Library Query
Problem Statement :
A giant library has just been inaugurated this week. It can be modeled as a sequence of N consecutive shelves with each shelf having some number of books. Now, being the geek that you are, you thought of the following two queries which can be performed on these shelves. Change the number of books in one of the shelves. Obtain the number of books on the shelf having the kth rank within the range of shelves. A shelf is said to have the kth rank if its position is k when the shelves are sorted based on the number of the books they contain, in ascending order. Can you write a program to simulate the above queries? Input Format The first line contains a single integer T, denoting the number of test cases. The first line of each test case contains an integer N denoting the number of shelves in the library. The next line contains N space separated integers where the ith integer represents the number of books on the ith shelf where 1<=i<=N. The next line contains an integer Q denoting the number of queries to be performed. Q lines follow with each line representing a query. Queries can be of two types: 1 x k - Update the number of books in the xth shelf to k (1 <= x <= N). 0 x y k - Find the number of books on the shelf between the shelves x and y (both inclusive) with the kth rank (1 <= x <= y <= N, 1 <= k <= y-x+1). Output Format For every test case, output the results of the queries in a new line. Constraints 1 <= T <= 5 1 <= N <= 104 1 <= Q <= 104 The number of books on each shelf is always guaranteed to be between 1 and 1000.
Solution :
Solution in C :
In C++ :
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <fstream>
#include <set>
#include <map>
#include <cmath>
#pragma comment(linker,"/STACK:16777216")
#define MAXN 100100
using namespace std;
string s1,s2;
int n,k,a[MAXN],x,q,l,r,cnt[1001],t;
int main()
{
cin>>t;
while(t--){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
scanf("%d",&q);
for(int i=0;i<q;i++)
{
scanf("%d",&x);
if(x==1){
scanf("%d %d",&l,&k);
a[l]=k;
}
else
{
scanf("%d %d %d",&l,&r,&k);
for(int j=1;j<1001;j++)
cnt[j]=0;
for(int i=l;i<=r;i++)
cnt[a[i]]++;
for(int j=1;j<1001;j++)
if(cnt[j]>=k)
{
printf("%d\n",j);
break;
}
else
k-=cnt[j];
}
}
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.math.*;
public class Solution
{
public static long MOD = 1000000007L;
public static void main(String[] args)
{
Scan scan = new Scan(System.in);
PrintWriter out = new PrintWriter(System.out);
int tc = scan.nextInt();
for (int t = 0; t < tc; t++)
{
int N = scan.nextInt();
int[] arr = new int[N];
for (int i = 0; i < N; i++)
arr[i] = scan.nextInt();
int Q = scan.nextInt();
for (int q = 0; q < Q; q++)
{
int op = scan.nextInt();
if (op == 1)
{
int x = scan.nextInt() - 1;
int k = scan.nextInt();
arr[x] = k;
}
else
{
int x = scan.nextInt() - 1;
int y = scan.nextInt() - 1;
int k = scan.nextInt();
int i;
int[] temp = new int[1001];
for (i = x; i <= y; i++)
temp[arr[i]]++;
for (i = 0; i < 1001 ; i++)
{
k -= temp[i];
if (k <= 0) break;
}
out.println(i);
}
}
}
out.close();
}
}
final class Scan
{
private InputStream in = null;
private int pos,count;
