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package g0301_0400.s0396_rotate_function;

// #Medium #Array #Dynamic_Programming #Math #2022_07_15_Time_4_ms_(81.33%)_Space_86_MB_(54.94%)

/**
 * 396 - Rotate Function\.
 *
 * Medium
 *
 * You are given an integer array `nums` of length `n`.
 *
 * Assume arrk to be an array obtained by rotating `nums` by `k` positions clock-wise. We define the **rotation function** `F` on `nums` as follow:
 *
 * *   F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1].
 *
 * Return _the maximum value of_ `F(0), F(1), ..., F(n-1)`.
 *
 * The test cases are generated so that the answer fits in a **32-bit** integer.
 *
 * **Example 1:**
 *
 * **Input:** nums = [4,3,2,6]
 *
 * **Output:** 26
 *
 * **Explanation:** 
 *
 * F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25 
 *
 * F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16 
 *
 * F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23 
 *
 * F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26 
 *
 * So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
 *
 * **Example 2:**
 *
 * **Input:** nums = [100]
 *
 * **Output:** 0
 *
 * **Constraints:**
 *
 * *   `n == nums.length`
 * *   1 <= n <= 105
 * *   `-100 <= nums[i] <= 100`
**/
public class Solution {
    public int maxRotateFunction(int[] nums) {
        int allSum = 0;
        int len = nums.length;
        int f = 0;
        for (int i = 0; i < len; i++) {
            f += i * nums[i];
            allSum += nums[i];
        }
        int max = f;
        for (int i = len - 1; i >= 1; i--) {
            f = f + allSum - len * nums[i];
            max = Math.max(f, max);
        }
        return max;
    }
}




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