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Leftover element after performing alternate Bitwise OR and Bitwise XOR operations on adjacent pairs

Given an array of N(always a power of 2) elements and Q queries.

Every Query consists of two elements index and value.. We need to write a program that assigns value to Aindex and prints the single element which is left after performing below operations for each query:

  • At alternate steps perform bitwise OR and bitwise XOR operations to the adjacent elements.
  • In first iteration select, select n/2 pairs moving from left to right, and do a bitwise OR of all the pair values. In second iteration select (n/2)/2 leftover pairs and do a bitwise XOR on them. In the third iteration select, select ((n/2)/2)/2 leftover pairs moving from left to right, and do a bitwise OR of all the pair values.
  • Continue the above steps till we are left with a single element.

Examples:

Input : n = 4   m = 2 
        arr = [1, 4, 5, 6] 
        Queries- 
        1st: index=0 value=2  
        2nd: index=3 value=5
Output : 1 
         3 

Explanation: 

1st query: 

Assigning 2 to index 0, the sequence is now 
[2, 4, 5, 6]. 
1st iteration: There are 4/2=2 pairs (2, 4) and (5, 6) 
2 OR 4 gives 6, and 5 OR 6 gives us 7. So the 
sequence is now [6, 7]. 

2nd iteration: There is 1 pair left now (6, 7) 
6^7=1. 

Hence the last element left is 1 which is the 
answer to our first query. 

2nd Query: 

Assigning 5 to index 3, the sequence is now 
[2, 4, 5, 5]. 
1st iteration: There are 4/2=2 pairs (2, 4) and (5, 5) 
2 OR 4 gives 6, and 5 OR 5 gives us 5. So the 
sequence is now [6, 5]. 

2nd iteration: There is 1 pair left now (6, 5) 
6^5=3. 

Hence the last element left is 3 which is the
answer to our second query.


Naive Approach: The naive approach is to perform every step till we are leftover with one element. Using 2-D vector we will store the resultant elements left after every step. V[steps-1][0..size] gives the number of elements at previous step. If the step number is odd, we perform a bitwise OR operation, else a bitwise XOR operation is done. Repeat the steps till we are left over with one element. The last element left will be our answer.

Below is the implementation of the naive approach:

// CPP program to print the Leftover element after
// performing alternate Bitwise OR and Bitwise XOR
// operations to the pairs.
#include <bits/stdc++.h>
using namespace std;
#define N 1000
  
int lastElement(int a[],int n)
{
    // count the step number
    int steps = 1;
    vector<int>v[N];
  
    // if one element is there, it will be the answer
    if (n==1) return a[0];
  
  
    // at first step we do a bitwise OR
    for (int i = 0 ; i < n ; i += 2)
        v[steps].push_back(a[i] | a[i+1]);
  
  
    // keep on doing bitwise operations till the
    // last element is left
    while (v[steps].size()>1)
    {
  
        steps += 1;
  
        // perform operations
        for (int i = 0 ; i < v[steps-1].size(); i+=2)
        {
            // if step is the odd step
            if (steps&1)
                v[steps].push_back(v[steps-1][i] | v[steps-1][i+1]);
            else  // even step
                v[steps].push_back(v[steps-1][i] ^ v[steps-1][i+1]);
        }
    }
  
    // answer when one element is left
    return v[steps][0];
}
  
// Driver Code
int main()
{
    int a[] = {1, 4, 5, 6};
    int n = sizeof(a)/sizeof(a[0]);
  
    // 1st query
    int index = 0;
    int value = 2;
    a[0] = 2;
    cout << lastElement(a,n) << endl;
  
    // 2nd query
    index = 3;
    value = 5;
    a[index] = value;
    cout << lastElement(a,n)  << endl;
  
    return 0;
}

Output:

1
3

Time Complexity: O(N * 2N )

Efficient Approach: The efficient approach is to use Segment tree. Below is the complete segment tree approach used to solve the problem.

Building the tree

The leaves of the segment tree will store the array values and their parent will store the OR of the leaves. Moving up-ward in the tree, with every alternate step, the parent stores either of bitwise XOR or bitwise OR of left and right child. At every odd-numbered iteration, we perform the bitwise OR of the pairs and similarly we perform bitwise XOR of pairs at every even-numbered operation. So the odd-numbered parent will store the bitwise OR of the left and right child. Similarly, the even-numbered parent stores the bitwise XOR of the left and right child. level[] is an array that stores levels of every parent starting from 1, to determine if the pair(right child and left child) below it performs an OR operation or an XOR operation. The root of the tree will be our answer to the given sequence after every update operation.

.

The image above explains the construction of the tree if the sequence was [1, 2, 3, 4, 5, 6, 7, 8], then after 3 iterations, we will be left over with 12 which is our answer and is stored at the root.

