The Strange Function


Problem Statement :


One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development.

You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting of n integers [ a1, a2, . . . , an ] .

Input Format

The first line contains a single integer n

The second  line contains n space-separated integers a1, a2, . . . , an .

Constraints

1  <=  n  <=  50000
- 10^6  <=   ai   <= 10^6

Output Format

Print a single integer denoting the answer



Solution :



title-img


                            Solution in C :

In C ++ :




#define _CRT_SECURE_NO_WARNINGS
#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>

#include <memory.h>
#include <assert.h>

#define y0 sdkfaslhagaklsldk

#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd
#define have adsgagshdshfhds
#define ends asdgahhfdsfshdshfd

#define eps 1e-8
#define M_PI 3.141592653589793
#define bsize 512

#define ldouble long double
using namespace std;

#define bs 1000000007

const int N = 600031;

int n,ar[N];
long long S[N];

long long sparse_gcd[20][N],sparse_s[20][N],sparse_max[20][N];

int gcd(int a,int b){
	b=abs(b);
	while (a&&b)a>b?a%=b:b%=a;
	return a+b;
}

vector<pair<long long, long long> > order;

long long suf_max[N];

int main(){
//	freopen("apache.in","r",stdin);
//	freopen("apache.out","w",stdout);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	ios_base::sync_with_stdio(0);
//	cin.tie(0);

	cin>>n;
	long long ttl=0;

	for (int i=1;i<=n;i++){
		cin>>ar[i];
		S[i]=S[i-1]+ar[i];
	}

	suf_max[n]=S[n];
	for (int i=n-1;i>=0;--i)
		suf_max[i]=max(suf_max[i+1],S[i]);

	long long ans=-1e18;

	order.clear();
	for (int i=1;i<=n;i++){
		order.push_back(make_pair(S[i],i));
	}

	sort(order.begin(),order.end());

	long long MX=order.back().first;

	srand(time(NULL));

	if (rand()%2)
		random_shuffle(order.begin(),order.end());

	for (int ii=0;ii<order.size();ii++){
		int i=order[ii].second;
		ttl=MX-order[ii].first;
		long long cur_gcd=0;
		int cur_max=-1e9;
		long long cur_s=0;
		for (int j=i;j<=n;j++){
			cur_gcd=gcd(cur_gcd,ar[j]);
			cur_max=max(cur_max,ar[j]);
			cur_s+=ar[j];
			long long here=cur_gcd*(cur_s-cur_max);
			if (cur_gcd*1ll*(suf_max[j]-order[ii].first)<=ans)
				break;
			if (here>ans)
				ans=here;
		}
		if (clock()*1.0/CLOCKS_PER_SEC>1.98)
			break;
	}
	cout<<ans<<endl;

	cin.get(); cin.get();
	return 0;
}








In Java : 




//package hackerrank;

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;
import java.util.TreeMap;

public class Solution98 {
    static final Scanner scanner = new Scanner(System.in);
    static class Segment {
        int left, right, val;
        Segment(int left, int right, int val) {
            this.left = left;
            this.right = right;
            this.val = val;
        }
    }
    private static int gcd(int x, int y) {
        if (y == 0) {
            return x;
        }
        return gcd(y, x % y);
    }

    public static void main(String[] args) {
        int n = scanner.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }

        long ans = 0;
        long sum = 0;
        TreeMap<Integer, Integer> tmap = new TreeMap<>();
        SegmentTree tree = new SegmentTree(n);
        Deque<Segment> dq = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            TreeMap<Integer, Integer> segmentTree = new TreeMap<>();
            for (int j : tmap.keySet()) {
                int k = gcd(j, Math.abs(a[i]));
                int idx = tmap.get(j);
                if (segmentTree.containsKey(k)) {
                    int cur = segmentTree.get(k);
                    segmentTree.put(k, Math.min(idx, cur));
                } else {
                    segmentTree.put(k, idx);
                }
            }
            if (segmentTree.containsKey(Math.abs(a[i]))) {
                int j = segmentTree.get(Math.abs(a[i]));
                segmentTree.put(Math.abs(a[i]), Math.min(i, j));
            } else {
                segmentTree.put(Math.abs(a[i]), i);
            }
            tmap = segmentTree;
            tree.add(i, i, sum);

