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

• Difficulty Level : Basic
• Last Updated : 24 May, 2021

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}.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 ``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);``    ``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

 ``
Output:
`3`

Time complexity: O(Q * log(N))

My Personal Notes arrow_drop_up