Sudo Placement | Range Queries
Last Updated :
22 Feb, 2023
Given Q queries, with each query consisting of two integers L and R, the task is to find the total numbers between L and R (Both inclusive), having almost three set bits in their binary representation.
Examples:
Input : Q = 2
L = 3, R = 7
L = 10, R = 16
Output : 5
6
For the first query, valid numbers are 3, 4, 5, 6, and 7.
For the second query, valid numbers are 10, 11, 12, 13, 14 and 16.
Prerequisites : Bit Manipulation and Binary Search
Method 1 (Simple): A naive approach is to traverse all the numbers between L and R and find the number of set bits in each of those numbers. Increment a counter variable if a number does not have more than 3 set bits. Return answer as counter. Note : This approach is very inefficient since the numbers L and R may have large values (upto 1018).
Method 2 (Efficient) : An efficient approach required here is precomputation. Since the values of L and R lie within the range [0, 1018] (both inclusive), thus their binary representation can have at most 60 bits. Now, since the valid numbers are those having almost 3 set bits, find them by generating all bit sequences of 60 bits with less than or equal to 3 set bits. This can be done by fixing, ith, jth and kth bits for all i, j, k from (0, 60). Once, all the valid numbers are generated in sorted order, apply binary search to find the count of those numbers that lie within the given range.
Below is the implementation of above approach.
C++
#include <bits/stdc++.h>
using namespace std;
#define LL long long int
void answerQueries(LL Q, vector<pair<LL, LL> > query)
{
set<LL> s;
s.insert(0);
for ( int i = 0; i <= 60; i++) {
for ( int j = i; j <= 60; j++) {
for ( int k = j; k <= 60; k++) {
if (j == i && i == k)
s.insert(1LL << i);
else if (j == k && i != j) {
LL x = (1LL << i) + (1LL << j);
s.insert(x);
}
else if (i == j && i != k) {
LL x = (1LL << i) + (1LL << k);
s.insert(x);
}
else if (i == k && i != j) {
LL x = (1LL << k) + (1LL << j);
s.insert(x);
}
else {
LL x = (1LL << i) + (1LL << j) + (1LL << k);
s.insert(x);
}
}
}
}
vector<LL> validNumbers;
for ( auto val : s)
validNumbers.push_back(val);
for ( int i = 0; i < Q; i++) {
LL L = query[i].first;
LL R = query[i].second;
if (R < L)
swap(L, R);
if (L == 0)
cout << (upper_bound(validNumbers.begin(), validNumbers.end(),
R) - validNumbers.begin()) << endl;
else
cout << (upper_bound(validNumbers.begin(), validNumbers.end(),
R) - upper_bound(validNumbers.begin(), validNumbers.end(),
L - 1)) << endl;
}
}
int main()
{
int Q = 2;
vector<pair<LL, LL> > query(Q);
query[0].first = 3;
query[0].second = 7;
query[1].first = 10;
query[1].second = 16;
answerQueries(Q, query);
return 0;
}
|
Java
import java.util.*;
import java.io.*;
public class RangeQueries {
static class Query {
long L;
long R;
}
public static int upperBound(ArrayList<Long> validNumbers,
Long value)
{
int low = 0 ;
int high = validNumbers.size()- 1 ;
while (low < high){
int mid = (low + high)/ 2 ;
if (value >= validNumbers.get(mid)){
low = mid+ 1 ;
} else {
high = mid;
}
}
return low;
}
public static void answerQueries(ArrayList<Query> queries){
Set<Long> allNum = new HashSet<>();
allNum.add(0L);
for ( int i= 0 ; i<= 60 ; i++){
for ( int j= 0 ; j<= 60 ; j++){
for ( int k= 0 ; k<= 60 ; k++){
if (i==j && j==k){
allNum.add(1L << i);
}
else if (i==j && j != k){
long toAdd = (1L << i) + (1L << k);
allNum.add(toAdd);
}
else if (i==k && k != j){
long toAdd = (1L << i) + (1L << j);
allNum.add(toAdd);
}
else if (j==k && k != i){
long toAdd = (1L << j) + (1L << i);
allNum.add(toAdd);
}
else {
long toAdd = (1L << i) + (1L << j) + (1L << k);
allNum.add(toAdd);
}
}
}
}
ArrayList<Long> validNumbers = new ArrayList<>();
for (Long num: allNum){
validNumbers.add(num);
}
Collections.sort(validNumbers);
for ( int i= 0 ; i<queries.size(); i++){
long L = queries.get(i).L;
long R = queries.get(i).R;
if (R < L){
long temp = L;
L = R;
R = temp;
}
if (L == 0 ){
int indxOfLastNum = upperBound(validNumbers, R);
System.out.println(indxOfLastNum+ 1 );
}
else {
int indxOfFirstNum = upperBound(validNumbers, L);
int indxOfLastNum = upperBound(validNumbers, R);
System.out.println((indxOfLastNum - indxOfFirstNum + 1 ));
}
}
}
public static void main(String[] args){
int Q = 2 ;
ArrayList<Query> queries = new ArrayList<>();
Query q1 = new Query();
q1.L = 3 ;
q1.R = 7 ;
