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Queries for the count of even digit sum elements in the given range using Segment Tree.

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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 } 
Output:
Explanation: 
Elements 19, 13 and 4 have even digit sum 
in the subarray {3, 9, 13, 5, 4}.
Input: arr[] = {0, 1, 2, 3, 4, 5, 6, 7} 
query = { 3, 5 } 
Output:
Explanation: 
Only 4 has even digit sum 
in the subarray {3, 4, 5}. 

Naive approach: 

  • Find the answer for each query by simply traversing the array from index L till R and keep adding 1 to the count whenever the array element has even digit sum. Time Complexity of this approach will be O(n * q)
     

Efficient approach: 
The idea is to build a Segment Tree

  1. Representation of Segment trees: 
    • Leaf Nodes are the elements of the input array..
    • Each internal node contains the number of leaves which has even digit sum of all leaves under it. 
       
  2. Construction of Segment Tree from given array: 
    • We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1) and then call the same procedure on both halves and for each such segment, we store the number of elements which has even digit sum of all nodes under it. 
       

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// 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;
}
 
// Procedure to build the segment tree
void buildTree(vector<int>& tree, int* arr,
               int index, int s, int e)
{
 
    // Reached the leaf node
    // of the segment tree
    if (s == e) {
        if (digitSum(arr[s]) & 1)
            tree[index] = 0;
        else
            tree[index] = 1;
        return;
    }
 
    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index,
              s, mid);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e);
 
    tree[index] = tree[2 * index]
                + tree[2 * index + 1];
}
 
// Query procedure to get the answer
// for each query l and r are
// query range
int query(vector<int> tree, int index,
          int s, int e, int l, int r)
{
 
    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;
 
    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r) {
        return tree[index];
    }
 
    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index, s,
                  mid, l, r)
            + query(tree, 2 * index + 1,
                    mid + 1, e, l, r));
}
 
// Driver code
int main()
{
    int arr[] = { 7, 3, 19, 13, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> tree(4 * n + 1);
 
    int L = 1, R = 5;
 
    buildTree(tree, arr, 1, 0, n - 1);
 
    cout << query(tree, 1, 0, n - 1, L, R)
         << endl;
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
class GFG{
 
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
    int sum = 0;
    while (num > 0)
    {
        sum += (num % 10);
        num /= 10;
    }
    return sum;
}
 
// Procedure to build the segment tree
static void buildTree(int []tree, int []arr,
                int index, int s, int e)
{
 
    // Reached the leaf node
    // of the segment tree
    if (s == e)
    {
        if (digitSum(arr[s]) % 2 == 1)
            tree[index] = 0;
        else
            tree[index] = 1;
        return;
    }
 
    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index,
              s, mid);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e);
 
    tree[index] = tree[2 * index] +
                    tree[2 * index + 1];
}
 
// Query procedure to get the answer
// for each query l and r are
// query range
static int query(int []tree, int index,
                 int s, int e,
                 int l, int r)
{
 
    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;
 
    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r)
    {
        return tree[index];
    }
 
    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index, s,
                  mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r));
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 7, 3, 19, 13, 5, 4 };
    int n = arr.length;
    int []tree = new int[4 * n + 1];
 
    int L = 1, R = 5;
 
    buildTree(tree, arr, 1, 0, n - 1);
 
    System.out.print(query(tree, 1, 0,
                           n - 1, L, R) + "\n");
}
}
 
// This code is contributed by gauravrajput1


Python3




# Python3 implementation of the above approach
 
# Function to find the digit sum
# for a number
def digitSum(num):
     
    sum = 0;
    while (num):
        sum += (num % 10)
        num //= 10
     
    return sum
 
# Procedure to build the segment tree
def buildTree(tree, arr, index, s, e):
 
    # Reached the leaf node
    # of the segment tree
    if (s == e):
        if (digitSum(arr[s]) & 1):
            tree[index] = 0
        else:
            tree[index] = 1
        return
 
    # Recursively call the buildTree
    # on both the nodes of the tree
    mid = (s + e) // 2
    buildTree(tree, arr, 2 * index,
              s, mid)
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e)
 
    tree[index] = (tree[2 * index] +
                   tree[2 * index + 1])
 
# Query procedure to get the answer
# for each query l and r are
# query range
def query(tree, index, s, e, l, r):
 
    # Out of bound or no overlap
    if (r < s or l > e):
        return 0
 
    # Complete overlap
    # Query range completely lies in
    # the segment tree node range
    if (s >= l and e <= r):
        return tree[index]
 
