#include <map>
#include <iostream>
#include <algorithm>
using namespace std;

int A[] = {10,20,30,30,20,10,10,20};

// return (i, j) s.t. A[i] + ... + A[j] is nearest to value c
pair<int, int> nearest_to_c(int c, int n, int A[]) {
    map<int, int> bst;
    bst[0] = -1;
    // barriers
    bst[-int(1e9)] = -2;
    bst[int(1e9)] = n;

    int sum = 0, start, end, ret = c;
    for (int i=0; i<n; ++i) {
            sum += A[i];
            // it->first >= sum-c, and with the minimal value in bst
            map<int, int>::iterator it = bst.lower_bound(sum - c);
            int tmp = -(sum - c - it->first);
            if (tmp < ret) {
                    ret = tmp;
                    start = it->second + 1;
                    end = i;
            }

            --it;
            // it->first < sum-c, and with the maximal value in bst
            tmp = sum - c - it->first;
            if (tmp < ret) {
                    ret = tmp;
                    start = it->second + 1;
                    end = i;
            }

            bst[sum] = i;
    }
    return make_pair(start, end);
}

// demo
int main() {
    int c;
    cin >> c;
    pair<int, int> ans = nearest_to_c(c, 8, A);

    cout << ans.first << ' ' << ans.second << endl;
    return 0;
}

import java.util.*;

public class FindSubarrayClosestToZero {

    void findSubarrayClosestToZero(int[] A) {
        int curSum = 0;
        List<Pair> list = new ArrayList<Pair>();

        // 1. create prefix array: curSum array
        for(int i = 0; i < A.length; i++) {
            curSum += A[i];
            Pair pair = new Pair(curSum, i);
            list.add(pair);
        }

        // 2. sort the prefix array by value
        Collections.sort(list, valueComparator);

        // printPairList(list);
        System.out.println();


        // 3. compute pair-wise value diff: Triple< diff, i, i+1>
        List<Triple> tList = new ArrayList<Triple>();
        for(int i=0; i < A.length-1; i++) {
            Pair p1 = list.get(i);
            Pair p2 = list.get(i+1);
            int valueDiff = p2.value - p1.value;

            Triple Triple = new Triple(valueDiff, p1.index, p2.index);
            tList.add(Triple);
        }

        // printTripleList(tList);
        System.out.println();

        // 4. Sort by min diff
        Collections.sort(tList, valueDiffComparator);
        // printTripleList(tList);

        Triple res = tList.get(0);

        int startIndex = Math.min(res.index1 + 1, res.index2);
        int endIndex = Math.max(res.index1 + 1, res.index2);

        System.out.println("nnThe subarray whose sum is closest to 0 is: ");
        for(int i= startIndex; i<=endIndex; i++) {
            System.out.print(" " + A[i]);
        }
    }

    class Pair {
        int value;
        int index;

        public Pair(int value, int index) {
            this.value = value;
            this.index = index;
        }
    }

    class Triple {
        int valueDiff;
        int index1;
        int index2;

        public Triple(int valueDiff, int index1, int index2) {
            this.valueDiff = valueDiff;
            this.index1 = index1;
            this.index2 = index2;
        }
    }

    public static Comparator<Pair> valueComparator = new Comparator<Pair>() {
        public int compare(Pair p1, Pair p2) {
            return p1.value - p2.value;
        }
    };

    public static Comparator<Triple> valueDiffComparator = new Comparator<Triple>() {
        public int compare(Triple t1, Triple t2) {
            return t1.valueDiff - t2.valueDiff;
        }
    };

    void printPairList(List<Pair> list) {
        for(Pair pair : list) {
            System.out.println("<" + pair.value + " : " + pair.index + ">");
        }
    }

    void printTripleList(List<Triple> list) {
        for(Triple t : list) {
            System.out.println("<" + t.valueDiff + " : " + t.index1 + " , " + t.index2 + ">");
        }
    }


    public static void main(String[] args) {
        int A1[] = {8, -3, 2, 1, -4, 10, -5};       // -3, 2, 1
        int A2[] = {-3, 2, 4, -6, -8, 10, 11};      // 2, 4, 6
        int A3[] = {10, -2, -7};                                // 10, -2, -7

