Given an array arr[] consisting of N integers an array Q[][] consisting of queries of the form {l, r}. For each query {l, r}, the task is to determine the minimum sum of all values that must be added to each array element in that range such that the Bitwise AND of all numbers in that range exceeds 0.
Note: Different values can be added to different integers in the given range.
Examples:
Input: arr[] = {1, 2, 4, 8}, Q[][] = {{1, 4}, {2, 3}, {1, 3}}
Output: 3 2 2
Explanation: Binary representation of array elements are as follows:
1 – 0001
2 – 0010
4 – 0100
8 – 1000
For first query {1, 4}, add 1 to all numbers present in the range except the first number. Therefore, Bitwise AND = (1 & 3 & 5 & 9) = 1 and minimum sum of elements added = 3.
For second query {2, 3}, add 2 to 3rd element. Therefore, Bitwise AND = (2 & 6) = 2 and minimum sum of elements added = 2.
For third query {1, 3}, add 1 to 2nd and 3rd elements. Therefore, Bitwise AND = (1 & 3 & 5) = 1 and minimum sum of elements added = 2.Input: arr[] = {4, 6, 5, 3}, Q[][] = {{1, 4}}
Output: 1
Explanation: Optimal way to make the Bitwise AND non-zero is to add 1 to the last element. Therefore, bitwise AND = (4 & 6 & 5 & 4) = 4 and minimum sum of elements added = 1.
Approach: The idea is to first observe that the Bitwise AND of all the integers in the range [l, r] can only be non-zero if a bit at a particular index is set for each integer in that range.
Follow the steps below to solve the problem:
- Initialize a 2D vector pre where pre[i][j] stores the minimum sum of integers to be added to all integers from index 0 to index i such that, for each of them, their jthbit is set.
- For each element at index i, check for each of its bit from j = 0 to 31.
- Initialize a variable sum with 0.
- If jthbit is set, update pre[i][j] as pre[i][j]=pre[i-1][j] and increment sum by 2j. Otherwise, update pre[i][j] = pre[i-1][j] + 2j – sum where (2j-sum) is the value that must be added to set the jthbit of arr[i].
- Now, for each query {l, r}, the minimum sum required to set jthbit of all elements in the given range is pre[r][j] – pre[l-1][j].
- For each query {l, r}, find the answer for each bit from j = 0 to 31 and print the minimum amongst them.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the min sum required // to set the jth bit of all integer void processing(vector< int > v,
vector<vector< long long int > >& pre)
{ // Number of elements
int N = v.size();
// Traverse all elements
for ( int i = 0; i < N; i++) {
// Binary representation
bitset<32> b(v[i]);
long long int sum = 0;
// Take previous values
if (i != 0) {
pre[i] = pre[i - 1];
}
// Processing each bit and
// store it in 2d vector
for ( int j = 0; j < 32; j++) {
if (b[j] == 1) {
sum += 1ll << j;
}
else {
pre[i][j]
+= (1ll << j) - sum;
}
}
}
} // Function to print the minimum // sum for each query long long int
process_query(vector<vector< int > > Q,
vector<vector< long long int > >& pre)
{ // Stores the sum for each query
vector< int > ans;
for ( int i = 0; i < Q.size(); i++) {
// Update wrt 0-based index
--Q[i][0], --Q[i][1];
// Initizlize answer
long long int min1 = INT_MAX;
// Find minimum sum for each bit
if (Q[i][0] == 0) {
for ( int j = 0; j < 32; j++) {
min1 = min(pre[Q[i][1]][j], min1);
}
}
else {
for ( int j = 0; j < 32; j++) {
min1 = min(pre[Q[i][1]][j]
- pre[Q[i][0] - 1][j],
min1);
}
}
// Store the answer for
// each query
ans.push_back(min1);
}
// Print the answer vector
for ( int i = 0; i < ans.size(); i++) {
cout << ans[i] << " " ;