            Segment s = new Segment(i, i, a[i]);
            while (!dq.isEmpty() && dq.getLast().val <= a[i]) {
                tree.add(dq.getLast().left, dq.getLast().right, -dq.getLast().val);
                s.left = dq.getLast().left;
                dq.removeLast();
            }
            tree.add(s.left, s.right, s.val);
            dq.addLast(s);
            sum += a[i];
            for (int j : tmap.keySet()) {
                int pos = tmap.get(j);
                ans = Math.max(ans, j * (sum - tree.get(pos, i)));
            }
        }

        System.out.println(ans);
    }
}

class SegmentTree {
    private int n;
    private long[] min, add;

    SegmentTree(int size) {
        n = 1;
        while (n < size) {
            n *= 2;
        }
        min = new long[2 * n];
        add = new long[2 * n];
    }

    private void clear(int i, int l, int r) {
        min[i] = 0;
        add[i] = 0;
        if (l == r - 1) {
            return;
        }
        int tm = (l + r) / 2;
        clear(2 * i + 1, l, tm);
        clear(2 * i + 2, tm, r);
    }

    void clear(int n) {
        this.n = n;
        clear(0, 0, n);
    }

    private void push(int i, int tl, int tr) {
        min[i] += add[i];
        if (tl != tr - 1) {
            add[2 * i + 1] += add[i];
            add[2 * i + 2] += add[i];
        }
        add[i] = 0;
    }

    private void add(int i, int lt, int rt, int l, int r, long diff) {
        push(i, lt, rt);
        if (l >= rt || r <= lt) {
            return;
        }

        if (l <= lt && rt <= r) {
            add[i] += diff;
            push(i, lt, rt);
            return;
        }

        int tm = (lt + rt) / 2;
        add(2 * i + 1, lt, tm, l, r, diff);
        add(2 * i + 2, tm, rt, l, r, diff);
        min[i] = Math.min(min[2 * i + 1], min[2 * i + 2]);
    }

    private long get(int i, int lt, int rt, int l, int r) {
        push(i, lt, rt);
        if (l >= rt || r <= lt) {
            return Long.MAX_VALUE;
        }

        if (l <= lt && rt <= r) {
            return min[i];
        }

        int tm = (lt + rt) / 2;
        return Math.min(get(2 * i + 1, lt, tm, l, r), get(2 * i + 2, tm, rt, l, r));
    }

    void add(int l, int r, long diff) {
        add(0, 0, n, l, r + 1, diff);
    }

    long get(int l, int r) {
        return get(0, 0, n, l, r + 1);
    }
}







In C :






#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define INF 200000
typedef struct _node{
  int gcd;
  int max;
  long long sum;
} node;
typedef struct _tree_node{
  enum {red,black} colour;
  int data;
  int real;
  struct _tree_node *left,*right,*parent;
}tree_node;
void solve(int start,int bs,int ns);
void copy_tree(tree_node **d,tree_node *r);
void build(int v,int tl,int tr);
long long query(int v,int tl,int tr,int l,int r,int *gcd,int *ma,long long *sum);
void getp(long long N,long long*prim);
long long CC(long long n, long long d);
int max(int x,int y);
int min(int x,int y);
int abss(int x);
int search(tree_node *root,int data);
void left_rotate(tree_node **root,tree_node *x);
void right_rotate(tree_node **root,tree_node *y);
void reconstruct(tree_node **root,tree_node *x);
int normal_insert(tree_node **root,tree_node *x);
void insert(tree_node **root,tree_node *x);
void sort_a(int*a,int size,int*new_size);
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size);
void init( int n );
int range_sum( int i, int j );
void update( int i, long long val );
int a[50000],b[50000],nonzero[50000],rp[30],rc[30],treei[200000],rs,s,N;
long long ans,p[1000],tree[200000];
node t[200000];
tree_node *map[50000];

