Queries for the count of even digit sum elements in the given range using Segment Tree.
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: 3
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: 1
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.

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.

Construction of Segment Tree from given array:
 We start with a segment arr[0 . . . n1]. 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++ 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; } 
3
Time complexity: O(Q * log(N))
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