Array Range Queries to find the Maximum Armstrong number with updates

Given an array arr[] of N integers, the task is to perform the following two queries: 

  • maximum(start, end): Print the maximum Armstrong number of elements in the sub-array from start to end
  • update(i, x): Add x to the array element referenced by array index i, that is: arr[i] = x

If there is no Armstrong number in the sub-array, print -1.


Input: arr = [192, 113, 535, 7, 19, 111] 
Query 1: maximum(start=1, end=3)
Query 2: update(i=1, x=153) i.e. arr[1]=153
Maximum Armstrong number in given range = 7 
Updated Maximum Armstrong number in given range = 153
In the Maximum Query, the sub-array [1…3] has 1 Armstrong number 7 viz. [113, 535, 7] 
Hence, 7 is the maximum Armstrong number in the given range.
In the Update Query, the value at index 1 is updated to 153, the array arr now is, [192, 153, 535, 7, 19, 111]
In Updated Maximum Query, the sub-array [1…3] has 2 Armstrong numbers 153 and 7 viz. [153, 535, 7] 
Hence, 153 is the maximum Armstrong number in the given range. 

Simple Solution:
A simple solution is to run a loop from l to r and calculate the maximum Armstrong number of elements in given range. To update a value, simply do arr[i] = x. The first operation takes O(N) time and the second operation takes O(1) time.

Efficient Approach: 

  • An efficient approach will be to build a Segment Tree where each node stores two values(value and max_set_bits), and do a range query on it to find the maximum Armstrong number. 
  • If we have a deep look into it, the maximum Armstrong number for any two range combining will either be the maximum Armstrong number from the left side or the maximum Armstrong number from the right side, whichever is maximum will be taken into account. 
  • Representation of Segment trees: 
    1. Leaf Nodes are the elements of the given array. 
    2. Each internal node represents the maximum Armstrong number of all of its child or -1 is no Armstrong number exists in the range. 
    3. An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at (i-1)/2
  • Construction of Segment Tree from given array: 
    1. 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 maximum Armstrong number value or -1 in a segment tree node. All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a full Binary Tree because we always divide segments into two halves at every level. Since the constructed tree is always a full binary tree with n leaves, there will be n-1 internal nodes. So total nodes will be 2*n – 1. Height of the segment tree will be log2n. Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be 2*( 2ceil(log2n) ) – 1
    2. A positive integer of n digits is called an Armstrong number of order n (order is number of digits) if 

      abcd… = pow(a, n) + pow(b, n) + pow(c, n) + pow(d, n) + ….

    3. In order to check for Armstrong numbers, the idea is to first count number digits (or find order). Let the number of digits be n. For every digit r in input number x, compute rn. If sum of all such values is equal to n, then return true else false.
    4. We then do a range query on the segment tree to find out the maximum Armstrong number for the given range and output the corresponding value.

Below is the implementation of above approach:






// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
// A utility function to get the
// middle index of given range.
int getMid(int s, int e)
    return s + (e - s) / 2;
// Function that return true
// if num is armstrong
// else return false
bool isArmstrong(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 true;
    return false;
/*  A recursive function to get 
    the sum of values in the 
    given range of the array. 
    The following are parameters 
    for this function.
st -> Pointer to segment tree
node -> Index of current node in 
        the segment tree .
ss & se -> Starting and ending indexes 
           of the segment represented
           by current node, 
           i.e., st[node]
l & r -> Starting and ending indexes 
         of range query */
int MaxUtil(
    int* st, int ss, int se, int l,
    int r, int node)
    // If segment of this node is
    // completely part of given range,
    // then return the max of segment.
    if (l <= ss && r >= se)
        return st[node];
    // If segment of this node does not
    // belong to given range
    if (se < l || ss > r)
        return -1;
    // If segment of this node is
    // partially the part of given
    // range
    int mid = getMid(ss, se);
    return max(
        MaxUtil(st, ss, mid,
                l, r, 2 * node + 1),
        MaxUtil(st, mid + 1, se,
                l, r, 2 * node + 2));
/* A recursive function to update
   the nodes which have the given 
   the index in their range. The
   following are parameters st, ss
   and se are same as defined above
   index -> index of the element
   to be updated.*/
void updateValue(int arr[], 
                 int* st,int ss, 
                 int se, int index,
                 int value, int node) {
    if (index < ss || index > se) {
        cout << "Invalid Input" << endl;
    if (ss == se) {
        // update value in array
        // and in segment tree
        arr[index] = value;
        if (isArmstrong(value))
            st[node] = value;
            st[node] = -1;
    else {
        int mid = getMid(ss, se);
        if (index >= ss && index <= mid)
            updateValue(arr, st, ss,
                        mid, index, value,
                        2 * node + 1);
            updateValue(arr, st, mid + 1,
                        se, index, value,
                        2 * node + 2);
        st[node] = max(st[2 * node + 1],
                       st[2 * node + 2]);
// Return max of elements in
// range from index
// l (query start) to r (query end).
int getMax(int* st, int n,
           int l, int r)
    // Check for erroneous input values
    if (l < 0 || r > n - 1
        || l > r) {
        printf("Invalid Input");
        return -1;
    return MaxUtil(st, 0, n - 1,
                   l, r, 0);
// A recursive function that
// constructs Segment Tree for
// array[]. si is index of
// current node in segment tree st
int constructSTUtil(int arr[],
                    int ss, int se,
                    int* st, int si)
    // If there is one
    // element in array, store
    // it in current node of
    // segment tree and return
    if (ss == se) {
        if (isArmstrong(arr[ss]))
            st[si] = arr[ss];
            st[si] = -1;
        return st[si];
    // If there are more than
    // one elements, then
    // recur for left and right
    // subtrees and store the
    // max of values in this node
    int mid = getMid(ss, se);
    st[si] = max(constructSTUtil(
                     arr, ss, mid, st,
                     si * 2 + 1),
                     arr, mid + 1, se,
                     st, si * 2 + 2));
    return st[si];
/* Function to construct a segment tree 
   from given array. This function 
   allocates memory for segment tree.*/
int* constructST(int arr[], int n)
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
    // Maximum size of segment tree
    int max_size
        = 2 * (int)pow(2, x) - 1;
    // Allocate memory
    int* st = new int[max_size];
    // Fill the allocated memory st
    constructSTUtil(arr, 0,
                    n - 1, st, 0);
    // Return the constructed
    // segment tree
    return st;
// Driver code
int main()
    int arr[] = { 192, 113,
                  535, 7, 19, 111 };
    int n = sizeof(arr) / sizeof(arr[0]);
    // Build segment tree from
    // given array
    int* st = constructST(arr, n);
    // Print max of values in array
    // from index 1 to 3
    cout << "Maximum armstrong "
         << "number in given range = "
         << getMax(st, n, 1, 3) << endl;
    // Update: set arr[1] = 153 and update
    // corresponding segment tree nodes.
    updateValue(arr, st, 0,
                n - 1, 1, 153, 0);
    // Find max after the value is updated
    cout << "Updated Maximum armstrong "
         << "number in given range = "
         << getMax(st, n, 1, 3) << endl;
    return 0;



Maximum armstrong number in given range = 7
Updated Maximum armstrong number in given range = 153

Time Complexity: The time complexity of each query and update is O(log N) and that of building the segment tree is O(N)


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