int main(){
  int n,x,ns,i,j,k,l;
  long long sum;
  tree_node *last_node,*p_node;
  getp(1000,p);
  scanf("%d",&n);
  for(i=0;i<n;i++)
    scanf("%d",a+i);
  build(1,0,n-1);
  init(n);
  for(i=sum=0;i<n;i++){
    sum+=a[i];
    update(i,sum);
  }
  for(i=n-1;i>=0;i--){
    if(i==n-1)
      last_node=NULL;
    else
      last_node=map[i+1];
    if(a[i]){
      nonzero[i]=i;
      for(j=rs=0,x=abss(a[i]);p[j] && p[j]*p[j]<=x;j++)
        if(x%p[j]==0){
          rp[rs]=p[j];
          rc[rs]=0;
          while(x%p[j]==0){
            rc[rs]++;
            x/=p[j];
          }
          rs++;
        }
      if(x!=1){
        rp[rs]=x;
        rc[rs]=1;
        rs++;
      }
      for(j=s=0;j<rs;j++)
        for(k=0,x=rp[j];k<rc[j];k++,x*=rp[j]){
          p_node=(tree_node*)malloc(sizeof(tree_node));
          p_node->data=x;
          p_node->left=p_node->right=p_node->parent=NULL;
          l=search(last_node,x);
          if(l!=INF){
            p_node->real=l;
            b[s++]=l+1;
          }
          else
            p_node->real=i;
          insert(&map[i],p_node);
        }
      b[s++]=i;
      sort_a(b,s,&ns);
      solve(i,ns,n);
    }
    else{
      nonzero[i]=INF;
      s=0;
      copy_tree(&map[i],last_node);
      if(i!=n-1 && nonzero[i+1]!=INF)
        b[s++]=nonzero[i+1];
      sort_a(b,s,&ns);
      solve(i,ns,n);
    }
    if(i!=n-1)
      nonzero[i]=min(nonzero[i],nonzero[i+1]);
  }
  printf("%lld",ans);
  return 0;
}
void solve(int start,int bs,int ns){
  int gcd,ma,i,j;
  long long t,sum;
  for(i=0;i<bs;i++){
    if(b[i]==ns)
      continue;
    if(i==bs-1)
      j=range_sum(b[i],ns-1);
    else
      j=range_sum(b[i],b[i+1]-1);
    t=query(1,0,ns-1,start,j,&gcd,&ma,&sum);
    if(t>ans)
      ans=t;
  }
  return;
}
void copy_tree(tree_node **d,tree_node *r){
  tree_node *p;
  if(!r)
    return;
  copy_tree(d,r->left);
  p=(tree_node*)malloc(sizeof(tree_node));
  p->data=r->data;
  p->real=r->real;
  p->left=p->right=p->parent=NULL;
  insert(d,p);
  b[s++]=r->real+1;
  copy_tree(d,r->right);
  return;
}
void build(int v,int tl,int tr){
  int tm;
  if(tl==tr){
    t[v].gcd=abss(a[tl]);
    t[v].max=t[v].sum=a[tl];
  }
  else{
    tm=(tl+tr)/2;
    build(2*v,tl,tm);
    build(2*v+1,tm+1,tr);
    t[v].gcd=CC(t[2*v].gcd,t[2*v+1].gcd);
    t[v].max=max(t[2*v].max,t[2*v+1].max);
    t[v].sum=t[2*v].sum+t[2*v+1].sum;
  }
  return;
}
long long query(int v,int tl,int tr,int l,int r,int *gcd,int *ma,long long *sum){
  int tm,g1,g2,m1,m2;
  long long s1,s2;
  if(tl>r || tr<l){
    *gcd=0;
    *ma=0;
    *sum=0;
    return 0;
  }
  if(tl>=l && tr<=r){
    *gcd=t[v].gcd;
    *ma=t[v].max;
    *sum=t[v].sum;
    return (*gcd)*((*sum)-(*ma));
  }
  tm=(tl+tr)/2;
  query(2*v,tl,tm,l,r,&g1,&m1,&s1);
  query(2*v+1,tm+1,tr,l,r,&g2,&m2,&s2);
  *gcd=CC(g1,g2);
  *ma=max(m1,m2);
  *sum=s1+s2;
  return (*gcd)*((*sum)-(*ma));
}
void getp(long long N,long long*prim)
{
  long long i,j,index=2,flag;
  if(N<=1){
    prim[0]=0;
    return;}
  if(N==2){
  prim[0]=2;
  prim[1]=0;
  return;}
  prim[0]=2;
  prim[1]=3;
  for(i=5;i<=N;i=i+2)
    {
      for(j=1,flag=1;prim[j]<=sqrt(i);j++)
        {
          if(i%prim[j]==0){
            flag=0;
            break;}
        }
      if(flag==1)
        {prim[index]=i;
          index++;}
    }
  prim[index]=0;
  return;
}
long long CC(long long n, long long d)
{
  if(n==0)
    return d;
  if(d==0)
    return n;
    while( 1 )
    {
        n = n % d;
	if( n == 0 )
