Given Q queries in the form of a 2D array arr[][] in which every row consists of two numbers L and R which denotes a range [L, R], the task is to find the sum of all Armstrong Numbers lying in the given range [L, R].
Examples:
Input: Q = 2, arr = [ [1, 13], [121, 211] ]
Output:
45
153
Explanation:
From 1 to 13, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are the Armstrong number. Therefore the sum is 45.
From 121 to 211 there is only one Armstrong number i.e., 153. So the sum is 153.
Input: Q = 4, arr = [ [ 10, 10 ], [ 258, 785 ], [45, 245], [ 1, 1000]]
Output:
0
1148
153
1346
Approach:
The idea is to use a Prefix Sum Array. The sum of all Armstrong Number till that particular index is precomputed and stored in an array pref[] so that every query can be answered in O(1) time.
- Initialise the prefix array pref[].
- Iterate from 1 to N and check if the number is Armstrong or not:
- If the number is Armstrong then, the current index of pref[] will store the count of Armstrong Numbers found so far calculated by (1 + the number at previous index of pref[]).
- Else the current index of pref[] is same as the value at previous index of pref[].
- For Q queries the sum of all Armstrong Numbers for range [L, R] can be calculated as follows:
sum = pref[R] - pref[L - 1]
Below is the implementation of the above approach
// C++ program to find the sum // of all armstrong numbers // in the given range #include <bits/stdc++.h> using namespace std;
// pref[] array to precompute // the sum of all armstrong // number int pref[100001] = {0};
// Function that return number // num if num is armstrong // else return 0 static int checkArmstrong( int x) {
int n = to_string(x).size();
int sum1 = 0;
int temp = x;
while (temp > 0) {
int digit = temp % 10;
sum1 += pow (digit, n);
temp /= 10;
}
if (sum1 == x)
return x;
return 0;
} // Function to precompute the // sum of all armstrong numbers // upto 100000 void preCompute() {
for ( int i = 1; i < 100001; i++) {
// checkarmstrong ()
// return the number i
// if i is armstrong
// else return 0
pref[i] = pref[i - 1] + checkArmstrong(i);
}
} // Function to print the sum // for each query void printSum( int L, int R) {
cout<<(pref[R] - pref[L - 1])<<endl;
} // Function to print sum of all // armstrong numbers between // [L, R] void printSumarmstrong( int arr[2][2], int Q) {
// Function that pre computes
// the sum of all armstrong
// numbers
preCompute();
// Iterate over all Queries
// to print the sum
for ( int i = 0; i < Q; i++) {
printSum(arr[i][0], arr[i][1]);
}
} int main()
{ // Queries
int Q = 2;
int arr[2][2] = { { 1, 13 }, { 121, 211 } };
// Function that print the
// the sum of all armstrong
// number in Range [L, R]
printSumarmstrong(arr, Q);
return 0;
} // This code is contributed by 29AjayKumar |
// Java program to find the sum // of all armstrong numbers // in the given range class GFG {
// pref[] array to precompute
// the sum of all armstrong
// number
static int [] pref = new int [ 100001 ];
// Function that return number
// num if num is armstrong
// else return 0
static int checkArmstrong( int x) {
int n = String.valueOf(x).length();
int sum1 = 0 ;
int temp = x;
while (temp > 0 ) {
int digit = temp % 10 ;
sum1 += Math.pow(digit, n);
temp /= 10 ;
}
if (sum1 == x)
return x;
return 0 ;
}
// Function to precompute the
// sum of all armstrong numbers
// upto 100000
static void preCompute() {
for ( int i = 1 ; i < 100001 ; i++) {
// checkarmstrong ()
// return the number i
// if i is armstrong
// else return 0
pref[i] = pref[i - 1 ] + checkArmstrong(i);
}
}
// Function to print the sum
// for each query
static void printSum( int L, int R) {
System.out.println(pref[R] - pref[L - 1 ]);
}
// Function to print sum of all
// armstrong numbers between
// [L, R]
static void printSumarmstrong( int [][] arr, int Q) {
// Function that pre computes
// the sum of all armstrong
// numbers
preCompute();
// Iterate over all Queries
// to print the sum
for ( int i = 0 ; i < Q; i++) {
printSum(arr[i][ 0 ], arr[i][ 1 ]);
}
}
// Driver code
public static void main(String[] args) {
// Queries
int Q = 2 ;
int [][] arr = { { 1 , 13 }, { 121 , 211 } };
// Function that print the
// the sum of all armstrong
// number in Range [L, R]
printSumarmstrong(arr, Q);
}
