[n-2] - (n-1)*arr[n-1]
After 2 rotations arr[n-2], becomes first element of array,
arr[n-1] becomes second element, arr[0] becomes third element
and so on.
R2 = 0*arr[n-2] + 1*arr[n-1] +...+ (n-1)*arr[n-3]
R2 - R1 = arr[0] + arr[1] + ... + arr[n-3] - (n-1)*arr[n-2] + arr[n-1]
If we take a closer look at above values, we can observe
below pattern
Rj - Rj-1 = arrSum - n * arr[n-j]
Where arrSum is sum of all array elements, i.e.,
arrSum = ∑ arr[i]
0<=i<=n-1 < pre>下面是完整的算法:
1) Compute sum of all array elements. Let this sum be 'arrSum'.
2) Compute R0 by doing i*arr[i] for given array. Let this value be currVal.
3) Initialize result: maxVal = currVal // maxVal is result.
// This loop computes Rj from Rj-1 4) Do following for j = 1 to n-1 ......a) currVal = currVal + arrSum-n*arr[n-j]; ......b) If (currVal > maxVal) maxVal = currVal
5) Return maxVal
以下是上述想法的实现:
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