		return d;
	d = d % n;
        if( d == 0 )
		return n;
    }
}
int max(int x,int y){
  return (x>y)?x:y;
}
int min(int x,int y){
  return (x<y)?x:y;
}
int abss(int x){
  return (x<0)?-x:x;
}
int search(tree_node *root,int data){
  if(!root)
    return INF;
  if(root->data==data)
    return root->real;
  if(data<root->data)
    return search(root->left,data);
  return search(root->right,data);
}
void left_rotate(tree_node **root,tree_node *x){
  tree_node *y;
  y=x->right;
  if(!y) return;
  x->right=y->left;
  if(y->left)
    y->left->parent=x;
  y->parent=x->parent;
  if(x->parent==NULL) *root=y;
  else
    if(x==x->parent->left)
      x->parent->left=y;
    else
      x->parent->right=y;
  y->left=x;
  x->parent=y;
  return;
}
void right_rotate(tree_node **root,tree_node *y){
  tree_node *x;
  x=y->left;
  if(!x) return;
  y->left=x->right;
  if(x->right)
    x->right->parent=y;
  x->parent=y->parent;
  if(y->parent==NULL) *root=x;
  else
    if(y==y->parent->right)
      y->parent->right=x;
    else
      y->parent->left=x;
  x->right=y;
  y->parent=x;
  return;
}
void reconstruct(tree_node **root,tree_node *x){
  tree_node *y,*z;
  y=x->parent;
  z=x->parent->parent;
  x->colour=black;
  z->colour=red;
  x->parent=z->parent;
  if(z->parent==NULL)
    *root=x;
  else if(z==z->parent->left)
    z->parent->left=x;
  else
    z->parent->right=x;
  if(z->left==y){
    x->left=y;
    x->right=z;
  }
  else{
    x->left=z;
    x->right=y;
  }
  y->parent=z->parent=x;
  y->left=y->right=z->left=z->right=NULL;
  return;
}
int normal_insert(tree_node **root,tree_node *x){
  if(*root==NULL)
    *root=x;
  else if((*root)->data==x->data)
    return 0;
  else{
    x->parent=*root;
    if((*root)->data>x->data)
      return normal_insert(&((*root)->left),x);
    else
      return normal_insert(&((*root)->right),x);
  }
  return 1;
}
void insert(tree_node **root,tree_node *x){
  if(!normal_insert(root,x))
    return;
  tree_node *y;
  x->colour=red;
  while(x!=*root && x->parent->colour==red){
    if(x->parent==x->parent->parent->left){
      y=x->parent->parent->right;
      if(!y)
        if(x==x->parent->left){
          x->parent->colour=black;
          x->parent->parent->colour=red;
          right_rotate(root,x->parent->parent);
        }
        else{
          y=x->parent;
          reconstruct(root,x);
          x=y;
        }
      else if(y->colour==red){
        x->parent->colour=black;
        y->colour=black;
        x->parent->parent->colour=red;
        x=x->parent->parent;
      }
      else{
        if(x==x->parent->right){
          x=x->parent;
          left_rotate(root,x);
        }
        x->parent->colour=black;
        x->parent->parent->colour=red;
        right_rotate(root,x->parent->parent);
      }
    }
    else{
      y=x->parent->parent->left;
      if(!y)
        if(x==x->parent->right){
          x->parent->colour=black;
          x->parent->parent->colour=red;
          left_rotate(root,x->parent->parent);
        }
        else{
          y=x->parent;
          reconstruct(root,x);
          x=y;
        }
      else if(y->colour==red){
        x->parent->colour=black;
        y->colour=black;
        x->parent->parent->colour=red;
        x=x->parent->parent;
      }
      else{
        if(x==x->parent->left){
          x=x->parent;
          right_rotate(root,x);
        }
        x->parent->colour=black;
        x->parent->parent->colour=red;
        left_rotate(root,x->parent->parent);
