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Space optimization using bit manipulations

There are many situations where we use integer values as index in array to see presence or absence, we can use bit manipulations to optimize space in such problems.

Let us consider below problem as an example.

Given two numbers say a and b, mark the multiples of 2 and 5 between a and b using less than O(|b – a|) space and output each of the multiples.
Note : We have to mark the multiples i.e save (key, value) pairs in memory such that each key either have value as 1 or 0 representing as multiple of 2 or 5 or not respectively.
Examples :

Input : 2 10
Output : 2 4 5 6 8 10

Input: 60 95
Output: 60 62 64 65 66 68 70 72 74 75 76 78 
        80 82 84 85 86 88 90 92 94 95


Approach 1 (Simple):
Hash the indices in an array from a to b and mark each of the indices as 1 or 0.
Space complexity : O(max(a, b))

Approach 2 (Better than simple):
Save memory, by translating a to 0th index and b to (b-a)th index.
Space complexity : O(|b-a|).

Simply hash |b – a| positions of an array as 0 and 1.

C++

// CPP program to mark numbers as multiple of 2 or 5
#include <bits/stdc++.h>
using namespace std;
  
// Driver code
int main()
{
    int a = 2, b = 10;
    int size = abs(b - a) + 1;
    int* array = new int[size];
  
    // Iterate through a to b, If it is a multiple
    // of 2 or 5 Mark index in array as 1
    for (int i = a; i <= b; i++)
        if (i % 2 == 0 || i % 5 == 0)
            array[i - a] = 1;
  
    cout << "MULTIPLES of 2 and 5: ";
    for (int i = a; i <= b; i++)
        if (array[i - a] == 1)
            cout << i << " ";
  
    return 0;
}

Java

// Java program to mark numbers as 
// multiple of 2 or 5
import java.lang.*;
  
class GFG {
      
    // Driver code
    public static void main(String[] args)
    {
        int a = 2, b = 10;
        int size = Math.abs(b - a) + 1;
        int array[] = new int[size];
      
        // Iterate through a to b, If 
        // it is a multiple of 2 or 5 
        // Mark index in array as 1
        for (int i = a; i <= b; i++)
            if (i % 2 == 0 || i % 5 == 0)
                array[i - a] = 1;
      
        System.out.println("MULTIPLES of 2"
                              + " and 5:");
        for (int i = a; i <= b; i++)
            if (array[i - a] == 1)
                System.out.printf(i + " ");
    }
}
  
// This code is contributed by
// Smitha Dinesh Semwal

Python 3

# Python 3 program to mark numbers
# as multiple of 2 or 5
import math
  
# Driver code
a = 2
b = 10
size = abs(b - a) + 1
array = [0] * size
  
# Iterate through a to b, 
# If it is a multiple of 2 
# or 5 Mark index in array as 1
for i in range(a, b + 1):
    if (i % 2 == 0 or i % 5 == 0):
            array[i - a] = 1
  
print("MULTIPLES of 2 and 5:")
for i in range(a, b + 1):
    if (array[i - a] == 1):
            print(i, end=" ")
  
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to mark numbers as 
// multiple of 2 or 5
using System;
  
class GFG {
      
    // Driver code
    static public void Main ()
    {
        int a = 2, b = 10;
        int size = Math.Abs(b - a) + 1;
        int[] array = new int[size];
      
        // Iterate through a to b, If 
        // it is a multiple of 2 or 5 
        // Mark index in array as 1
        for (int i = a; i <= b; i++)
            if (i % 2 == 0 || i % 5 == 0)
                array[i - a] = 1;
      
        Console.WriteLine("MULTIPLES of 2" +
                          " and 5:");
        for (int i = a; i <= b; i++)
            if (array[i - a] == 1)
                Console.Write(i + " ");
    }
}
  
// This code is contributed by Ajit.

