Sum of all armstrong numbers lying in the range [L, R] for Q queries Last Updated : 28 Jul, 2025 Comments Improve Suggest changes Like Article Like Report 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++ // 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 // 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 # 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# // 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 JavaScript <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. 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