Range Queries to count the number of even parity values with updates

Given an array arr[] of N integers, the task is to perform the following two queries:

query(L, R): Print the number of Even Parity numbers in the subarray from L to R.
update(i, x): Update the array element reference by index i to x.

Examples:

Input: arr[] = {18, 15, 8, 9, 14, 5}
Query 1: query(L = 0, R = 4)
Query 2: update(i = 3, x = 11)
Query 3: query(L = 0, R = 4)
Output:
3
2
Explanation:
Query 1: Subarray is {18, 15, 8, 9, 14}
Binary Representation of these elements –
18 => 10010, Parity = 2
15 => 1111, Parity = 4
8 => 1000, Parity = 1
9 => 1001, Parity = 2
14 => 1110, Parity = 3
Subarray[0-4] have 3 elements with even parity.
Query 2: Update arr[3] = 11
Updated array, {18, 15, 8, 11, 14, 5}
Query 3: Subarray is {18, 15, 8, 11, 14}
Binary Representation of these elements –
18 => 10010, Parity = 2
15 => 1111, Parity = 4
8 => 1000, Parity = 1
11 => 1011, Parity = 3
14 => 1110, Parity = 3
Subarray[0-4] have 2 elements with even parity.

Approach: The idea is to use segment tree to query the count of the even parity elements in a array range and update simultaneously.

We can find the parity for the current value by iterating through each bit of the binary representation of the number and counting the number of set bits. Then check if the parity is even or not. If it has even parity the set it to 1 else to 0.

Building the segment tree:

Leaf nodes of the segment tree is represented as either 0 (if it is odd parity number) or 1 (if it is even parity number).
The internal nodes of the segment tree equal to the sum of its child nodes, thus a node represent the total Even Parity numbers in the range from L to R with range [L, R] falling under this node and the sub-tree underneath it.

Handling Queries:

Query(L, R): Whenever we receive a query from start to end, we can query the segment tree for the sum of nodes in the range from start to end, which in turn represents the number of Even Parity numbers in the range start to end.
Update(i, x): To perform a update query to update the value at index i to x, we check for the following cases:
- Case 1: If previous value and new value both are Even Parity numbers
  Count of Even Parity numbers in the subarray does not change so we just update array and do not modify the segment tree
- Case 2: If previous value and new value both are not Even Parity numbers
  Count of Even Parity numbers in the subarray does not change so we just update array and do not modify the segment tree
- Case 3: If previous value is a Even Parity number but new value is not a Even Parity number
  Count of Even Parity numbers in the subarray decreases so we update array and add -1 to every range. The index i which is to be updated is a part of in the segment tree
- Case 4: If previous value is not a Even Parity number but new value is a Even Parity number
  Count of Even Parity numbers in the subarray increases so we update array and add 1 to every range. The index i which is to be updated is a part of in the segment tree

Below is the implementation of the above approach:

C++

// C++ implementation to find 
// number of Even Parity numbers 
// in a subarray and performing updates 
  
#include <bits/stdc++.h> 
using namespace std; 
  
#define MAX 1000 
  
// Function that returns true if count 
// of set bits in x is even 
bool isEvenParity(int x) 
{ 
    // parity will store the 
    // count of set bits 
    int parity = 0; 
    while (x != 0) { 
        if (x & 1) 
            parity++; 
        x = x >> 1; 
    } 
  
    if (parity % 2 == 0) 
        return true; 
    else
        return false; 
} 
  
// A utility function to get 
// the middle index 
int getMid(int s, int e) 
{ 
    return s + (e - s) / 2; 
} 
  
// Recursive function to get the number 
// of Even Parity numbers in a given range 
int queryEvenParityUtil( 
    int* segmentTree, int segmentStart, 
    int segmentEnd, int queryStart, 
    int queryEnd, int index) 
{ 
    // If segment of this node is a part 
    // of given range, then return 
    // the number of Even Parity numbers 
    // in the segment 
    if (queryStart <= segmentStart 
        && queryEnd >= segmentEnd) 
        return segmentTree[index]; 
  
    // If segment of this node 
    // is outside the given range 
    if (segmentEnd < queryStart 
        || segmentStart > queryEnd) 
        return 0; 
  
    // If a part of this segment 
    // overlaps with the given range 
    int mid = getMid(segmentStart, segmentEnd); 
    return queryEvenParityUtil( 
               segmentTree, segmentStart, mid, 
               queryStart, queryEnd, 2 * index + 1) 
           + queryEvenParityUtil( 
                 segmentTree, mid + 1, segmentEnd, 
                 queryStart, queryEnd, 2 * index + 2); 
} 
  
// Recursive function to update 
// the nodes which have the given 
// index in their range 
void updateValueUtil( 
    int* segmentTree, int segmentStart, 
    int segmentEnd, int i, int diff, int si) 
{ 
    // Base Case: 
    if (i < segmentStart || i > segmentEnd) 
        return; 
  