      }
    }
  }
  (*root)->colour=black;
  return;
}
void sort_a(int*a,int size,int*new_size){
  if (size < 2){
    (*new_size)=size;
    return;
  }
  int m = (size+1)/2,i;
  int *left,*right;
  left=(int*)malloc(m*sizeof(int));
  right=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++)
    left[i]=a[i];
  for(i=0;i<size-m;i++)
    right[i]=a[i+m];
  int new_l_size=0,new_r_size=0;
  sort_a(left,m,&new_l_size);
  sort_a(right,size-m,&new_r_size);
  merge(a,left,right,new_l_size,new_r_size,new_size);
  free(left);
  free(right);
  return;
}
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size){
  int i = 0, j = 0,index=0;
  while (i < left_size|| j < right_size) {
    if (i == left_size) {
      a[index++] = right[j];
      j++;
    } else if (j == right_size) {
      a[index++] = left[i];
      i++;
    } else if (left[i] <= right[j]) {
      a[index++] = left[i];
      i++;
    } else {
      a[index++] = right[j];
      j++;
    }
    if(index>1&&a[index-2]==a[index-1])
      index--;
  }
  (*new_size)=index;
  return;
}
void init( int n )
{
  N = 1;
  while( N < n ) N *= 2;
  int i;
  for( i = 1; i < N + n; i++ ) tree[i] = -1000000000000000000LL;
}
int range_sum( int i, int j )
{
  int ansi;
  long long ans = -1000000000000000000LL;
  for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
  {
    if( i % 2 == 1 )
      if(tree[i]>ans){
        ans=tree[i];
        ansi=treei[i];
      }
    if( j % 2 == 0 )
      if(tree[j]>ans){
        ans=tree[j];
        ansi=treei[j];
      }
  }
  return ansi;
}
void update( int i, long long val )
{
  int j;
  for( j = i + N; j; j /= 2 )
    if(val>tree[j]){
      tree[j]=val;
      treei[j]=i;
    }
}









In Python3 :





from math import gcd

def parseInput(f):
    return [f(x) for x in input().split()]

n=int(input())
array=parseInput(int)
stack=[]
answer=float('-inf')
for number in array:
    for i in range(len(stack)):
        stack[i][0]=gcd(abs(stack[i][0]),abs(number))
        stack[i][1]+=number
        if number > stack[i][2]:
            stack[i][1]-=number-stack[i][2]
            stack[i][2]=number

    stack.append([number,0,number])
    newStack=[]
    for i in range(len(stack)):
        if newStack and newStack[-1][0] == stack[i][0]:
            if newStack[-1][1] <= stack[i][1]:
                if newStack[-1][1]+newStack[-1][2] > stack[i][1]+stack[i][2]:
                    newStack.append(stack[i])
                    continue
                newStack[-1][1]=stack[i][1]
                newStack[-1][2]=stack[i][2]
        else:
            newStack.append(stack[i])
    stack = newStack[:]
    answer=max(answer,max(abs(stack[i][0])*stack[i][1] for i in range(len(stack))))
print(answer)
                        








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Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

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Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

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QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

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Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

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Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

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Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

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