}
} // Driver Code int main()
{ // Given array
vector< int > arr = { 1, 2, 4, 8 };
// Given Queries
vector<vector< int > > Q
= { { 1, 4 }, { 2, 3 }, { 1, 3 } };
// 2d Prefix vector
vector<vector< long long int > > pre(
100001, vector< long long int >(32, 0));
// Preprocessing
processing(arr, pre);
// Function call for queries
process_query(Q, pre);
return 0;
} |
// Java code to implement the approach import java.util.Arrays;
class Main {
public static void main(String[] args)
{
// Given array
long [] arr = { 1 , 2 , 4 , 8 };
// Given Queries
long [][] Q = { new long [] { 1 , 4 }, new long [] { 2 , 3 }, new long [] { 1 , 3 } };
// 2d Prefix vector
long [][] pre = new long [ 100001 ][];
Arrays.setAll(pre, i -> new long [ 32 ]);
// Preprocessing
processing(arr, pre);
// Function call for queries
long [] ans = processQuery(Q, pre);
// Print the answer vector
System.out.println(String.join( ", " , Arrays.toString(ans)));
}
private static void processing( long [] v, long [][] pre) {
// Number of elements
long N = v.length;
// Traverse all elements
for ( long i = 0 ; i < N; i++) {
// Binary representation
String b = Long.toBinaryString(v[( int ) i]);
b = "0" .repeat(( int ) ( 32 - b.length())) + b;
b = new StringBuilder(b).reverse().toString();
long sum = 0 ;
// Take previous values
if (i != 0 ) {
pre[( int ) i] = Arrays.copyOf(pre[( int ) (i - 1 )], pre[( int ) (i - 1 )].length);
}
// Processing each bit and store it in 2d vector
for ( int j = 0 ; j < 32 ; j++) {
if (b.charAt(j) == '1' ) {
sum += ( long ) Math.pow( 2 , j);
} else {
pre[( int ) i][j] += ( long ) Math.pow( 2 , j) - sum;
}
}
}
}
private static long [] processQuery( long [][] Q, long [][] pre) {
// Stores the sum for each query
long [] ans = new long [Q.length];
for ( long i = 0 ; i < Q.length; i++) {
// Update wrt 0-based index
Q[( int ) i][ 0 ] -= 1 ;
Q[( int ) i][ 1 ] -= 1 ;
// Initialize answer
long min1 = Long.MAX_VALUE;
// Find minimum sum for each bit
if (Q[( int ) i][ 0 ] == 0 ) {
for ( long j = 0 ; j < 32 ; j++) {
min1 =( long ) Math.min(pre[( int ) Q[( int ) i][ 1 ]][( int ) j], min1);
}
} else {
for ( long j = 0 ; j < 32 ; j++) {
min1 =( long ) Math.min(
pre[( int ) Q[( int ) i][ 1 ]][( int ) j] - pre[( int ) (Q[( int ) i][ 0 ] - 1 )][( int ) j], min1);
}
}
// Store the answer for each query
ans[( int )i] = min1;
}
return ans;
}
} // This code is contributed by phasing17 |
from typing import List , Tuple
def processing(v: List [ int ], pre: List [ List [ int ]]) - > None :
# Number of elements
N = len (v)
# Traverse all elements
for i in range (N):
# Binary representation
b = bin (v[i])[ 2 :]
b = "0" * ( 32 - len (b)) + b
b = b[:: - 1 ]
sum = 0
# Take previous values
if i ! = 0 :
pre[i] = pre[i - 1 ][:]
# Processing each bit and store it in 2d vector
for j in range ( 32 ):
if b[j] = = "1" :
sum + = 1 << j
else :
pre[i][j] + = ( 1 << j) - sum
def process_query(Q: List [ Tuple [ int , int ]], pre: List [ List [ int ]]) - > List [ int ]:
# Stores the sum for each query
ans = []
for i in range ( len (Q)):
# Update wrt 0-based index
Q[i] = (Q[i][ 0 ] - 1 , Q[i][ 1 ] - 1 )
# Initialize answer
min1 = float ( "inf" )
# Find minimum sum for each bit
if Q[i][ 0 ] = = 0 :
for j in range ( 32 ):
min1 = min (pre[Q[i][ 1 ]][j], min1)
else :
for j in range ( 32 ):
min1 = min (pre[Q[i][ 1 ]][j] - pre[Q[i][ 0 ] - 1 ][j], min1)
# Store the answer for each query
ans.append(min1)
return ans
