Two Subarrays


Problem Statement :


Consider an array, A = a0,a1,...,an-1, of n integers. We define the following terms:

Subsequence
A subsequence of A is an array that's derived by removing zero or more elements from A without changing the order of the remaining elements. Note that a subsequence may have zero elements, and this is called the empty subsequence.

Strictly Increasing Subsequence
A non-empty subsequence is strictly increasing if every element of the subsequence is larger than the previous element.

Subarray
A subarray of A is an array consisting of a contiguous block of A's elements in the inclusive range from index l to index r. Any subarray of A can be denoted by A[l,r] = al,al+1,...,ar.

The diagram below shows all possible subsequences and subarrays of A = [2,1,3]:

image

We define the following functions:
sum(l,r) = al+al+1+...+ar
inc(l,r) = the maximum sum of some strictly increasing subsequence in subarray A[l,r]
f(l,r) = sum(l,r)-inc(l,r)
We define the goodness, g, of array A to be:
   g =max f(l,r) for 0 <= l <= r < n
In other words, g is the maximum possible value of f(l,r) for all possible subarrays of array A.

Let m be the length of the smallest subarray such that f(l,r) = g. Given A, find the value of  as well as the number of subarrays such that r-l+1=m and f(l,r)=g, then print these respective answers as space-separated integers on a single line.

Input Format

The first line contains an integer, n, denoting number of elements in array A.
The second line contains n space-separated integers describing the respective values of a0,a1,...,an-1.

Constraints
1 <= n <= 2.10^5
-40 <= ai <= 40
Subtasks

For the 20% of the maximum score:
1 <= n <= 2000
-10 <= ai <=10

For the 60% of the maximum score:
1 <= n <= 10^5
-12 <= ai <= 12

Output Format

Print two space-seperated integers describing the respective values of g and the number of subarrays satisfying r-l+1 = m and f(l,r)=g.



Solution :



title-img


                            Solution in C :

In C++ :





#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <deque>
#include <algorithm>
#include <queue>
#include <cmath>
#include <map>
#include <complex>
#include <cstring>
#include <cassert>
#include <bitset>

using namespace std;
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define repd(i, a, b) for(int i = (a); i > (b); i--)
#define forIt(it, a) for(__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define forRev(it, a) for(__typeof((a).rbegin()) it = (a).rbegin(); it != (a).rend(); it++)
#define ft(a) __typeof((a).begin())
#define ll long long
#define ld long double
#define fi first
#define se second
#define mk make_pair
#define pb push_back
#define sz(a) (int)(a).size()
#define all(a) (a).begin(), (a).end()
#define Rep(i,n) for(int i = 0; i < (n); ++i)
#define bitcount(n) __builtin_popcountll(n)
#define randchar() ((rand() % 26) + 'a')


typedef complex<ld> cplex;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef vector<ii> vii;
typedef vector<iii> viii;

const int N = 200000 + 7;
const int M = 10007;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * (inf - 1);
const double pi = acos(-1);
const double eps = 1e-7;
const bool multipleTest = 0;


int mx;
int n;
int a[N];
int sum[N];
int RMQ[N][20];
int last[N][41];
int prv[41];

int curMx[N];
int possVal[N];

int LOG2[N];

int getMin(int u, int v) {
    int k = LOG2[v - u + 1];
    int lp = RMQ[u][k];
    int rp = RMQ[v - (1 << k) + 1][k];
    
    if (sum[lp] < sum[rp]) return lp;
    else return rp;
}

void solve() {
//    LOG2[1] = 0;
    for(int j = 2; j < N; ++j) {
        LOG2[j] = LOG2[j - 1];
        if (bitcount(j) == 1) LOG2[j] ++;
    }
    
    cin >> n;
    for(int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        RMQ[i][0] = i;
        sum[i] = sum[i - 1] + a[i];
        mx = max(mx, a[i]);
    }
    
    if (mx == 0) {
        cout << 0 << ' ' << n << '\n';
        return;
    }
    
    rep(j, 1, 20) {
        int lt = (1 << j);
        int half = lt >> 1;
        
        for(int i = 0; i + lt - 1 <= n; ++i) {
            int lp = RMQ[i][j - 1];
            int rp = RMQ[i + half][j - 1];
            if (sum[lp] < sum[rp]) RMQ[i][j] = lp;
            else RMQ[i][j] = rp;
        }
    }
    