PHP

<?php
// PHP program to mark 
// numbers as multiple 
// of 2 or 5
  
// Driver Code
$a = 2;
$b = 10;
$size = abs($b - $a) + 1;
$array = array_fill(0, $size, 0);
  
// Iterate through a to b, 
// If it is a multiple of
// 2 or 5 Mark index in
// array as 1
for ($i = $a; $i <= $b; $i++)
    if ($i % 2 == 0 || $i % 5 == 0)
        $array[$i - $a] = 1;
  
echo "MULTIPLES of 2 and 5: ";
for ($i = $a; $i <= $b; $i++)
    if ($array[$i - $a] == 1)
        echo $i . " ";
  
// This code is contributed by mits.
?>

Output :

MULTIPLES of 2 and 5:
2 4 5 6 8 10

Approach 3 (Using Bit Manipulations):

Here is a space optimized which uses bit manipulation technique that can be applied to problems mapping binary values in arrays.
Size of int variable in 64-bit compiler is 4 bytes. 1 byte is represented by 8 bit positions in memory. So, an integer in memory is represented by 32 bit positions(4 Bytes) these 32 bit positions can be used instead of just one index to hash binary values.

C++

// CPP code to for marking multiples
#include <bits/stdc++.h>
using namespace std;
  
// index >> 5 corresponds to dividing index by 32
// index & 31 corresponds to modulo operation of 
// index by 32
  
// Function to check value of bit position whether 
// it is zero or one
bool checkbit(int array[], int index)
{
    return array[index >> 5] & (1 << (index & 31));
}
  
// Sets value of bit for corresponding index
void setbit(int array[], int index)
{
    array[index >> 5] |= (1 << (index & 31));
}
  
/* Driver program to test above functions*/
int main()
{
    int a = 2, b = 10;
    int size = abs(b - a);
  
    // Size that will be used is actual_size/32
    // ceil is used to initialize the array with
    // positive number
    size = ceil(size / 32);
      
    // Array is dynamically initialized as
    // we are calculating size at run time
    int* array = new int[size];
  
    // Iterate through every index from a to b and
    // call setbit() if it is a multiple of 2 or 5
    for (int i = a; i <= b; i++)
        if (i % 2 == 0 || i % 5 == 0)
            setbit(array, i - a);
  
    cout << "MULTIPLES of 2 and 5: ";
    for (int i = a; i <= b; i++)
        if (checkbit(array, i - a))
            cout << i << " ";
  
    return 0;
}

Java

// Java code to for marking multiples
import java.io.*;
import java.util.*;
  
class GFG 
{
    // index >> 5 corresponds to dividing index by 32 
    // index & 31 corresponds to modulo operation of 
    // index by 32 
  
    // Function to check value of bit position whether 
    // it is zero or one 
    static boolean checkbit(int array[], int index) 
    
            int val = array[index >> 5] & (1 << (index & 31));
            if (val == 0)
                return false;
            return true
    
  
    // Sets value of bit for corresponding index 
    static void setbit(int array[], int index) 
    
            array[index >> 5] |= (1 << (index & 31)); 
    }
  
    // Driver code
    public static void main(String args[])
    {
            int a = 2, b = 10
            int size = Math.abs(b-a);
  
            // Size that will be used is actual_size/32 
            // ceil is used to initialize the array with 
            // positive number
            size = (int)Math.ceil((double)size / 32);
  
            // Array is dynamically initialized as 
            // we are calculating size at run time 
            int[] array = new int[size];
  
            // Iterate through every index from a to b and 
            // call setbit() if it is a multiple of 2 or 5 
            for (int i = a; i <= b; i++) 
                if (i % 2 == 0 || i % 5 == 0
                    setbit(array, i - a);
  
            System.out.println("MULTIPLES of 2 and 5:");
            for (int i = a; i <= b; i++) 
                if (checkbit(array, i - a)) 
                    System.out.print(i + " "); 
    }
}
  
// This code is contributed by rachana soma

Output:

MULTIPLES of 2 and 5:
2 4 5 6 8 10


This article is attributed to GeeksforGeeks.org

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