    # Partially overlap
    # Query range partially lies in
    # the segment tree node range
    mid = (s + e) // 2
    return (query(tree, 2 * index,
                  s, mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r))
 
# Driver code
arr = [ 7, 3, 19, 13, 5, 4 ]
n = len(arr)
 
tree = [0] * (4 * n + 1)
 
L = 1
R = 5
 
buildTree(tree, arr, 1, 0, n - 1);
 
print(query(tree, 1, 0, n - 1, L, R))
 
# This code is contributed by Apurvaraj


C#




// C# implementation of the approach
using System;
class GFG{
 
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
    int sum = 0;
    while (num > 0)
    {
        sum += (num % 10);
        num /= 10;
    }
    return sum;
}
 
// Procedure to build the segment tree
static void buildTree(int []tree, int []arr,
                      int index, int s, int e)
{
 
    // Reached the leaf node
    // of the segment tree
    if (s == e)
    {
        if (digitSum(arr[s]) % 2 == 1)
            tree[index] = 0;
        else
            tree[index] = 1;
        return;
    }
 
    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index,
              s, mid);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e);
 
    tree[index] = tree[2 * index] +
                  tree[2 * index + 1];
}
 
// Query procedure to get the answer
// for each query l and r are
// query range
static int query(int []tree, int index,
                 int s, int e,
                 int l, int r)
{
 
    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;
 
    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r)
    {
        return tree[index];
    }
 
    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index, s,
                  mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r));
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 7, 3, 19, 13, 5, 4 };
    int n = arr.Length;
    int []tree = new int[4 * n + 1];
 
    int L = 1, R = 5;
 
    buildTree(tree, arr, 1, 0, n - 1);
 
    Console.Write(query(tree, 1, 0,
                        n - 1, L, R) + "\n");
}
}
 
// This code is contributed by gauravrajput1


Javascript




<script>
      // JavaScript implementation of the approach
       
      // Function to find the digit sum
      // for a number
      function digitSum(num) {
        var sum = 0;
        while (num > 0) {
          sum += parseInt(num % 10);
          num = parseInt(num / 10);
        }
        return sum;
      }
 
      // Procedure to build the segment tree
      function buildTree(tree, arr, index, s, e) {
        // Reached the leaf node
        // of the segment tree
        if (s === e) {
          if (digitSum(arr[s]) % 2 === 1) tree[index] = 0;
          else tree[index] = 1;
          return;
        }
 
        // Recursively call the buildTree
        // on both the nodes of the tree
        var mid = parseInt((s + e) / 2);
        buildTree(tree, arr, 2 * index, s, mid);
        buildTree(tree, arr, 2 * index + 1, mid + 1, e);
 
        tree[index] = tree[2 * index] + tree[2 * index + 1];
      }
 
      // Query procedure to get the answer
      // for each query l and r are
      // query range
      function query(tree, index, s, e, l, r) {
        // Out of bound or no overlap
        if (r < s || l > e) return 0;
 
        // Complete overlap
        // Query range completely lies in
        // the segment tree node range
        if (s >= l && e <= r) {
          return tree[index];
        }
 
        // Partially overlap
        // Query range partially lies in
        // the segment tree node range
        var mid = (s + e) / 2;
        return (
          query(tree, 2 * index, s, mid, l, r) +
          query(tree, 2 * index + 1, mid + 1, e, l, r)
        );
      }
 
      // Driver code
      var arr = [7, 3, 19, 13, 5, 4];
      var n = arr.length;
      var tree = new Array(4 * n + 1).fill(0);
 
      var L = 1,
        R = 5;
 
      buildTree(tree, arr, 1, 0, n - 1);
 
      document.write(query(tree, 1, 0, n - 1, L, R) + "<br>");
       
</script>


Output

3


Time complexity: O(Q * log(N))
 

using  Brute Force:

Approach:

We can solve this problem by iterating through each element in the given subarray and checking if its digit sum is even. We can count the number of such elements and return the count as the answer.

  • The get_digit_sum function returns the digit sum of a given number. 
  • The count_even_digit_sum function takes an array arr and the left and right indices left and right of a subarray and counts the number of elements in the subarray that have an even digit sum. 
  • The count_even_digit_sum_queries function takes an array arr and a list of queries, where each query is a tuple of left and right indices, and returns a list of the number of elements with an even digit sum in each subarray specified by the queries.

