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 . . . 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.
- 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 numberint digitSum(int num){ int sum = 0; while (num) { sum += (num % 10); num /= 10; } return sum;}// Procedure to build the segment treevoid 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 rangeint 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 codeint 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 approachimport java.util.*;class GFG{// Function to find the digit sum// for a numberstatic int digitSum(int num){ int sum = 0; while (num > 0) { sum += (num % 10); num /= 10; } return sum;}// Procedure to build the segment treestatic 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 rangestatic 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 codepublic 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 numberdef digitSum(num): sum = 0; while (num): sum += (num % 10) num //= 10 return sum# Procedure to build the segment treedef 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 rangedef 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 codearr = [ 7, 3, 19, 13, 5, 4 ]n = len(arr)tree = [0] * (4 * n + 1)L = 1R = 5buildTree(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 approachusing System;class GFG{// Function to find the digit sum// for a numberstatic int digitSum(int num){ int sum = 0; while (num > 0) { sum += (num % 10); num /= 10; } return sum;}// Procedure to build the segment treestatic 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 rangestatic 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 codepublic 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 namespaceint 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 Sonigaraimport 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_sumdef 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 countdef 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 resultsarr = [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 Sonigarafunction 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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