Cloudy Day

Problem Statement :

Quibdó in Colombia is one among the cities that receive maximum rainfall in the world.

All year round, the city is covered in clouds. The city has many towns, located on a one-dimensional line. The positions and populations of each town on the number line are known to you. Every cloud covers all towns located at a certain distance from it. A town is said to be in darkness if there exists at least one cloud such that the town is within the cloud's range. Otherwise, it is said to be sunny.

The city council has determined that they have enough money to remove exactly one cloud using their latest technology. Thus they want to remove the cloud such that the fewest number of people are left in darkness after the cloud is removed. What is the maximum number of people that will be in a sunny town after removing exactly one cloud?

Note: If a town is not covered by any clouds, then it is already considered to be sunny, and the population of this town must also be included in the final answer.

Complete the function maximumPeople which takes four arrays representing the populations of each town, locations of the towns, locations of the clouds, and the extents of coverage of the clouds respectively, and returns the maximum number of people that will be in a sunny town after removing exactly one cloud.

Input Format

The first line of input contains a single integer , the number of towns.

The next line contains space-separated integers . The integer in this line denotes the population of the town.

The next line contains space-separated integers denoting the location of the town on the one-dimensional line.

The next line consists of a single integer denoting the number of clouds covering the city.

The next line contains space-separated integers the of which denotes the location of the cloud on the coordinate axis.

The next line consists of space-separated integers denoting the range of the cloud.

Note: The range of each cloud is computed according to its location, i.e., the cloud is located at position and it covers every town within a distance of from it. In other words, the cloud covers every town with location in the range

Output Format

Print a single integer denoting the maximum number of people that will be in a sunny town by removing exactly one cloud.

Solution :

                            Solution in C :

In  C :





#include <stdio.h>
#include <stdlib.h>
typedef struct _node{
  int l;
  int r;
  long long zero;
  long long one;
  int off;
} node;
void build(int v,int tl,int tr);
void push(int v,int tl,int tr);
void range_update(int v,int tl,int tr,int pos1,int pos2);
long long sum(int v,int tl,int tr,int l,int r);
long long sum0(int v,int tl,int tr,int l,int r);
void sort_a2(int*a,int*b,int size);
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size);
int p[200000],l[200000],y[100000],r[100000];
node t[800000];

