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Count set bits in index range [L, R] in given Array for Q queries

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  • Last Updated : 14 Dec, 2021

Given an array arr[] containing N integers and an array queries[] containing Q queries in the form of {L, R}, the task is to count the total number of set bits from L to R in array arr for each query.

Example:

Input: arr[]={1, 2, 3, 4, 5, 6}, queries[]={{0, 2}, {1, 1}, {3, 5}}
Output:
4
2
5
Explanation:
Query 1 (0, 2): Elements {1, 2, 3} have set bits {1, 1, 2} respectively. So sum=1+1+2=4
Query 1 (1, 1): Element {2} have set bit {1}. So sum=1
Query 1 (3, 5): Elements {4, 5, 6} have set bits {1, 2, 2} respectively. So sum=1+2+2=5

Input: arr[]={2, 4, 3, 5, 6}, queries[]={{0, 1}, {2, 4}}
Output:
2
6

 

Approach: To solve this question, follow the below steps:

  1. Create a vector, say onBits, in which each ith element will represent the number of set bits from arr[0] to arr[i].
  2. Run a loop from i=0 to i<N, and on each iteration, find the number of set bits in arr[i], say bits and make onBits[i]=onBits[i-1] + bits.
  3. Now traverse the array queries and for each query {L, R}, the number of set bits from L to R is onBits[R]-onBits[L-1].
  4. Print the answer according to the above observation.

Below is the implementation of the above approach:

C++




// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number
// of set bits from L to R
int setBitsinLtoR(int L, int R,
                  vector<int>& onBits)
{
    if (L == 0) {
        return onBits[R];
    }
 
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
void totalSetBits(vector<int>& arr,
                  vector<pair<int, int> >& queries)
{
    int N = arr.size();
    vector<int> onBits(N, 0);
    onBits[0] = __builtin_popcount(arr[0]);
 
    for (int i = 1; i < N; ++i) {
        onBits[i]
            = onBits[i - 1]
              + __builtin_popcount(arr[i]);
    }
 
    // Finding for Q queries
    for (auto x : queries) {
        int L = x.first;
        int R = x.second;
 
        cout << setBitsinLtoR(L, R, onBits)
             << endl;
    }
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    vector<pair<int, int> > queries
        = { { 0, 2 }, { 1, 1 }, { 3, 5 } };
 
    totalSetBits(arr, queries);
}


Java




// Java code for the above approach
import java.util.*;
class GFG{
 
static class pair
{
    int first, second;
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
   
// Function to return the number
// of set bits from L to R
static int setBitsinLtoR(int L, int R,int []onBits)
{
    if (L == 0) {
        return onBits[R];
    }
 
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
static void totalSetBits(int[] arr,
                  pair[] queries)
{
    int N = arr.length;
     
    int []onBits = new int[N];
    onBits[0] = Integer.bitCount(arr[0]);
 
    for (int i = 1; i < N; ++i) {
        onBits[i]
            = onBits[i - 1]
              + Integer.bitCount(arr[i]);
    }
 
    // Finding for Q queries
    for (pair x : queries) {
        int L = x.first;
        int R = x.second;
 
        System.out.print(setBitsinLtoR(L, R, onBits)
             +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    pair []queries
        = { new pair( 0, 2 ), new pair( 1, 1 ), new pair( 3, 5 ) };
 
    totalSetBits(arr, queries);
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python code for the above approach
def  __builtin_popcount(n):
    count = 0
    while (n):
        count += n & 1
        n >>= 1
    return count
 
# Function to return the number
# of set bits from L to R
def setBitsinLtoR(L, R, onBits):
    if (L == 0):
        return onBits[R];
    return onBits[R] - onBits[L - 1];
 
# Function to find the number
# of set bits for Q queries
def totalSetBits(arr, queries):
    N = len(arr);
    onBits = [0] * N
    onBits[0] = __builtin_popcount(arr[0]);
 
    for i in range(1, N):
        onBits[i] = onBits[i - 1] +  __builtin_popcount(arr[i]);
 
    # Finding for Q queries
    for x in queries:
        L = x[0];
        R = x[1];
 
        print(setBitsinLtoR(L, R, onBits))
 
# Driver Code
arr = [ 1, 2, 3, 4, 5, 6 ];
queries = [ [ 0, 2 ], [ 1, 1 ], [ 3, 5 ] ];
 
totalSetBits(arr, queries);
 
# This code is contributed by gfgking


C#




// C# code for the above approach
using System;
 
class GFG{
 
class pair
{
    public int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
   
// Function to return the number
// of set bits from L to R
static int setBitsinLtoR(int L, int R, int []onBits)
{
    if (L == 0)
    {
        return onBits[R];
    }
     
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
static void totalSetBits(int[] arr,
                         pair[] queries)
{
    int N = arr.Length;
    int []onBits = new int[N];
    onBits[0] = bitCount(arr[0]);
 
    for(int i = 1; i < N; ++i)
    {
        onBits[i] = onBits[i - 1] +
                    bitCount(arr[i]);
    }
 
    // Finding for Q queries
    foreach (pair x in queries)
    {
        int L = x.first;
        int R = x.second;
 
        Console.Write(setBitsinLtoR(L, R, onBits) +
                      "\n");
    }
}
 
static int bitCount(int x)
{
    int setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    pair []queries = { new pair(0, 2),
                       new pair(1, 1),
                       new pair(3, 5) };
 
    totalSetBits(arr, queries);
}
}
 
// This code is contributed by shikhasingrajput


Javascript




<script>
 
// JavaScript code for the above approach
function __builtin_popcount(x)
{
    return((x).toString(2).split('').
           filter(x => x == '1').length);
}
 
// Function to return the number
// of set bits from L to R
function setBitsinLtoR(L, R, onBits)
{
    if (L == 0)
    {
        return onBits[R];
    }
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
function totalSetBits(arr, queries)
{
    let N = arr.length;
    let onBits = new Array(N).fill(0);
    onBits[0] = __builtin_popcount(arr[0]);
 
    for(let i = 1; i < N; ++i)
    {
        onBits[i] = onBits[i - 1] +
                    __builtin_popcount(arr[i]);
    }
 
    // Finding for Q queries
    for(let x of queries)
    {
        let L = x[0];
        let R = x[1];
 
        document.write(
            setBitsinLtoR(L, R, onBits) + '<br>')
    }
}
 
// Driver Code
let arr = [ 1, 2, 3, 4, 5, 6 ];
let queries = [ [ 0, 2 ], [ 1, 1 ], [ 3, 5 ] ];
 
totalSetBits(arr, queries);
 
// This code is contributed by Potta Lokesh
 
</script>


Output

4
1
5

Time Complexity: O(Q*N*logK), where N is the size of array arr, Q is the number of queries and K is the number of bits
Auxiliary Space: O(N)

 


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