} // This code is contributed by 29AjayKumar |
# Python3 program to find the sum # of all armstrong numbers # in the given range # pref[] array to precompute # the sum of all armstrong # number pref = [ 0 ] * 100001
# Function that return number # num if num is armstrong # else return 0 def checkArmstrong(x):
n = len ( str (x))
sum1 = 0 temp = x
while temp > 0 :
digit = temp % 10
sum1 + = digit * * n
temp / / = 10
if sum1 = = x:
return x
return 0
# Function to precompute the # sum of all armstrong numbers # upto 100000 def preCompute():
for i in range ( 1 , 100001 ):
# checkarmstrong ()
# return the number i
# if i is armstrong
# else return 0
pref[i] = pref[i - 1 ] + checkArmstrong(i)
# Function to print the sum # for each query def printSum(L, R):
print (pref[R] - pref[L - 1 ])
# Function to print sum of all # armstrong numbers between # [L, R] def printSumarmstrong (arr, Q):
# Function that pre computes
# the sum of all armstrong
# numbers
preCompute()
# Iterate over all Queries
# to print the sum
for i in range (Q):
printSum(arr[i][ 0 ], arr[i][ 1 ])
# Driver code # Queries Q = 2
arr = [[ 1 , 13 ], [ 121 , 211 ]]
# Function that print the # the sum of all armstrong # number in Range [L, R] printSumarmstrong (arr, Q) |
// C# program to find the sum // of all armstrong numbers // in the given range using System;
class GFG
{ // pref[] array to precompute
// the sum of all armstrong
// number
static int [] pref = new int [100001];
// Function that return number
// num if num is armstrong
// else return 0
static int checkArmstrong( int x) {
int n = x.ToString().Length;
int sum1 = 0;
int temp = x;
while (temp > 0) {
int digit = temp % 10;
sum1 += ( int )Math.Pow(digit, n);
temp /= 10;
}
if (sum1 == x)
return x;
return 0;
}
// Function to precompute the
// sum of all armstrong numbers
// upto 100000
static void preCompute()
{
for ( int i = 1; i < 100001; i++)
{
// checkarmstrong ()
// return the number i
// if i is armstrong
// else return 0
pref[i] = pref[i - 1] + checkArmstrong(i);
}
}
// Function to print the sum
// for each query
static void printSum( int L, int R) {
Console.WriteLine(pref[R] - pref[L - 1]);
}
// Function to print sum of all
// armstrong numbers between
// [L, R]
static void printSumarmstrong( int [,] arr, int Q) {
// Function that pre computes
// the sum of all armstrong
// numbers
preCompute();
// Iterate over all Queries
// to print the sum
for ( int i = 0; i < Q; i++) {
printSum(arr[i, 0], arr[i, 1]);
}
}
// Driver code
public static void Main( string [] args)
{
// Queries
int Q = 2;
int [,] arr = { { 1, 13 }, { 121, 211 } };
// Function that print the
// the sum of all armstrong
// number in Range [L, R]
printSumarmstrong(arr, Q);
}
} // This code is contributed by AnkitRai01 |
<script> // Javascript program to find the sum
// of all armstrong numbers
// in the given range
// pref[] array to precompute
// the sum of all armstrong
// number
let pref = new Array(100001);
pref.fill(0);
// Function that return number
// num if num is armstrong
// else return 0
function checkArmstrong(x) {
let n = (x.toString()).length;
let sum1 = 0;
let temp = x;
while (temp > 0) {
let digit = temp % 10;
sum1 += Math.pow(digit, n);
temp = parseInt(temp / 10, 10);
}
if (sum1 == x)
return x;
return 0;
}
// Function to precompute the
// sum of all armstrong numbers
// upto 100000
function preCompute() {
for (let i = 1; i < 100001; i++) {
// checkarmstrong ()
// return the number i
// if i is armstrong
// else return 0
pref[i] = pref[i - 1] + checkArmstrong(i);
}
}
// Function to print the sum
// for each query
function printSum(L, R) {
document.write((pref[R] - pref[L - 1]) + "</br>" );
}
// Function to print sum of all
// armstrong numbers between
// [L, R]
function printSumarmstrong(arr, Q) {
// Function that pre computes
// the sum of all armstrong
// numbers
preCompute();
// Iterate over all Queries
// to print the sum
for (let i = 0; i < Q; i++) {
printSum(arr[i][0], arr[i][1]);
}
}
// Queries
let Q = 2;
let arr = [ [ 1, 13 ], [ 121, 211 ] ];
// Function that print the
// the sum of all armstrong
// number in Range [L, R]
printSumarmstrong(arr, Q);
// This code is contributed by divyesh072019.
</script> |
Output:
45 153
Time Complexity: O(N), where N is the maximum element in the query.
Auxiliary Space: O(105)