Sub-string Divisibility by 11 Queries

Given a large number, n (having number digits up to 10^6) and various queries of the below form :
Query(l, r) : find if the sub-string between the
indices l and r (Both inclusive)
are divisible by 11.
Examples:
Input: n = 122164154695
Queries: l = 0 r = 3, l = 1 r = 2, l = 5 r = 9,
l = 0 r = 11
Output:
True
False
False
True
Explanation:
In the first query, 1221 is divisible by 11
In the second query, 22 is divisible by 11 and so on.
We know that any number is divisible by 11 if the difference between the sum of odd indexed digits and the sum of even indexed digits is divisible by 11, i.e.,
Sum(digits at odd places) – Sum(digits at even places) should be divisible by 11.
Hence, the idea is to pre-process an auxiliary array that would store the sum of digits at odd and even places.
To evaluate a query we can use the auxiliary array to answer it in O(1).
C++
// C++ program to check divisibility by 11 in// substrings of a number string#include <iostream>using namespace std;const int MAX = 1000005;// To store sums of even and odd digitsstruct OddEvenSums{ // Sum of even placed digits int e_sum; // Sum of odd placed digits int o_sum;};// Auxiliary arrayOddEvenSums sum[MAX];// Utility function to evaluate a character's// integer valueint toInt(char x){ return int(x) - 48;}// This function receives the string representation// of the number and precomputes the sum arrayvoid preCompute(string x){ // Initialize everb sum[0].e_sum = sum[0].o_sum = 0; // Add the respective digits depending on whether // they're even indexed or odd indexed for (int i=0; i<x.length(); i++) { if (i%2==0) { sum[i+1].e_sum = sum[i].e_sum+toInt(x[i]); sum[i+1].o_sum = sum[i].o_sum; } else { sum[i+1].o_sum = sum[i].o_sum+toInt(x[i]); sum[i+1].e_sum = sum[i].e_sum; } }}// This function receives l and r representing// the indices and prints the required outputbool query(int l,int r){ int diff = (sum[r+1].e_sum - sum[r+1].o_sum) - (sum[l].e_sum - sum[l].o_sum); return (diff%11==0);}//driver function to check the programint main(){ string s = "122164154695"; preCompute(s); cout << query(0, 3) << endl; cout << query(1, 2) << endl; cout << query(5, 9) << endl; cout << query(0, 11) << endl; return 0;} |
Java
// Java program to check divisibility by 11 in// subStrings of a number Stringclass GFG{ static int MAX = 1000005; // To store sums of even and odd digitsstatic class OddEvenSums{ // Sum of even placed digits int e_sum; // Sum of odd placed digits int o_sum;}; // Auxiliary arraystatic OddEvenSums []sum = new OddEvenSums[MAX]; // Utility function to evaluate a character's// integer valuestatic int toInt(char x){ return x - 48;} // This function receives the String representation// of the number and precomputes the sum arraystatic void preCompute(String x){ // Initialize everb sum[0].e_sum = sum[0].o_sum = 0; // Add the respective digits depending on whether // they're even indexed or odd indexed for (int i = 0; i < x.length(); i++) { if (i % 2 == 0) { sum[i + 1].e_sum = sum[i].e_sum + toInt(x.charAt(i)); sum[i + 1].o_sum = sum[i].o_sum; } else { sum[i + 1].o_sum = sum[i].o_sum + toInt(x.charAt(i)); sum[i + 1].e_sum = sum[i].e_sum; } }} // This function receives l and r representing// the indices and prints the required outputstatic boolean query(int l, int r){ int diff = (sum[r + 1].e_sum - sum[r + 1].o_sum) - (sum[l].e_sum - sum[l].o_sum); return (diff % 11 == 0);} //driver function to check the programpublic static void main(String[] args){ for (int i = 0; i < MAX; i++) { sum[i] = new OddEvenSums(); } String s = "122164154695"; preCompute(s); System.out.println(query(0, 3) ? 1 : 0); System.out.println(query(1, 2) ? 1 : 0); System.out.println(query(5, 9) ? 1 : 0); System.out.println(query(0, 11) ? 