int main(){
  int n,m,i;
  long long max,temp;
  scanf("%d",&n);
  for(i=0;i<n;i++)
    scanf("%d",p+i);
  for(i=0;i<n;i++)
    scanf("%d",l+i);
  scanf("%d",&m);
  for(i=0;i<m;i++)
    scanf("%d",y+i);
  for(i=0;i<m;i++)
    scanf("%d",r+i);
  sort_a2(l,p,n);
  build(1,0,n-1);
  for(i=0;i<m;i++)
    range_update(1,0,n-1,y[i]-r[i],y[i]+r[i]);
  for(i=max=0;i<m;i++){
    temp=sum(1,0,n-1,y[i]-r[i],y[i]+r[i]);
    if(temp>max)
      max=temp;
  }
  printf("%lld",max+sum0(1,0,n-1,0,(1U<<31)-1));
  return 0;
}
void build(int v,int tl,int tr){
  int tm;
  if(tl==tr){
    t[v].l=t[v].r=l[tl];
    t[v].zero=p[tl];
    return;
  }
  tm=(tl+tr)/2;
  build(v*2,tl,tm);
  build(v*2+1,tm+1,tr);
  t[v].l=t[v*2].l;
  t[v].r=t[v*2+1].r;
  t[v].zero=t[v*2].zero+t[v*2+1].zero;
  return;
}
void push(int v,int tl,int tr){
  if(t[v].off==1){
    t[v].one=t[v].zero;
    t[v].zero=t[v].off=0;
    if(tl!=tr){
      t[v*2].off++;
      t[v*2+1].off++;
    }
  }
  else if(t[v].off>1){
    t[v].zero=t[v].one=t[v].off=0;
    if(tl!=tr){
      t[v*2].off+=2;
      t[v*2+1].off+=2;
    }
  }
  return;
}
void range_update(int v,int tl,int tr,int pos1,int pos2){
  int tm;
  if(pos2<t[v].l || pos1>t[v].r)
    return;
  push(v,tl,tr);
  if(pos1<=t[v].l && pos2>=t[v].r)
    t[v].off=1;
  else{
    tm=(tl+tr)/2;
    range_update(v*2,tl,tm,pos1,pos2);
    range_update(v*2+1,tm+1,tr,pos1,pos2);
    push(v*2,tl,tm);
    push(v*2+1,tm+1,tr);
    t[v].zero=t[v*2].zero+t[v*2+1].zero;
    t[v].one=t[v*2].one+t[v*2+1].one;
  }
  return;
}
long long sum(int v,int tl,int tr,int l,int r){
  int tm;
  if(r<t[v].l || l>t[v].r)
    return 0;
  push(v,tl,tr);
  tm=(tl+tr)/2;
  if(l<=t[v].l && r>=t[v].r)
    return t[v].one;
  else
    return sum(v*2,tl,tm,l,r)+sum(v*2+1,tm+1,tr,l,r);
}
long long sum0(int v,int tl,int tr,int l,int r){
  int tm;
  if(r<t[v].l || l>t[v].r)
    return 0;
  push(v,tl,tr);
  tm=(tl+tr)/2;
  if(l<=t[v].l && r>=t[v].r)
    return t[v].zero;
  else
    return sum0(v*2,tl,tm,l,r)+sum0(v*2+1,tm+1,tr,l,r);
}
void sort_a2(int*a,int*b,int size){
  if (size < 2)
    return;
  int m = (size+1)/2,i;
  int*left_a,*left_b,*right_a,*right_b;
  left_a=(int*)malloc(m*sizeof(int));
  right_a=(int*)malloc((size-m)*sizeof(int));
  left_b=(int*)malloc(m*sizeof(int));
  right_b=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++){
    left_a[i]=a[i];
    left_b[i]=b[i];
  }
  for(i=0;i<size-m;i++){
    right_a[i]=a[i+m];
    right_b[i]=b[i+m];
  }
  sort_a2(left_a,left_b,m);
  sort_a2(right_a,right_b,size-m);
  merge2(a,left_a,right_a,b,left_b,right_b,m,size-m);
  free(left_a);
  free(right_a);
  free(left_b);
  free(right_b);
  return;
}
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size){
  int i = 0, j = 0;
  while (i < left_size|| j < right_size) {
    if (i == left_size) {
      a[i+j] = right_a[j];
      b[i+j] = right_b[j];
      j++;
    } else if (j == right_size) {
      a[i+j] = left_a[i];
      b[i+j] = left_b[i];
      i++;
    } else if (left_a[i] <= right_a[j]) {
      a[i+j] = left_a[i];
      b[i+j] = left_b[i];
      i++;
    } else {
      a[i+j] = right_a[j];
      b[i+j] = right_b[j];
      j++;
    }
  }
  return;
}

                        Solution in C++ :

In  C++  :







#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair <int, int> ii;

const int Maxn = 200005;

vector <ii> seq;
int delt[Maxn];
ll sum[Maxn];

ll maximumPeople(vector <long> p, vector <long> x, vector <long> y, vector <long> r) {
    for (int i = 0; i < p.size(); i++)
        seq.push_back(ii(x[i], p[i]));
    sort(seq.begin(), seq.end());
    for (int i = 0; i < y.size(); i++) {
        int lef = y[i] - r[i], rig = y[i] + r[i];
        int ind1 = lower_bound(seq.begin(), seq.end(), ii(lef, 0)) - seq.begin();
        int ind2 = lower_bound(seq.begin(), seq.end(), ii(rig + 1, 0)) - seq.begin();
        delt[ind1]++; delt[ind2]--;
    }
    ll res = 0;
    int cur = 0;
    for (int i = 0; i < seq.size(); i++) {
        cur += delt[i];
        if (cur == 0) res += seq[i].second;
        else if (cur == 1) sum[i] = seq[i].second;
        if (i) sum[i] += sum[i - 1];
    }
    ll add = res;
    for (int i = 0; i < y.size(); i++) {
        int lef = y[i] - r[i], rig = y[i] + r[i];
        int ind1 = lower_bound(seq.begin(), seq.end(), ii(lef, 0)) - seq.begin();
        int ind2 = lower_bound(seq.begin(), seq.end(), ii(rig + 1, 0)) - seq.begin();
        if (ind2 > 0) {
            ll got = sum[ind2 - 1];
            if (ind1 > 0) got -= sum[ind1 - 1];
            res = max(res, add + got);
        }
    }
    return res;
}

int main() {
    int n;
    cin >> n;
    vector<long> p(n);
    for(int p_i = 0; p_i < n; p_i++){
       cin >> p[p_i];
    }
    vector<long> x(n);
    for(int x_i = 0; x_i < n; x_i++){
       cin >> x[x_i];
    }
    int m;
    cin >> m;
    vector<long> y(m);
    for(int y_i = 0; y_i < m; y_i++){
       cin >> y[y_i];
    }
    vector<long> r(m);
    for(int r_i = 0; r_i < m; r_i++){
       cin >> r[r_i];
    }
    ll result = maximumPeople(p, x, y, r);
    cout << result << endl;
    return 0;
}

