Queries for count of even digit sum elements in given range using MO’s Algorithm

Given an array arr[] of N elements, the task is to answer Q queries each having two integers L and R. For each query, the task is to find the number of elements in the subarray arr[L…R] whose digit sum is even.

Examples:

Input:arr[] = {7, 3, 19, 13, 5, 4}
query = { {1, 5}, {0, 1} }

Output: 3
Explanation:
Query 1: Elements 19, 13 and 4 have even digit sum in the subarray: {3, 9, 13, 5, 4}.
Query 2: No elements have even sum in the subarray: {7, 3}

Input:arr[] = {0, 1, 2, 3, 4, 5, 6, 7}
query = { 3, 5 }
Output:1



We have already discussed this approach: Queries for the count of even digit sum elements in the given range using Segment Tree

Approach: (Using MO’s Algorithm)
The idea is to pre-process all queries so that result of one query can be used in the next query.

  1. Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, followed by queries from √n to 2×√n – 1, and so on. All queries within a block are sorted in increasing order of R values.
  2. Process all queries one by one and increase the count of even digit sum elements and store the result in the structure.
    • Let count_even store the count of even digits sum elements in the previous query.
    • Remove extra elements of previous query and add new elements for the current query. For example, if previous query was [0, 8] and the current query is [3, 9], then remove the elements arr[0], arr[1] and arr[2] and add arr[9].
  3. In order to display the results, sort the queries in the order they were provided.

Helper functions:

Adding elements()

  • If the current element has even digit sum then increase the count of count_even.
  • Removing elements()

  • If the current element has even digit sum then decrease the count of count_even.
  • Below code is the implementation of the above approach:

    C++

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    // C++ program to count of even
    // digit sum elements in the given
    // range using MO's algorithm
      
    #include <bits/stdc++.h>
    using namespace std;
      
    #define MAX 100000
      
    // Variable to represent block size.
    // This is made global so compare()
    // of sort can use it.
    int block;
      
    // Structure to represent a query range
    struct Query {
      
        // Starting index
        int L;
      
        // Ending index
        int R;
      
        // Index of query
        int index;
      
        // Count of even
        // even digit sum
        int even;
    };
      
    // To store the count of
    // even digit sum
    int count_even;
      
    // Function used to sort all queries so that
    // all queries of the same block are arranged
    // together and within a block, queries are
    // sorted in increasing order of R values.
    bool compare(Query x, Query y)
    {
        // Different blocks, sort by block.
        if (x.L / block != y.L / block)
            return x.L / block < y.L / block;
      
        // Same block, sort by R value
        return x.R < y.R;
    }
      
    // Function used to sort all queries
    // in order of their index value so that
    // results of queries can be printed
    // in same order as of input
    bool compare1(Query x, Query y)
    {
        return x.index < y.index;
    }
      
    // Function to find the digit sum
    // for a number
    int digitSum(int num)
    {
        int sum = 0;
        while (num) {
            sum += (num % 10);
            num /= 10;
        }
      
        return sum;
    }
      
    // Function to Add elements
    // of current range
    void add(int currL, int a[])
    {
        // If digit sum of a[currL]
        // is even then increment
        if (digitSum(a[currL]) % 2 == 0)
            count_even++;
    }
      
    // Function to remove elements
    // of previous range
    void remove(int currR, int a[])
    {
      
        // If digit sum of a[currL]
        // is even then decrement
        if (digitSum(a[currR]) % 2 == 0)
            count_even--;
    }
      
    // Function to generate
    // the result of queries
    void queryResults(int a[], int n,
                      Query q[], int m)
    {
      
        // Initialize number of
        // even digit sum to 0
        count_even = 0;
      
        // Find block size
        block = (int)sqrt(n);
      
        // Sort all queries so that queries of
        // same blocks are arranged together.
        sort(q, q + m, compare);
      
        // Initialize current L, current R and
        // current result
        int currL = 0, currR = 0;
      
        for (int i = 0; i < m; i++) {
            // L and R values of current range
            int L = q[i].L, R = q[i].R;
      
            // Add Elements of current range
            while (currR <= R) {
                add(currR, a);
                currR++;
            }
            while (currL > L) {
                add(currL - 1, a);
                currL--;
            }
      
            // Remove element of previous range
            while (currR > R + 1)
      
            {
                remove(currR - 1, a);
                currR--;
            }
            while (currL < L) {
                remove(currL, a);
                currL++;
            }
      
            q[i].even = count_even;
        }
    }
    // Function to display the results of
    // queries in their initial order
    void printResults(Query q[], int m)
    {
        sort(q, q + m, compare1);
        for (int i = 0; i < m; i++) {
            cout << q[i].even << endl;
        }
    }
      
    // Driver Code
    int main()
    {
      
        int arr[] = { 5, 2, 3, 1, 4, 8, 10, 12 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        Query q[] = { { 1, 3, 0, 0 },
                      { 0, 4, 1, 0 },
                      { 4, 7, 2, 0 } };
      
        int m = sizeof(q) / sizeof(q[0]);
      
        queryResults(arr, n, q, m);
      
        printResults(q, m);
      
        return 0;
    }

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    Output:

    1
    2
    2
    

    Time Complexity: O(Q x √N)

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