    // If the input index is in range 
    // of this node, then update the value 
    // of the node and its children 
    segmentTree[si] = segmentTree[si] + diff; 
    if (segmentEnd != segmentStart) { 
  
        int mid = getMid( 
            segmentStart, segmentEnd); 
        updateValueUtil( 
            segmentTree, segmentStart, 
            mid, i, diff, 2 * si + 1); 
        updateValueUtil( 
            segmentTree, mid + 1, segmentEnd, 
            i, diff, 2 * si + 2); 
    } 
} 
  
// Function to update a value in the 
// input array and segment tree 
void updateValue(int arr[], int* segmentTree, 
                 int n, int i, int new_val) 
{ 
    // Check for erroneous input index 
    if (i < 0 || i > n - 1) { 
        printf("Invalid Input"); 
        return; 
    } 
  
    int diff, oldValue; 
  
    oldValue = arr[i]; 
  
    // Update the value in array 
    arr[i] = new_val; 
  
    // Case 1: Old and new values 
    // both are Even Parity numbers 
    if (isEvenParity(oldValue) 
        && isEvenParity(new_val)) 
        return; 
  
    // Case 2: Old and new values 
    // both not Even Parity numbers 
    if (!isEvenParity(oldValue) 
        && !isEvenParity(new_val)) 
        return; 
  
    // Case 3: Old value was Even Parity, 
    // new value is non Even Parity 
    if (isEvenParity(oldValue) 
        && !isEvenParity(new_val)) { 
        diff = -1; 
    } 
  
    // Case 4: Old value was non Even Parity, 
    // new_val is Even Parity 
    if (!isEvenParity(oldValue) 
        && !isEvenParity(new_val)) { 
        diff = 1; 
    } 
  
    // Update the values of 
    // nodes in segment tree 
    updateValueUtil(segmentTree, 0, 
                    n - 1, i, diff, 0); 
} 
  
// Return number of Even Parity numbers 
void queryEvenParity(int* segmentTree, 
                     int n, int queryStart, 
                     int queryEnd) 
{ 
    int EvenParityInRange 
        = queryEvenParityUtil( 
            segmentTree, 0, n - 1, 
            queryStart, queryEnd, 0); 
  
    cout << EvenParityInRange << "\n"; 
} 
  
// Recursive function that constructs 
// Segment Tree for the given array 
int constructSTUtil(int arr[], 
                    int segmentStart, 
                    int segmentEnd, 
                    int* segmentTree, 
                    int si) 
{ 
    // If there is one element in array, 
    // check if it is Even Parity number 
    // then store 1 in the segment tree 
    // else store 0 and return 
    if (segmentStart == segmentEnd) { 
  
        // if arr[segmentStart] is 
        // Even Parity number 
        if (isEvenParity(arr[segmentStart])) 
            segmentTree[si] = 1; 
        else
            segmentTree[si] = 0; 
  
        return segmentTree[si]; 
    } 
  
    // If there are more than one elements, 
    // then recur for left and right subtrees 
    // and store the sum of the 
    // two values in this node 
    int mid = getMid(segmentStart, 
                     segmentEnd); 
    segmentTree[si] 
        = constructSTUtil( 
              arr, segmentStart, mid, 
              segmentTree, si * 2 + 1) 
          + constructSTUtil( 
                arr, mid + 1, segmentEnd, 
                segmentTree, si * 2 + 2); 
    return segmentTree[si]; 
} 
  
// Function to construct a segment 
// tree from given array 
int* constructST(int arr[], int n) 
{ 
    // Height of segment tree 
    int x = (int)(ceil(log2(n))); 
  
    // Maximum size of segment tree 
    int max_size = 2 * (int)pow(2, x) - 1; 
  
    int* segmentTree = new int[max_size]; 
  
    // Fill the allocated memory st 
    constructSTUtil(arr, 0, n - 1, 
                    segmentTree, 0); 
  
    // Return the constructed segment tree 
    return segmentTree; 
} 
  
// Driver Code 
int main() 
{ 
  
    int arr[] = { 18, 15, 8, 9, 14, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    // Build segment tree from given array 
    int* segmentTree = constructST(arr, n); 
  
    // Query 1: Query(start = 0, end = 4) 
    int start = 0; 
    int end = 4; 
    queryEvenParity(segmentTree, n, start, end); 
  
    // Query 2: Update(i = 3, x = 11), 
    // i.e Update a[i] to x 
    int i = 3; 
    int x = 11; 
    updateValue(arr, segmentTree, n, i, x); 
  
    // Query 3: Query(start = 0, end = 4) 
    start = 0; 
    end = 4; 
    queryEvenParity( 
        segmentTree, n, start, end); 
  
    return 0; 
} 

Output:

3
2

My Personal Notes

C++

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