C++




// Added by: Nikunj Sonigara
 
#include <bits/stdc++.h>
using namespace std; // Using the std namespace
 
int get_digit_sum(int num) {
    // Returns the digit sum of the given number
    int digit_sum = 0;
    while (num > 0) {
        digit_sum += num % 10;
        num /= 10;
    }
    return digit_sum;
}
 
int count_even_digit_sum(const vector<int>& arr, int left, int right) {
    // Counts the number of elements with even digit sum in the subarray [left, right]
    int count = 0;
    for (int i = left; i <= right; i++) {
        if (get_digit_sum(arr[i]) % 2 == 0) {
            count++;
        }
    }
    return count;
}
 
vector<int> count_even_digit_sum_queries(const vector<int>& arr, const vector<pair<int, int>>& queries) {
    // Counts the number of elements with even digit sum for each query
    vector<int> results;
    for (const auto& query : queries) {
        int left = query.first;
        int right = query.second;
        results.push_back(count_even_digit_sum(arr, left, right));
    }
    return results;
}
 
int main() {
    vector<int> arr = {7, 3, 19, 13, 5, 4};
    vector<pair<int, int>> queries = {{1, 5}};
    vector<int> results = count_even_digit_sum_queries(arr, queries);
 
    cout << "[";
    for (int i = 0; i < results.size(); i++) {
        cout << results[i];
        if (i < results.size() - 1) {
            cout << ", ";
        }
    }
    cout << "]" << endl;
 
    arr = {0, 1, 2, 3, 4, 5, 6, 7};
    queries = {{3, 5}};
    results = count_even_digit_sum_queries(arr, queries);
 
    cout << "[";
    for (int i = 0; i < results.size(); i++) {
        cout << results[i];
        if (i < results.size() - 1) {
            cout << ", ";
        }
    }
    cout << "]" << endl;
 
    return 0;
}


Java




// Added by: Nikunj Sonigara
 
import java.util.*;
 
public class Main {
    static int getDigitSum(int num) {
        // Returns the digit sum of the given number
        int digitSum = 0;
        while (num > 0) {
            digitSum += num % 10;
            num /= 10;
        }
        return digitSum;
    }
 
    static int countEvenDigitSum(List<Integer> arr, int left, int right) {
        // Counts the number of elements with even digit sum in the subarray [left, right]
        int count = 0;
        for (int i = left; i <= right; i++) {
            if (getDigitSum(arr.get(i)) % 2 == 0) {
                count++;
            }
        }
        return count;
    }
 
    static List<Integer> countEvenDigitSumQueries(List<Integer> arr, List<Pair<Integer, Integer>> queries) {
        // Counts the number of elements with even digit sum for each query
        List<Integer> results = new ArrayList<>();
        for (Pair<Integer, Integer> query : queries) {
            int left = query.getKey();
            int right = query.getValue();
            results.add(countEvenDigitSum(arr, left, right));
        }
        return results;
    }
 
    public static void main(String[] args) {
        List<Integer> arr = List.of(7, 3, 19, 13, 5, 4);
        List<Pair<Integer, Integer>> queries = List.of(new Pair<>(1, 5));
 
        List<Integer> results = countEvenDigitSumQueries(arr, queries);
 
        System.out.print("[");
        for (int i = 0; i < results.size(); i++) {
            System.out.print(results.get(i));
            if (i < results.size() - 1) {
                System.out.print(", ");
            }
        }
        System.out.println("]");
 
        arr = List.of(0, 1, 2, 3, 4, 5, 6, 7);
        queries = List.of(new Pair<>(3, 5));
 
        results = countEvenDigitSumQueries(arr, queries);
 
        System.out.print("[");
        for (int i = 0; i < results.size(); i++) {
            System.out.print(results.get(i));
            if (i < results.size() - 1) {
                System.out.print(", ");
            }
        }
        System.out.println("]");
    }
}
 
class Pair<K, V> {
    private final K key;
    private final V value;
 
    public Pair(K key, V value) {
        this.key = key;
        this.value = value;
    }
 
    public K getKey() {
        return key;
    }
 
    public V getValue() {
        return value;
    }
}


Python3




def get_digit_sum(num):
    # Returns the digit sum of the given number
    digit_sum = 0
    while num > 0:
        digit_sum += num % 10
        num //= 10
    return digit_sum
 