        FindSubarrayClosestToZero f = new FindSubarrayClosestToZero();
        f.findSubarrayClosestToZero(A1);
        f.findSubarrayClosestToZero(A2);
        f.findSubarrayClosestToZero(A3);
    }
}

#include<bits/stdc++.h>
#define M 1000010
#define REP(i,n) for (int i=1;i<=n;i++)
using namespace std;
typedef long long ll;
ll a[M],n,cum[M],ans=numeric_limits<ll>::max(); //cum->cumulative array
int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>n; REP(i,n) cin>>a[i],cum[i]=cum[i-1]+a[i];
    sort(cum+1,cum+n+1);
    REP(i,n-1) ans=min(ans,cum[i+1]-cum[i]);
    cout<<ans; //min +ve difference from 0 we can get
}

#include <map>
#include <stdio.h>
#include <algorithm>
#include <limits.h>

using namespace std;

#define IDX_LOW_BOUND -2

// Return [i..j] range of A
pair<int, int> nearest_to_c(int A[], int n, int t) {
  map<int, int> bst;
  int presum, subsum, closest, i, j, start, end;
  bool unset;
  map<int, int>::iterator it;

  bst[0] = -1;
  // Barriers. Assume that no prefix sum is equal to INT_MAX or INT_MIN.
  bst[INT_MIN] = IDX_LOW_BOUND;
  bst[INT_MAX] = n;
  unset = true;
  // This initial value is always overwritten afterwards.
  closest = 0;
  presum = 0;
  for (i = 0; i < n; ++i) {
    presum += A[i];
    for (it = bst.lower_bound(presum - t), j = 0; j < 2; --it, j++) {
      if (it->first == INT_MAX || it->first == INT_MIN)
        continue;
      subsum = presum - it->first;
      if (unset || abs(closest - t) > abs(subsum - t)) {
        closest = subsum;
        start = it->second + 1;
        end = i;
        if (closest - t == 0)
          goto ret;
        unset = false;
      }
    }
    bst[presum] = i;
  }
ret:
  return make_pair(start, end);
}

int main() {
  int A[] = {10, 20, 30, 30, 20, 10, 10, 20};
  int t;
  scanf("%d", &t);
  pair<int, int> ans = nearest_to_c(A, 8, t);
  printf("[%d:%d]n", ans.first, ans.second);
  return 0;
}

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number
     *          and the index of the last number
     */
    public ArrayList<Integer> subarraySumClosest(int[] nums) {
        // write your code here
        int len = nums.length;
        ArrayList<Integer> result = new ArrayList<Integer>();
        int[] sum = new int[len];
        HashMap<Integer,Integer> mapHelper = new HashMap<Integer,Integer>();
        int min = Integer.MAX_VALUE;
        int curr1 = 0;
        int curr2 = 0;
        sum[0] = nums[0];
        if(nums == null || len < 2){
            result.add(0);
            result.add(0);
            return result;
        }
        for(int i = 1;i < len;i++){
            sum[i] = sum[i-1] + nums[i];
        }
        for(int i = 0;i < len;i++){
            if(mapHelper.containsKey(sum[i])){
                result.add(mapHelper.get(sum[i])+1);
                result.add(i);
                return result;
            }
            else{
                mapHelper.put(sum[i],i);
            }
        }
        Arrays.sort(sum);
        for(int i = 0;i < len-1;i++){
            if(Math.abs(sum[i] - sum[i+1]) < min){
                min = Math.abs(sum[i] - sum[i+1]);
                curr1 = sum[i];
                curr2 = sum[i+1];
            }
        }
        if(mapHelper.get(curr1) < mapHelper.get(curr2)){
            result.add(mapHelper.get(curr1)+1);
            result.add(mapHelper.get(curr2));
        }
        else{
            result.add(mapHelper.get(curr2)+1);
            result.add(mapHelper.get(curr1));
        }
        return result;
    }
}