1 : 0); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to check divisibility by # 11 in subStrings of a number StringMAX = 1000005# To store sums of even and odd digitsclass OddEvenSums: def __init__(self, e_sum, o_sum): # Sum of even placed digits self.e_sum = e_sum # Sum of odd placed digits self.o_sum = o_sumsum = [OddEvenSums(0, 0) for i in range(MAX)]# This function receives the String# representation of the number and# precomputes the sum arraydef preCompute(x): # Initialize everb sum[0].e_sum = sum[0].o_sum = 0 # Add the respective digits # depending on whether # they're even indexed or # odd indexed for i in range(len(x)): if (i % 2 == 0): sum[i + 1].e_sum = (sum[i].e_sum + int(x[i])) sum[i + 1].o_sum = sum[i].o_sum else: sum[i + 1].o_sum = (sum[i].o_sum + int(x[i])) sum[i + 1].e_sum = sum[i].e_sum # This function receives l and r representing# the indices and prints the required outputdef query(l, r): diff = ((sum[r + 1].e_sum - sum[r + 1].o_sum) - (sum[l].e_sum - sum[l].o_sum)) if (diff % 11 == 0): return True else: return False# Driver codeif __name__=="__main__": s = "122164154695" preCompute(s) print(1 if query(0, 3) else 0) print(1 if query(1, 2) else 0) print(1 if query(5, 9) else 0) print(1 if query(0, 11) else 0)# This code is contributed by rutvik_56 |
C#
// C# program to check // divisibility by 11 in// subStrings of a number Stringusing System;class GFG{ static int MAX = 1000005; // To store sums of even // and odd digits public class OddEvenSums{ // Sum of even placed digits public int e_sum; // Sum of odd placed digits public int o_sum;}; // Auxiliary arraystatic OddEvenSums []sum = new OddEvenSums[MAX]; // Utility function to // evaluate a character's// integer valuestatic int toInt(char x){ return x - 48;} // This function receives the // String representation of the // number and precomputes the sum arraystatic void preCompute(String x){ // Initialize everb sum[0].e_sum = sum[0].o_sum = 0; // Add the respective digits // depending on whether they're // even indexed or odd indexed for (int i = 0; i < x.Length; i++) { if (i % 2 == 0) { sum[i + 1].e_sum = sum[i].e_sum + toInt(x[i]); sum[i + 1].o_sum = sum[i].o_sum; } else { sum[i + 1].o_sum = sum[i].o_sum + toInt(x[i]); sum[i + 1].e_sum = sum[i].e_sum; } }} // This function receives l and r // representing the indices and // prints the required outputstatic bool query(int l, int r){ int diff = (sum[r + 1].e_sum - sum[r + 1].o_sum) - (sum[l].e_sum - sum[l].o_sum); return (diff % 11 == 0);} // Driver function to check the programpublic static void Main(String[] args){ for (int i = 0; i < MAX; i++) { sum[i] = new OddEvenSums(); } String s = "122164154695"; preCompute(s); Console.WriteLine(query(0, 3) ? 1 : 0); Console.WriteLine(query(1, 2) ? 1 : 0); Console.WriteLine(query(5, 9) ? 1 : 0); Console.WriteLine(query(0, 11) ? 1 : 0);}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program to check divisibility by 11 in// subStrings of a number Stringlet MAX = 1000005;// To store sums of even and odd digitsclass OddEvenSums{ constructor() { this.e_sum = 0; this.o_sum = 0; }}// Auxiliary arraylet sum = new Array(MAX);// Utility function to evaluate a character's// integer valuefunction toInt(x){ return x.charCodeAt(0) - 48;}// This function receives the String representation// of the number and precomputes the sum arrayfunction preCompute(x){ // Initialize everb sum[0].e_sum = sum[0].o_sum = 0; // Add the respective digits depending on whether // they're even indexed or odd indexed for (let i = 0; i < x.length; i++) { if (i % 2 == 0) { sum[i + 1].e_sum = sum[i].e_sum + parseInt(x[i]); sum[i + 1].o_sum = sum[i].o_sum; } else { sum[i + 1].o_sum = sum[i].o_sum + parseInt(x[i]); sum[i + 1].e_sum = sum[i].e_sum; } }}// This function receives l and r representing// the indices and prints the required outputfunction query(l,r){ let diff = (sum[r + 1].e_sum - sum[r + 1].o_sum) - (sum[l].e_sum - sum[l].o_sum); return (diff % 11 == 0);}// driver function to check the programfor (let i = 0; i < MAX; i++) { sum[i] = new OddEvenSums();}let s = "122164154695";preCompute(s);document.write((query(0, 3) ? 1 : 0)+"<br>");document.write((query(1, 2) ? 1 : 0)+"<br>");document.write((query(5, 9) ? 1 : 0)+"<br>");document.write((query(0, 11) ? 1 :0)+"<br>");// This code is contributed by unknown2108</script> |
Output:
1 1 0 1
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