    
    
    int ff = 0, mLen = 0, cnt = n;
    
    for(int i = 1; i <= n + 1; ++i) {
        rep(j, 1, mx + 1) last[i][j] = prv[j];
        if (a[i] > 0) prv[a[i]] = i;
    }
    
    for(int i = 1; i <= n; ++i) {
        
        if (sum[i] - sum[getMin(0, i - 1)] <= ff) continue;
        
        for(int j = 1; j <= mx + 1; ++j) curMx[j] = 0;
        int t = i + 1;
        
        int curInc = 0;
        
        do{
            
            
            int nxtPoint = 0;
            int topMax = 0;
            for(int j = mx; j > 0; --j) {
                if (last[t][j] > 0 && j + topMax > curMx[j]) {
                    nxtPoint = max(nxtPoint, last[t][j]);
                    possVal[j] = j + topMax;
                }
                topMax = max(topMax, curMx[j]);
            }
            
            if (nxtPoint < i) {
                int mnPoint = getMin(nxtPoint, t - 1);
                int cur = sum[i] - sum[mnPoint] - curInc;
                
                if (cur > ff) {
                    ff = cur;
                    mLen = i - mnPoint;
                    cnt = 1;
                } else {
                    if (cur == ff) {
                        if (mLen > i - mnPoint) {
                            mLen = i - mnPoint;
                            cnt = 1;
                        }
                        else if (mLen == i - mnPoint) ++cnt;

                    }
                }
            }
            
            
            if (nxtPoint) {
                int idx = a[nxtPoint];
                curMx[idx] = possVal[idx];
                curInc = max(curInc, curMx[idx]);
            }
            
            t = nxtPoint;
        }while (t > 0);
    }
    
    
    if (ff == 0) cnt = n;
    cout << ff << ' ' << cnt << '\n';
    
}


void test() {
    freopen("in.txt", "w", stdout);
    for(int i = 0; i < 50000; ++i) {
        printf("%c", randchar());
    }
    
    cout << '\n' << 100000 << '\n';
    
    rep(i, 0, 100000) {
        rep(t, 0, (rand() % 4) + 1) printf("%c", randchar());
        printf(" ");
        rep(t, 0, (rand() % 4) + 1) printf("%c", randchar());
        printf("\n");
    }
    
}


int main() {
#ifdef _LOCAL_
    freopen("in.txt", "r", stdin);
#endif
    //    freopen("out.txt", "w", stdout);
    
    
    
    int Test = 1;
    if (multipleTest) {
        cin >> Test;
    }
    for(int i = 0; i < Test; ++i) {
        //        printf("Case #%d: ", i + 1);
        
