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396. Rotate Function

2024-08-23 13:58:38   216人阅读

Given an array of integers A and let n to be its length.

Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].

Calculate the maximum value of F(0), F(1), ..., F(n-1).

Note:
n is guaranteed to be less than 105.

Example:

A = [4, 3, 2, 6]

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.

public class Solution {
    // Math way  F[k+1] - F[k] = sum - n * Bk[n-1]; F[k+1] = F[k] + sum - n * Bk[n-1];
    public int maxRotateFunction(int[] A) {
        if(A == null || A.length == 0) return 0;
        int sum = 0;
        int rotateSum = 0;
        for(int i = 0 ; i < A.length ; i++){
           sum += A[i]; 
           rotateSum += i * A[i];
        }
        int max = rotateSum;
        for(int i = 1; i < A.length  ; i++){
            rotateSum += sum - A.length * A[A.length-i];
            max = Math.max(max, rotateSum);
        }
        return max;
    }
    /** 
    // brute way but TLE
    public int maxRotateFunction(int[] A) {
        if(A == null || A.length == 0) return 0;
        int len = A.length;
        int max = Integer.MIN_VALUE;
        for(int i = 0; i < A.length ; i++){
            max = Math.max(max, getValue(A, i));
        }
        return max;
    }
    
    public int getValue(int[] A, int k){
        int res = 0;
        int len = A.length;
        for(int i = 0 ; i < A.length; i++){
            res += i * A[(k+i) % len];
        }
        return res;
    }
    */
}

 

396. Rotate Function


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