Array range queries to find the number of perfect square elements with updates

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

  • query(start, end): Print the number of perfect square numbers 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

Note: 0 based indexing is followed in the below example.

Example: 



Input: arr = [ 16, 15, 8, 9, 14, 25 ]; 
Query 1: query(start = 0, end = 4) 
Query 2: update(i = 3, x = 11) i.e. arr[3]=11  
Query 3: query(start = 0, end = 4) 
Output: 2 1 
Explanation:
In Query 1, the sub-array [0…4] has 2 perfect square numbers 16 and 9 viz. [ 16, 15, 8, 9, 14 ]
In Query 2, the value at index 3 is updated to 11,  
the array arr now is, [ 16, 15, 8, 11, 14, 25 ]
In Query 3, the sub-array [0…4] has 1 perfect square number 16 viz. [ 16, 15, 8, 11, 14 ]
 

Approach: 

To handle both point updates and range queries, a segment tree is optimal for this purpose.
In order to check for perfect square numbers, the idea is to first compute the square root of the number and if the square root is an integer then the current element is a perfect square otherwise not. If the current element is a perfect square, then set it to 1 else to 0.

Building the segment tree: 

  • The problem is now reduced to the subarray sum using segment tree problem.
  • Now, we can build the segment tree where a leaf node is represented as either 0 (if it is not an perfect square number) or 1 (if it is perfect square number).
  • The internal nodes of the segment tree equal to the sum of its child nodes, thus a node represent the total perfect square numbers in the range from L to R with range [L, R] falling under this node and the sub-tree underneath it.

Handling Queries and Point Updates: 

  • Whenever we receive a query from beginning to end, we can query the segment tree for the sum of nodes in range from start to end, which in turn represent the number of perfect square numbers in the range start to end. 
     
  • To perform a point update and to update the value at index i to x, we check for the following cases: 
    Let the old value of arr[i] be y and the new value be x. 
    1. Case 1: If x and y both are perfect square numbers 
      Count of perfect square numbers in the subarray does not change so we just update array and do not modify the segment tree
    2. Case 2: If x and y both are not perfect square numbers 
      Count of perfect square numbers in the subarray does not change so we just update array and do not modify the segment tree
    3. Case 3: If y is a perfect square number but x is not 
      Count of perfect square numbers in the subarray decreases so we update array and add -1 to every range. The index i which is to be updated is a part of in the segment tree
    4. Case 4: If y is not an perfect square number but x is an perfect square number 
      Count of perfect square numbers in the subarray increases so we update array and add 1 to every range. The index i which is to be updated is a part of in the segment tree

Below is the implementation of the above approach:

C++

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// C++ program to find number of
// perfect square numbers in a
// subarray and performing updates
  
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 1000
  
// Function to check if a number is
// a perfect square or not
bool isPerfectSquare(long long int x)
{
    // Find floating point value of
    // square root of x.
    long double sr = sqrt(x);
  
    // If square root is an integer
    return ((sr - floor(sr)) == 0)
               ? true
               : false;
}
  
// A utility function to get the middle
// index from corner indexes.
int getMid(int s, int e)
{
    return s + (e - s) / 2;
}
  
// Recursive function to get the number
// of perfect square numbers in a given
// range
/* where
  st    --> Pointer to segment tree
  index --> Index of current node in the
            segment tree. Initially 0 is 
            passed as root is always 
            at index 0
  ss & se  --> Starting and ending indexes 
              of the segment represented by 
              current node i.e. st[index]
  qs & qe  --> Starting and ending indexes
               of query range   */
int queryUtil(int* st, int ss,
              int se, int qs,
              int qe, int index)
{
    // If segment of this node is a part
    // of given range, then return
    // the number of perfect square numbers
    // in the segment
    if (qs <= ss && qe >= se)
        return st[index];
  
    // If segment of this node
    // is outside the given range
    if (se < qs || ss > qe)
        return 0;
  
    // If a part of this segment
    // overlaps with the given range
    int mid = getMid(ss, se);
    return queryUtil(st, ss, mid, qs,
                     qe, 2 * index + 1)
           + queryUtil(st, mid + 1, se,
                       qs, qe, 2 * index + 2);
}
  