Query q2 = new Query();
q2.L = 10 ;
q2.R = 16 ;
queries.add(q1);
queries.add(q2);
answerQueries(queries);
}
}
|
Python3
import bisect
def answerQueries(Q, query):
s = set ()
s.add( 0 )
for i in range ( 61 ):
for j in range (i, 61 ):
for k in range (j, 61 ):
if (j = = i and i = = k):
s.add( 1 << i)
elif (j = = k and i ! = j):
x = ( 1 << i) + ( 1 << j)
s.add(x)
elif (i = = j and i ! = k):
x = ( 1 << i) + ( 1 << k)
s.add(x)
elif (i = = k and i ! = j):
x = ( 1 << k) + ( 1 << j)
s.add(x)
else :
x = ( 1 << i) + ( 1 << j) + ( 1 << k)
s.add(x);
validNumbers = []
for val in sorted (s):
validNumbers.append(val)
for i in range (Q):
L = query[i][ 0 ]
R = query[i][ 1 ]
if (R < L):
L, R = R, L
if (L = = 0 ):
print (bisect.bisect_right(validNumbers, R))
else :
print (bisect.bisect_right(validNumbers, R) - bisect.bisect_right(validNumbers, L - 1 ))
Q = 2
query = [[ 3 , 7 ], [ 10 , 16 ]]
answerQueries(Q, query)
|
C#
using System;
using System.Collections.Generic;
public class Query {
public long L;
public long R;
}
public class RangeQueries {
public static int upperBound(List< long > validNumbers,
long value)
{
int low = 0;
int high = validNumbers.Count - 1;
while (low < high) {
int mid = (low + high) / 2;
if (value >= validNumbers[mid]) {
low = mid + 1;
}
else {
high = mid;
}
}
return low;
}
public static void answerQueries(List<Query> queries)
{
HashSet< long > allNum = new HashSet< long >();
allNum.Add(0L);
for ( int i = 0; i <= 60; i++) {
for ( int j = 0; j <= 60; j++) {
for ( int k = 0; k <= 60; k++) {
if (i == j && j == k) {
allNum.Add(1L << i);
}
else if (i == j && j != k) {
long toAdd = (1L << i) + (1L << k);
allNum.Add(toAdd);
}
else if (i == k && k != j) {
long toAdd = (1L << i) + (1L << j);
allNum.Add(toAdd);
}
else if (j == k && k != i) {
long toAdd = (1L << j) + (1L << i);
allNum.Add(toAdd);
}
else {
long toAdd = (1L << i) + (1L << j)
+ (1L << k);
allNum.Add(toAdd);
}
}
}
}
List< long > validNumbers = new List< long >();
foreach ( long num in allNum)
{
validNumbers.Add(num);
}
validNumbers.Sort();
for ( int i = 0; i < queries.Count; i++) {
long L = queries[i].L;
long R = queries[i].R;
if (R < L) {
long temp = L;
L = R;
R = temp;
}
if (L == 0) {
int indxOfLastNum
= upperBound(validNumbers, R);
Console.WriteLine(indxOfLastNum + 1);
}
else {
int indxOfFirstNum
= upperBound(validNumbers, L);
int indxOfLastNum
= upperBound(validNumbers, R);
Console.WriteLine(
(indxOfLastNum - indxOfFirstNum + 1));
}
}
}
public static void Main( string [] args)
{
List<Query> queries = new List<Query>();
Query q1 = new Query();
q1.L = 3;
q1.R = 7;
Query q2 = new Query();
q2.L = 10;
q2.R = 16;
queries.Add(q1);
queries.Add(q2);
answerQueries(queries);
}
}
|
Javascript
class Query {
constructor(L, R) {
this .L = L;
this .R = R;
}
}
function upperBound(validNumbers, value) {
let low = 0;
let high = validNumbers.length - 1;
while (low < high) {
let mid = Math.floor((low + high) / 2);
if (value >= validNumbers[mid]) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
function answerQueries(queries) {
let allNum = new Set();
allNum.add(0);
for (let i = 0; i <= 60; i++) {
for (let j = 0; j <= 60; j++) {
for (let k = 0; k <= 60; k++) {
if (i == j && j == k) {
allNum.add(1 << i);
}
else if (i == j && j != k) {
let toAdd = (1 << i) + (1 << k);
allNum.add(toAdd);
} else if (i == k && k != j) {
let toAdd = (1 << i) + (1 << j);
allNum.add(toAdd);
} else if (j == k && k != i) {
let toAdd = (1 << j) + (1 << i);
allNum.add(toAdd);
}
else {
let toAdd = (1 << i) + (1 << j) + (1 << k);
allNum.add(toAdd);
}
}
}
}
let validNumbers = Array.from(allNum);
validNumbers.sort((a, b) => a - b);
for (let i = 0; i < queries.length; i++) {
let L = queries[i].L;
let R = queries[i].R;
if (R < L) {
let temp = L;
L = R;
R = temp;
}
if (L == 0) {
let indxOfLastNum = upperBound(validNumbers, R);
console.log(indxOfLastNum + 1);
} else {
let indxOfFirstNum = upperBound(validNumbers, L);
let indxOfLastNum = upperBound(validNumbers, R);
console.log(indxOfLastNum - indxOfFirstNum + 1);
}
}
}
let Q = 2;
let queries = [];
let q1 = new Query(3, 7);
let q2 = new Query(10, 16);
queries.push(q1);
queries.push(q2);
answerQueries(queries);
|
Time Complexity : O((Maximum Number of Bits)3 + Q * logN), where Q is the number of queries and N is the size of set containing all valid numbers. l valid numbers.
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