# Given array arr = [ 1 , 2 , 4 , 8 ]
# Given Queries Q = [( 1 , 4 ), ( 2 , 3 ), ( 1 , 3 )]
# 2d Prefix vector pre = [[ 0 ] * 32 for _ in range ( 100001 )]
# Preprocessing processing(arr, pre) # Function call for queries ans = process_query(Q, pre)
# Print the answer vector print (ans)
# This code is contributed by phasing17. |
// C# code to implement the approach using System;
using System.Linq;
class Program {
static void Main( string [] args)
{
// Given array
long [] arr = { 1, 2, 4, 8 };
// Given Queries
long [][] Q
= { new long [] { 1, 4 }, new long [] { 2, 3 },
new long [] { 1, 3 } };
// 2d Prefix vector
long [][] pre = new long [100001][]
.Select(x => new long [32])
.ToArray();
// Preprocessing
Processing(arr, pre);
// Function call for queries
long [] ans = ProcessQuery(Q, pre);
// Prlong the answer vector
Console.WriteLine( string .Join( ", " , ans));
}
private static void Processing( long [] v, long [][] pre)
{
// Number of elements
long N = v.Length;
// Traverse all elements
for ( long i = 0; i < N; i++) {
// Binary representation
string b = Convert.ToString(v[i], 2);
b = "0" .PadLeft(32 - b.Length, '0' ) + b;
b = new string (b.Reverse().ToArray());
long sum = 0;
// Take previous values
if (i != 0) {
pre[i] = pre[i - 1].ToArray();
}
// Processing each bit and store it in 2d vector
for ( int j = 0; j < 32; j++) {
if (b[j] == '1' ) {
sum += ( long )Math.Pow(2, j);
}
else {
pre[i][j] += ( long )Math.Pow(2, j) - sum;
}
}
}
}
private static long [] ProcessQuery( long [][] Q,
long [][] pre)
{
// Stores the sum for each query
long [] ans = new long [Q.Length];
for ( long i = 0; i < Q.Length; i++) {
// Update wrt 0-based index
Q[i][0] -= 1;
Q[i][1] -= 1;
// Initialize answer
long min1 = long .MaxValue;
// Find minimum sum for each bit
if (Q[i][0] == 0) {
for ( long j = 0; j < 32; j++) {
min1 = Math.Min(pre[Q[i][1]][j], min1);
}
}
else {
for ( long j = 0; j < 32; j++) {
min1 = Math.Min(
pre[Q[i][1]][j]
- pre[Q[i][0] - 1][j],
min1);
}
}
// Store the answer for each query
ans[i] = min1;
}
return ans;
}
} // This code is contributed by phasing17 |
// JS program to implement the above approach const processing = (v, pre) => { // Number of elements
const N = v.length;
// Traverse all elements
for (let i = 0; i < N; i++) {
// Binary representation
let b = v[i].toString(2);
b = "0" .repeat(32 - b.length) + b;
b = b.split( "" ).reverse().join( "" );
let sum = 0;
// Take previous values
if (i !== 0) {
pre[i] = pre[i - 1].slice();
}
// Processing each bit and store it in 2d vector
for (let j = 0; j < 32; j++) {
if (b[j] === "1" ) {
sum += 2 ** j;
} else {
pre[i][j] += 2 ** j - sum;
}
}
}
}; const processQuery = (Q, pre) => { // Stores the sum for each query
const ans = [];
for (let i = 0; i < Q.length; i++) {
// Update wrt 0-based index
Q[i] = [Q[i][0] - 1, Q[i][1] - 1];
// Initialize answer
let min1 = Number.POSITIVE_INFINITY;
// Find minimum sum for each bit
if (Q[i][0] === 0) {
for (let j = 0; j < 32; j++) {
min1 = Math.min(pre[Q[i][1]][j], min1);
}
} else {
for (let j = 0; j < 32; j++) {
min1 = Math.min(pre[Q[i][1]][j] - pre[Q[i][0] - 1][j], min1);
}
}
// Store the answer for each query
ans.push(min1);
}
return ans;
}; // Given array const arr = [1, 2, 4, 8]; // Given Queries const Q = [[1, 4], [2, 3], [1, 3]]; // 2d Prefix vector const pre = new Array(100001).fill(0).map(() => new Array(32).fill(0));
// Preprocessing processing(arr, pre); // Function call for queries const ans = processQuery(Q, pre); // Print the answer vector console.log(ans); // This code is implemented by Phasing17 |
3 2 2
Time Complexity: O(N*32 + sizeof(Q)*32)
Auxiliary Space: O(N*32)