Problem:
Score : 600 points

Problem Statement

Snuke has an integer sequence A of length N.

He will make three cuts in A and divide it into four (non-empty) contiguous subsequences B, C, D and E.
The positions of the cuts can be freely chosen.

Let P,Q,R,S be the sums of the elements in B,C,D,E, respectively.
Snuke is happier when the absolute difference of the maximum and the minimum among P,Q,R,S is smaller.
Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

Constraints

4 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

Sample Input 1

5
3 2 4 1 2

Sample Output 1

2

If we divide A as B,C,D,E=(3),(2),(4),(1,2), then P=3,Q=2,R=4,S=1+2=3.
Here, the maximum and the minimum among P,Q,R,S are 4 and 2, with the absolute difference of 2.
We cannot make the absolute difference of the maximum and the minimum less than 2, so the answer is 2.

Sample Input 2

10
10 71 84 33 6 47 23 25 52 64

Sample Output 2

36

Sample Input 3

7
1 2 3 1000000000 4 5 6

Sample Output 3

999999994

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

class Main{
    static int N;
    static long[] sum;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        sum = new long[N+1];
        long[] A = new long[N+1];
        for(int i=1;i<=N;i++) A[i]=sc.nextLong();
        for(int i=1;i<=N;i++) sum[i] = sum[i-1]+A[i];
        long ans = Long.MAX_VALUE;
        for(int i=2;i<=N-2;i++){
            long difLeft = halfClo(1,i);
            long P = (sum[i]+difLeft)/2;
            long Q = (sum[i]-difLeft)/2;
            long difRight = halfClo(i+1,N);
            long R = (sum[N]-sum[i]+difRight)/2;
            long S = (sum[N]-sum[i]-difRight)/2;
            long[] cur = new long[]{P,Q,R,S};
            Arrays.sort(cur);
            ans = Math.min(ans,cur[3]-cur[0]);
        }
        System.out.println(ans);
    }
    static long halfClo(int left, int right){
        long hi = sum[right]-sum[left-1];
        long lo = 0;
        while(hi-lo>1){
            long mid = (hi+lo)/2;
            if(check(left,right,mid)) hi = mid;
            else lo = mid;
        }
        return hi;
    }
    static boolean check(int left, int right, long target){
        long all = sum[right]-sum[left-1];
        int hi = right-1;
        int lo = left;
        while(hi-lo>1){
            int mid = (hi+lo)/2;
            long cur = sum[mid]-sum[left-1];
            if(2*cur>=all-target&&2*cur<=all+target) return true;
            if(2*cur<all-target) lo = mid;
            else hi = mid;
        }
        boolean b1 = 2*(sum[hi]-sum[left-1])>=all-target&&2*(sum[hi]-sum[left-1])<=all+target;
        boolean b2 = 2*(sum[lo]-sum[left-1])>=all-target&&2*(sum[lo]-sum[left-1])<=all+target;
        return b1||b2;
    }
}