def count_even_digit_sum(arr, left, right):
    # Counts the number of elements with even digit sum in the subarray [left, right]
    count = 0
    for i in range(left, right+1):
        if get_digit_sum(arr[i]) % 2 == 0:
            count += 1
    return count
 
def count_even_digit_sum_queries(arr, queries):
    # Counts the number of elements with even digit sum for each query
    results = []
    for query in queries:
        left, right = query
        results.append(count_even_digit_sum(arr, left, right))
    return results
 
arr = [7, 3, 19, 13, 5, 4]
queries = [(1, 5)]
print(count_even_digit_sum_queries(arr, queries)) # Output: [3]
 
arr = [0, 1, 2, 3, 4, 5, 6, 7]
queries = [(3, 5)]
print(count_even_digit_sum_queries(arr, queries)) # Output: [1]


C#




using System;
using System.Collections.Generic;
 
class Program
{
    static int GetDigitSum(int num)
    {
        // Returns the digit sum of the given number
        int digitSum = 0;
        while (num > 0)
        {
            digitSum += num % 10;
            num /= 10;
        }
        return digitSum;
    }
 
    static int CountEvenDigitSum(List<int> arr, int left, int right)
    {
        // Counts the number of elements with an even digit sum in the subarray [left, right]
        int count = 0;
        for (int i = left; i <= right; i++)
        {
            if (GetDigitSum(arr[i]) % 2 == 0)
            {
                count++;
            }
        }
        return count;
    }
 
    static List<int> CountEvenDigitSumQueries(List<int> arr, List<Tuple<int, int>> queries)
    {
        // Counts the number of elements with an even digit sum for each query
        List<int> results = new List<int>();
        foreach (var query in queries)
        {
            int left = query.Item1;
            int right = query.Item2;
            results.Add(CountEvenDigitSum(arr, left, right));
        }
        return results;
    }
 
    static void Main()
    {
        List<int> arr = new List<int> { 7, 3, 19, 13, 5, 4 };
        List<Tuple<int, int>> queries = new List<Tuple<int, int>> { Tuple.Create(1, 5) };
        List<int> results = CountEvenDigitSumQueries(arr, queries);
 
        Console.Write("[");
        for (int i = 0; i < results.Count; i++)
        {
            Console.Write(results[i]);
            if (i < results.Count - 1)
            {
                Console.Write(", ");
            }
        }
        Console.WriteLine("]");
 
        arr = new List<int> { 0, 1, 2, 3, 4, 5, 6, 7 };
        queries = new List<Tuple<int, int>> { Tuple.Create(3, 5) };
        results = CountEvenDigitSumQueries(arr, queries);
 
        Console.Write("[");
        for (int i = 0; i < results.Count; i++)
        {
            Console.Write(results[i]);
            if (i < results.Count - 1)
            {
                Console.Write(", ");
            }
        }
        Console.WriteLine("]");
    }
}


Javascript




// Added by: Nikunj Sonigara
 
function getDigitSum(num) {
    // Returns the digit sum of the given number
    let digitSum = 0;
    while (num > 0) {
        digitSum += num % 10;
        num = Math.floor(num / 10);
    }
    return digitSum;
}
 
function countEvenDigitSum(arr, left, right) {
    // Counts the number of elements with even digit sum in the subarray [left, right]
    let count = 0;
    for (let i = left; i <= right; i++) {
        if (getDigitSum(arr[i]) % 2 === 0) {
            count++;
        }
    }
    return count;
}
 
function countEvenDigitSumQueries(arr, queries) {
    // Counts the number of elements with even digit sum for each query
    const results = [];
    for (const query of queries) {
        const [left, right] = query;
        results.push(countEvenDigitSum(arr, left, right));
    }
    return results;
}
 
const arr1 = [7, 3, 19, 13, 5, 4];
const queries1 = [[1, 5]];
 
const results1 = countEvenDigitSumQueries(arr1, queries1);
 
console.log(`[${results1}]`);
 
const arr2 = [0, 1, 2, 3, 4, 5, 6, 7];
const queries2 = [[3, 5]];
 
const results2 = countEvenDigitSumQueries(arr2, queries2);
 
console.log(`[${results2}]`);


Output

[3]
[1]


 The time complexity of this approach would be O(n*q), where n is the size of the input array and q is the number of queries. 
The Auxiliary space would be O(1).


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Last Updated : 28 Oct, 2023
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