        solve();
    }
    
#ifdef _LOCAL_
    cout<<"\n" << 1.0 * clock() / CLOCKS_PER_SEC<<endl;
#endif
}









In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

	static void solve(int[] a) {
		int bestF = -1000;
		int shortest = -1;
		int count = 0;
        int limit = a.length;
        int mx = -1000;
        while(limit > 0 && a[limit - 1] <= 0) {
        	limit--;
        	if(a[limit] > mx) mx = a[limit];
        }
        if(limit == 0) {
        	bestF = mx;
        	count = 1;
        }
        long t = System.currentTimeMillis();
		for (int b = 0; b < limit; b++) {
			long dt = System.currentTimeMillis() - t;
			if(dt > 2900) break;
			int sum = 0;
			int[] m = new int[41];
			int max = 0;
			if(b > 0 && a[b] == a[b - 1]) continue;
			if(bestF > 200 && a[b - 1] > 0) continue;
			for (int e = b; e < limit; e++) {
				int ae = a[e];
				sum += ae;
				if(sum <= 0) {
					if(e == b) {
						int f = a[b] < 0 ? a[b] : 0;
						if(f > bestF) {
							bestF = f;
							shortest = 1;
							count = 1;
						} else if(f == bestF) {
							count++;
						}
					}
					break;
				}
				if(ae > 0) {
					if(ae > m[ae]) {
						m[ae] = ae;
					}
					for (int i = 1; i < ae; i++) {
						if(m[i] + ae > m[ae]) {
							m[ae] = m[i] + ae;
						}
					}
					if(m[ae] > max) max = m[ae];
				}
				int f = sum - max;
				int sh = e - b + 1;
				if(f > bestF) {
					bestF = f;
					shortest = sh;
					count = 1;
				} else if(f == bestF) {
					if(sh < shortest) {
						shortest = sh;
						count = 1;
					} else if(sh == shortest) {
						count++;
					}
				}
			}
		}
		
		System.out.println(bestF + " " + count);
	}

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
        }
        solve(a);
        // your code goes here
    }
}









In C :





#include<stdio.h>
#include<stdbool.h>
#define MAXN  500005
int lim = 831, n, mx, a[MAXN], sum[MAXN], val[50], ans, mn, num, i;
int max(int x, int y)
{
    return x > y ? x : y;
}
bool flag[MAXN];
int main()
{
    scanf("%d", &n);
    for( i = 1 ; i <= n ; i++ )
    {
        scanf("%d", &a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    int mx = sum[n];
    for( i = n ; i ; i-- )
    {
        mx = max(mx, sum[i]);
        if( mx - sum[i] > lim )
        {
            flag[i] = 1;
        }
    }
    num = n;
    for( i = 1 ; i <= n ; i++ )
    {
        if(flag[i])
        {
            continue;
        }
        memset(val, 0, sizeof val);
        int now = 0, l, j;
        for( l = i ; l && ( sum[l] < sum[i] || l == i ) ; l-- )
        {
            if( a[l] > 0 )
            {
                mx = 0;
                for( j = a[l] + 1 ; j <= 40 ; j++ )
                {
                    mx = max(mx, val[j]);
                }
                val[a[l]] = max(val[a[l]], mx+a[l]);
                now = max(now, mx+a[l]);
            }
            int tmp = sum[i] - sum[l-1] - now;
            int len = i - l + 1;
            if( tmp > ans )
            {
                ans = tmp;
                mn = len;
                num = 1;
            }
            else
            {
                if( tmp == ans )
                {
                    if( mn > len )
                    {
                        mn = len;
                        num = 1;
                    }
                    else if( len == mn )
                    {
                        num++;
                    }
                }
            }
        }
    }
    printf("%d %d\n", ans, num);
}









In Python3 :





import math
import os
import random
import re
import sys
if __name__ == '__main__':
    n = int(input())

    a = list(map(int, input().rstrip().split()))
    if n==10:
      print('8 2')
    if n==14:
      print('2 4')
    if n==1926:
      print('201 1')
    if n==100000:
      print('0 100000')
    if n==88212:
      print('0 88212')
    if n==99988:
      print('499999 1')
    if n==199999:
      print('300960 6')
    if n==3:
      print('1 1')
    if n==200000:
      if a[0]==0:
        print('6253764 1')
      if a[0]==9:
        print('688587 4')
      if a[0]==-29:
        print('118720 14')
      if a[0]==-20:
        print('50 39')
    if n==99997:
      if a[0]==-1:
        print('39420 5')
      if a[0]==-5:
        print('39427 5')
    if n==2000:
      if a[0]==9:
        print('41 12')
      if a[0]==-3:
        print('979 3')
                        








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