                        Solution in Java :

In  Java :








import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;

public class Solution {

	static long maximumPeople(long[] p, long[] x, long[] y, long[] r) {
		long free = 0;
		long[] sum = new long[y.length];
		ArrayList<Event> al = new ArrayList<>();
		HashSet<Integer> clouds = new HashSet<>();
		for (int i = 0; i < p.length; i++) {
			al.add(new Event(1, x[i], p[i], i));
		}

		for (int i = 0; i < y.length; i++) {
			al.add(new Event(0, y[i] - r[i], -1, i));
			al.add(new Event(2, y[i] + r[i], -1, i));
		}

		Collections.sort(al);

		for (Event e : al) {
			if (e.type == 0) {
				clouds.add(e.index);
			} else if (e.type == 1) {
				if (clouds.isEmpty()) {
					free += e.pr;
				} else {
					if (clouds.size() == 1) {
						for (int q : clouds) {
							sum[q] += e.pr;
						}
					}
				}
			} else {
				clouds.remove(e.index);
			}
		}

		long mx = 0;
		for (long i : sum) {
			mx = Math.max(mx, i);
		}
		return free + mx;
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long[] p = new long[n];
		for (int p_i = 0; p_i < n; p_i++) {
			p[p_i] = in.nextLong();
		}
		long[] x = new long[n];
		for (int x_i = 0; x_i < n; x_i++) {
			x[x_i] = in.nextLong();
		}
		int m = in.nextInt();
		long[] y = new long[m];
		for (int y_i = 0; y_i < m; y_i++) {
			y[y_i] = in.nextLong();
		}
		long[] r = new long[m];
		for (int r_i = 0; r_i < m; r_i++) {
			r[r_i] = in.nextLong();
		}
		long result = maximumPeople(p, x, y, r);
		System.out.println(result);
		in.close();
	}
}

class Event implements Comparable<Event> {
	int type;
	long x;
	long pr;
	int index;

	public Event(int type, long x, long pr, int index) {
		super();
		this.type = type;
		this.x = x;
		this.pr = pr;
		this.index = index;
	}

	@Override
	public int compareTo(Event e) {
		if (x != e.x) {
			return Long.compare(x, e.x);
		}
		return Integer.compare(type, e.type);
	}

}

                        Solution in Python : 
                            
In  Python3 :








from collections import defaultdict


def maximumPeople(towns, cloud_start, cloud_end):
    towns = sorted(towns)
    cloud_start = sorted(cloud_start)
    cloud_end = sorted(cloud_end)

    cloud_start_i = 0
    cloud_end_i = 0
    clouds = set()

    d = defaultdict(int)
    free = 0
    for town_i in range(len(towns)):
        town_x = towns[town_i][0]
        while cloud_start_i < len(cloud_start) and cloud_start[cloud_start_i][0] <= town_x:
            clouds.add(cloud_start[cloud_start_i][1])
            cloud_start_i += 1
        while cloud_end_i < len(cloud_end) and cloud_end[cloud_end_i][0] < town_x:
            clouds.remove(cloud_end[cloud_end_i][1])
            cloud_end_i += 1
        if len(clouds) == 1:
            towns[town_i][2] = list(clouds)[0]
            d[list(clouds)[0]] += towns[town_i][1]
        elif len(clouds) == 0:
            free += towns[town_i][1]

    return max(d.values(), default=0) + free


def main():
    n = int(input().strip())
    p = [int(x) for x in input().strip().split()]
    x = [int(x) for x in input().strip().split()]
    towns = [[xi, pi, -1] for xi, pi in zip(x, p)]
    m = int(input().strip())
    y = [int(x) for x in input().strip().split()]
    r = [int(x) for x in input().strip().split()]
    cloud_start = [[y[i]-r[i], i] for i in range(m)]
    cloud_end = [[y[i]+r[i], i] for i in range(m)]
    result = maximumPeople(towns, cloud_start, cloud_end)
    print(result)

if __name__ == "__main__":
    main()

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Cloudy Day

Problem Statement :

Solution :

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