private byte[] buf = new byte[1<<16];
public Scan(InputStream in)
{
this.in = in;
pos = 0;
count = 0;
}
public int nextInt()
{
int c=read(), sign = 1;
while (c <= ' ')
c = read();
if (c == '-')
{
sign = -1;
c = read();
}
int n = c - '0';
while((c = read() - '0') >= 0)
n = n * 10 + c;
return n * sign;
}
public String next()
{
StringBuilder sb = new StringBuilder();
int c = read();
while (c <= ' ')
c = read();
while (c >= 33)
{
sb.append((char) c );
c = read();
}
return sb.toString();
}
public long nextLong()
{
int c = read(), sign = 1;
while (c <= ' ')
c = read();
if (c == '-')
{
sign = -1;
c = read();
}
long n = 1L * (c - '0');
while((c = read() - '0') >= 0)
n = n * 10 + c;
return n * sign;
}
public int read()
{
if( pos == count)
fillBuffer();
return buf[pos++];
}
private void fillBuffer()
{
try
{
count = in.read(buf, pos = 0 , buf.length);
if (count == -1)
buf[0] = -1;
}
catch(Exception e)
{
}
}
}
In C :
#include<stdio.h>
struct node{
int v,i;
};
struct node a[10005],temp;
int part(int lo,int hi)
{int i=lo,j=hi,pivot=a[(i+j)/2].v;
while (i<=j)
{while (a[i].v<pivot) i++;
while (a[j].v>pivot) j--;
if (i<=j)
{temp=a[j]; a[j]=a[i]; a[i]=temp; i++; j--;}
}
return i;
}
void sort(int lo,int hi)
{int x=part(lo,hi);
if (lo<x-1) sort(lo,x-1);
if (x<hi) sort(x,hi);
}
int main(){
int t,k,i,j,n,q,x,y,f,l,count;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(j=1;j<=n;j++){
scanf("%d",&a[j].v);
a[j].i=j;
}
sort(1,n);/*
for(j=1;j<=n;j++){
printf("%d %d\n",a[j].v,a[j].i);
}
printf("\n");*/
scanf("%d",&q);
while(q--){
scanf("%d",&f);
if(f==0){
scanf("%d%d%d",&x,&y,&k);
count=0;
for(j=1;count!=k;j++){
if( x <=a[j].i && a[j].i<=y ) count++;
}
printf("%d\n",a[j-1].v);
}
else{
scanf("%d%d",&l,&k);
for(j=1;j<=n;j++){
if(a[j].i == l){
a[j].v=k;
}
temp=a[j];
while(temp.v<a[j-1].v && j>0){
a[j]=a[j-1];
j--;
}
a[j]=temp;
while(temp.v>a[j+1].v && j<n){
a[j]=a[j+1];
j++;
}
a[j]=temp;
}/*
for(j=1;j<=n;j++){
printf("%d %d\n",a[j].v,a[j].i);
}
printf("\n");*/
}
}
}
return 0;
}
In Python3 :
class FenwickTree2D:
def __init__(self, x,y):
self.size_x = x
self.size_y = y
self.data = [[0]*self.size_y for _ in range(self.size_x)]
def sum(self, x1, y1):
if min(x1,y1) <= 0: return 0
s = 0
x = x1
while x >= 0:
y = y1
while y >= 0:
s += self.data[x][y]
y = (y & (y+1)) - 1
x = (x & (x+1)) - 1
return s
def sumrange(self, x1, y1, x2, y2):
return self.sum(x2, y2) \
- self.sum(x1-1, y2) - self.sum(x2, y1-1) \
+ self.sum(x1-1,y1-1)
def add(self, x1, y1, w):
assert min(x1,y1) > 0
x = x1
while x < self.size_x:
y = y1
while y < self.size_y:
self.data[x][y] += w
y |= y + 1
x |= x + 1
for t in range(int(input())):
N = int(input())
arr = list(map(int,input().split()))
t = FenwickTree2D(10001,1001)
for i in range(len(arr)):
t.add(i+1,arr[i],1)
Q = int(input())
for q in range(Q):
c = list(map(int,input().split()))
if c[0]==1:
t.add(c[1],arr[c[1]-1],-1)
arr[c[1]-1] = c[2]
t.add(c[1],arr[c[1]-1],1)
else:
def select(l,r,k):
lo,hi=1,1000
while lo < hi:
med = (hi+lo)//2
a = t.sumrange(l,0,r,med)
if a>=k:
hi = med
else:
lo = med+1
if not t.sumrange(l,0,r,lo) >= k:
raise ValueError
return lo
print(select(c[1],c[2],c[3]))
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