Answering Query

There is no need to rebuild the complete tree to perform an update operation. To do an update, we should find a path from the root to the corresponding leaf and recalculate the values only at the parents that are lying on the found path.

Level of parent:
Using DP on trees, we can easily store the level of every parent. Initialize the leaf nodes level to 0, and keep adding as we move up to every parent.

The recurrence relation for calculating the level of parent is:

level[parent] = level[child] + 1
Here, child is 2*pos + 1 or 2*pos + 2

Below is the implementation of the above approach:

C++

// CPP program to print the Leftover element after
// performing alternate Bitwise OR and
// Bitwise XOR operations to the pairs.
#include <bits/stdc++.h>
using namespace std;
#define N 1000
  
// array to store the tree
int tree[N];
  
// array to store the level of every parent
int level[N];
  
// fucntion to construct the tree
void constructTree(int low, int high, int pos, int a[])
{
    if (low == high)
    {
        // level of child is always 0
        level[pos] = 0;
        tree[pos] = a[high];
        return;
    }
    int mid = (low + high) / 2;
  
    // recursive call
    constructTree(low, mid, 2 * pos + 1, a);
    constructTree(mid + 1, high, 2 * pos + 2, a);
  
    // increase the level of every parent, which is
    // level of child + 1
    level[pos] = level[2 * pos + 1] + 1;
  
    // if the parent is at odd level, then do a
    // bitwise OR
    if (level[pos] & 1)
        tree[pos] = tree[2 * pos + 1] | tree[2 * pos + 2];
  
    // if the parent is at even level, then
    // do a bitwise XOR
    else
        tree[pos] = tree[2 * pos + 1] ^ tree[2 * pos + 2];
}
  
// fucntion that updates the tree
void update(int low, int high, int pos, int index, int a[])
{
    // if it is a leaf and the leaf which is 
    // to be updated
    if (low == high and low == index)
    {
        tree[pos] = a[low];
        return;
    }
  
    // out of range
    if (index < low || index > high)
        return;
  
    // not a leaf then recurse
    if (low != high)
    {
        int mid = (low + high) / 2;
  
        // recursive call
        update(low, mid, 2 * pos + 1, index, a);
        update(mid + 1, high, 2 * pos + 2, index, a);
  
        // check if the parent is at odd or even level
        // and perform OR or XOR according to that
        if (level[pos] & 1)
            tree[pos] = tree[2 * pos + 1] | tree[2 * pos + 2];
        else
            tree[pos] = tree[2 * pos + 1] ^ tree[2 * pos + 2];
    }
}
  
// fucntion that assigns value to a[index]
// and calls update function to update the tree
void updateValue(int index, int value, int a[], int n)
{
    a[index] = value;
    update(0, n - 1, 0, index, a);
}
  
// Driver Code
int main()
{
    int a[] = { 1, 4, 5, 6 };
    int n = sizeof(a) / sizeof(a[0]);
  
    // builds the tree
    constructTree(0, n - 1, 0, a);
  
    // 1st query
    int index = 0;
    int value = 2;
    updateValue(index, value, a, n);
    cout << tree[0] << endl;
  
    // 2nd query
    index = 3;
    value = 5;
    updateValue(index, value, a, n);
    cout << tree[0] << endl;
  
    return 0;
}

Java

// java program to print the Leftover
// element after performing alternate
// Bitwise OR and Bitwise XOR operations
// to the pairs.
import java .io.*;
  
public class GFG {
  
    static int N = 1000;
      
    // array to store the tree
    static int []tree = new int[N];
      
    // array to store the level of
    // every parent
    static int []level = new int[N];
      
    // fucntion to construct the tree
    static void constructTree(int low, int high,
                               int pos, int []a)
    {
        if (low == high)
        {
              
            // level of child is
            // always 0
            level[pos] = 0;
            tree[pos] = a[high];
            return;
        }
        int mid = (low + high) / 2;
      
        // recursive call
        constructTree(low, mid, 2 * pos + 1, a);
          
        constructTree(mid + 1, high, 
                                2 * pos + 2, a);
      
        // increase the level of every parent,
        // which is level of child + 1
        level[pos] = level[2 * pos + 1] + 1;
      
        // if the parent is at odd level, then
        // do a bitwise OR
        if ((level[pos] & 1) > 0)
            tree[pos] = tree[2 * pos + 1] |
                              tree[2 * pos + 2];
      
        // if the parent is at even level, then
        // do a bitwise XOR
        else
            tree[pos] = tree[2 * pos + 1] ^ 
                              tree[2 * pos + 2];
    }
      
    // fucntion that updates the tree
    static void update(int low, int high, int pos,
                               int index, int []a)
    {
          