// Recursive function to update
// the nodes which have the given
// index in their range.
/* where
   st, si, ss & se are same as getSumUtil()
   i --> index of the element to be updated. 
         This index is in input array.
   diff --> Value to be added to all nodes
          which have i in range 
*/
void updateValueUtil(int* st, int ss,
                     int se, int i,
                     int diff, int si)
{
    // Base Case:
    // If the input index lies outside
    // the range of this segment
    if (i < ss || i > se)
        return;
  
    // If the input index is in range
    // of this node, then update the value
    // of the node and its children
    st[si] = st[si] + diff;
    if (se != ss) {
  
        int mid = getMid(ss, se);
        updateValueUtil(st, ss, mid, i,
                        diff, 2 * si + 1);
        updateValueUtil(st, mid + 1, se,
                        i, diff, 2 * si + 2);
    }
}
  
// Function to update a value in the
// input array and segment tree.
// It uses updateValueUtil() to update
// the value in segment tree
void updateValue(int arr[], int* st,
                 int n, int i,
                 int new_val)
{
    // Check for erroneous input index
    if (i < 0 || i > n - 1) {
        printf("Invalid Input");
        return;
    }
  
    int diff, oldValue;
  
    oldValue = arr[i];
  
    // Update the value in array
    arr[i] = new_val;
  
    // Case 1: Old and new values
    // both are perfect square numbers
    if (isPerfectSquare(oldValue)
        && isPerfectSquare(new_val))
        return;
  
    // Case 2: Old and new values
    // both not perfect square numbers
    if (!isPerfectSquare(oldValue)
        && !isPerfectSquare(new_val))
        return;
  
    // Case 3: Old value was perfect square,
    // new value is not a perfect square
    if (isPerfectSquare(oldValue)
        && !isPerfectSquare(new_val)) {
        diff = -1;
    }
  
    // Case 4: Old value was
    // non-perfect square,
    // new_val is perfect square
    if (!isPerfectSquare(oldValue)
        && !isPerfectSquare(new_val)) {
        diff = 1;
    }
  
    // Update values of nodes in segment tree
    updateValueUtil(st, 0, n - 1, i, diff, 0);
}
  
// Return no. of perfect square numbers
// in range from index qs (query start)
// to qe (query end).
// It mainly uses queryUtil()
void query(int* st, int n,
           int qs, int qe)
{
  
    int perfectSquareInRange
        = queryUtil(
            st, 0, n - 1, qs, qe, 0);
  
    cout << perfectSquareInRange << "\n";
}
  
// Recursive function that constructs
// Segment Tree for array[ss..se].
// 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,
    // check if it is perfect square number
    // then store 1 in the segment tree
    // else store 0 and return
    if (ss == se) {
  
        // if arr[ss] is a perfect
        // square number
        if (isPerfectSquare(arr[ss]))
            st[si] = 1;
        else
            st[si] = 0;
  
        return st[si];
    }
  
    // If there are more than one
    // elements, then recur for
    // left and right subtrees
    // and store the sum of the
    // two values in this node
    int mid = getMid(ss, se);
    st[si]
        = constructSTUtil(arr, ss,
                          mid, st, si * 2 + 1)
          + constructSTUtil(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 and calls
// constructSTUtil() to fill
// the allocated memory
int* constructST(int arr[], int n)
{
  
    // Allocate memory for segment tree
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
  
    // Maximum size of segment tree
    int max_size = 2 * (int)pow(2, x) - 1;
  
    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[] = { 16, 15, 8, 9, 14, 25 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Build segment tree from given array
    int* st = constructST(arr, n);
  
    // Query 1: Query(start = 0, end = 4)
    int start = 0;
    int end = 4;
    query(st, n, start, end);
  
    // Query 2: Update(i = 3, x = 11),
    // i.e Update a[i] to x
    int i = 3;
    int x = 11;
    updateValue(arr, st, n, i, x);
  
    // uncomment to see array after update
    // for(int i = 0; i < n; i++)
    // cout << arr[i] << " ";
  
    // Query 3: Query(start = 0, end = 4)
    start = 0;
    end = 4;
    query(st, n, start, end);
  
    return 0;
}

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Output:

2
1

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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