        // if it is a leaf and the leaf which is 
        // to be updated
        if (low == high && low == index)
        {
            tree[pos] = a[low];
            return;
        }
      
        // out of range
        if (index < low || index > high)
            return;
      
        // not a leaf then recurse
        if (low != high)
        {
            int mid = (low + high) / 2;
      
            // recursive call
            update(low, mid, 2 * pos + 1, index, a);
              
            update(mid + 1, high, 2 * pos + 2,
                                          index, a);
      
            // check if the parent is at odd or
            // even level and perform OR or XOR
            // according to that
            if ((level[pos] & 1) > 0)
                tree[pos] = tree[2 * pos + 1] |
                                  tree[2 * pos + 2];
            else
                tree[pos] = tree[2 * pos + 1] ^
                                  tree[2 * pos + 2];
        }
    }
      
    // fucntion that assigns value to a[index]
    // and calls update function to update the
    // tree
    static void updateValue(int index, int value,
                                   int []a, int n)
    {
        a[index] = value;
        update(0, n - 1, 0, index, a);
    }
      
    // Driver Code
    static public void main (String[] args)
    {
        int []a = { 1, 4, 5, 6 };
        int n = a.length;;
      
        // builds the tree
        constructTree(0, n - 1, 0, a);
      
        // 1st query
        int index = 0;
        int value = 2;
        updateValue(index, value, a, n);
        System.out.println(tree[0]);
      
        // 2nd query
        index = 3;
        value = 5;
        updateValue(index, value, a, n);
        System.out.println(tree[0]);
    }
}
  
// This code is contributed by vt_m.

C#

// C# program to print the Leftover
// element after performing alternate
// Bitwise OR and Bitwise XOR
// operations to the pairs.
using System;
  
public class GFG {
  
    static int N = 1000;
      
    // array to store the tree
    static int []tree = new int[N];
      
    // array to store the level of
    // every parent
    static int []level = new int[N];
      
    // fucntion to construct the
    // tree
    static void constructTree(int low, int high,
                               int pos, int []a)
    {
        if (low == high)
        {
              
            // level of child is always 0
            level[pos] = 0;
            tree[pos] = a[high];
            return;
        }
        int mid = (low + high) / 2;
      
        // recursive call
        constructTree(low, mid, 2 * pos + 1, a);
          
        constructTree(mid + 1, high,
                                2 * pos + 2, a);
      
        // increase the level of every parent,
        // which is level of child + 1
        level[pos] = level[2 * pos + 1] + 1;
      
        // if the parent is at odd level,
        // then do a bitwise OR
        if ((level[pos] & 1) > 0)
            tree[pos] = tree[2 * pos + 1] |
                           tree[2 * pos + 2];
      
        // if the parent is at even level,
        // then do a bitwise XOR
        else
            tree[pos] = tree[2 * pos + 1] ^
                          tree[2 * pos + 2];
    }
      
    // fucntion that updates the tree
    static void update(int low, int high,
               int pos, int index, int []a)
    {
          
        // if it is a leaf and the leaf
        // which is to be updated
        if (low == high && low == index)
        {
            tree[pos] = a[low];
            return;
        }
      
        // out of range
        if (index < low || index > high)
            return;
      
        // not a leaf then recurse
        if (low != high)
        {
            int mid = (low + high) / 2;
      
            // recursive call
            update(low, mid, 2 * pos + 1,
                                   index, a);
                                     
            update(mid + 1, high, 2 * pos + 2,
                                    index, a);
      
            // check if the parent is at odd
            // or even level and perform OR
            // or XOR according to that
            if ((level[pos] & 1) > 0)
                tree[pos] = tree[2 * pos + 1] |
                              tree[2 * pos + 2];
            else
                tree[pos] = tree[2 * pos + 1]
                            ^ tree[2 * pos + 2];
        }
    }
      
    // fucntion that assigns value to a[index]
    // and calls update function to update
    // the tree
    static void updateValue(int index, int value,
                                 int []a, int n)
    {
        a[index] = value;
        update(0, n - 1, 0, index, a);
    }
      
    // Driver Code
    static public void Main ()
    {
        int []a = { 1, 4, 5, 6 };
        int n = a.Length;;
      
        // builds the tree
        constructTree(0, n - 1, 0, a);
      
        // 1st query
        int index = 0;
        int value = 2;
        updateValue(index, value, a, n);
        Console.WriteLine(tree[0]);
      
        // 2nd query
        index = 3;
        value = 5;
        updateValue(index, value, a, n);
        Console.WriteLine(tree[0]);
    }
}
  
// This code is contributed by vt_m.


Output:

1
3

Time Complexity:

  • Tree construction: O(N)
  • Answering Query: O(log2N)


This article is attributed to GeeksforGeeks.org

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