# Count of elements having odd number of divisors in index range [L, R] for Q queries

Given an array arr[] of N positive integers and the number of queries Q, each query contains two numbers L and R. The task is to count the number of elements in the array having odd number of divisors from index L to R.

Examples:

Input: arr[] = [2, 4, 5, 6, 9], Q = 3, Query[][] = { {0, 2}, {1, 3}, {1, 4} }
Output: 1 1 2
Explanation:
1st query: in 2 4 5 only 4 has an odd number of divisors.
2nd query: in 4 5 6 only 4 has an odd number of divisors.
3rd query: in 4 5 6 9 only 4, 9 has an odd number of divisors.

Input: arr[] = [1, 16, 5, 4, 9], Q = 2, Query[][] = { {1, 3}, {0, 2} }
Output: 2 1

Naive Approach: The naive approach is to iterate over the array from L to R for each query and count the element in the range [L, R] having odd numbers of divisors. If yes then count that element for that query.

Time Complexity: O(Q * N * sqrt(N))
Auxiliary Space: O(1)

Efficient Approach: We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution is to check if the given number is a perfect square or not in the range [L, R]. Below are the steps:

1. Initialize the array dp[] of size N with value 0.
2. Traverse the given array arr[] and if any element in the it is a perfect square the update the value at that index in dp[] as 1.
3. To calculate the answer for each query efficiently we will precompute the answer.
4. We will do the prefix sum of the array dp[] and for each query in the range [L, R] the answer will be given by:
```OddDivisorCount(L, R) = DP[R] - DP[L-1]
```

Below is the implementation of the above approach:

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function count the number of elements` `// having odd number of divisors` `void` `OddDivisorsCount(` `    ``int` `n, ``int` `q, ``int` `a[],` `    ``vector > Query)` `{` `    ``// Initialise dp[] array` `    ``int` `DP[n] = { 0 };`   `    ``// Precomputation` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `x = ``sqrt``(a[i]);`   `        ``if` `(x * x == a[i])` `            ``DP[i] = 1;` `    ``}`   `    ``// Find the Prefix Sum` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``DP[i] = DP[i - 1] + DP[i];` `    ``}`   `    ``int` `l, r;`   `    ``// Iterate for each query` `    ``for` `(``int` `i = 0; i < q; i++) {` `        ``l = Query[i].first;` `        ``r = Query[i].second;`   `        ``// Find the answer for each query` `        ``if` `(l == 0) {` `            ``cout << DP[r] << endl;` `        ``}` `        ``else` `{` `            ``cout << DP[r] - DP[l - 1]` `                 ``<< endl;` `        ``}` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 5;` `    ``int` `Q = 3;`   `    ``// Given array arr[]` `    ``int` `arr[] = { 2, 4, 5, 6, 9 };`   `    ``// Given Query` `    ``vector > Query` `        ``Query` `        ``= { { 0, 2 }, { 1, 3 }, { 1, 4 } };`   `    ``// Function Call` `    ``OddDivisorsCount(N, Q, arr, Query);`   `    ``return` `0;` `}`

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{` `    `  `static` `class` `pair` `{ ` `    ``int` `first, second; ` `    ``public` `pair(``int` `first, ``int` `second) ` `    ``{ ` `        ``this``.first = first; ` `        ``this``.second = second; ` `    ``} ` `} `   `// Function count the number of elements` `// having odd number of divisors` `static` `void` `OddDivisorsCount(``int` `n, ``int` `q,` `                             ``int` `a[], ` `                             ``pair []Query)` `{` `    `  `    ``// Initialise dp[] array` `    ``int` `DP[] = ``new` `int``[n];`   `    ``// Precomputation` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{` `       ``int` `x = (``int``)Math.sqrt(a[i]);` `       `  `       ``if` `(x * x == a[i])` `           ``DP[i] = ``1``;` `    ``}` `    `  `    ``// Find the Prefix Sum` `    ``for``(``int` `i = ``1``; i < n; i++)` `    ``{` `       ``DP[i] = DP[i - ``1``] + DP[i];` `    ``}` `    `  `    ``int` `l, r;`   `    ``// Iterate for each query` `    ``for``(``int` `i = ``0``; i < q; i++)` `    ``{` `       ``l = Query[i].first;` `       ``r = Query[i].second;` `       `  `       ``// Find the answer for each query` `       ``if` `(l == ``0``)` `       ``{` `           ``System.out.print(DP[r] + ``"\n"``);` `       ``}` `       ``else` `       ``{` `           ``System.out.print(DP[r] - ` `                            ``DP[l - ``1``] + ``"\n"``);` `       ``}` `    ``}` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``5``;` `    ``int` `Q = ``3``;`   `    ``// Given array arr[]` `    ``int` `arr[] = { ``2``, ``4``, ``5``, ``6``, ``9` `};`   `    ``// Given Query` `    ``pair []Query = { ``new` `pair(``0``, ``2``),` `                     ``new` `pair(``1``, ``3``), ` `                     ``new` `pair(``1``, ``4``) };`   `    ``// Function Call` `    ``OddDivisorsCount(N, Q, arr, Query);` `}` `}`   `// This code is contributed by amal kumar choubey`

 `# Python3 program for the above approach` `import` `math`   `# Function count the number of elements` `# having odd number of divisors` `def` `OddDivisorsCount(n, q, a, Query):` `    `  `    ``# Initialise dp[] array` `    ``DP ``=` `[``0` `for` `i ``in` `range``(n)]` ` `  `    ``# Precomputation` `    ``for` `i ``in` `range``(n):` `        ``x ``=` `int``(math.sqrt(a[i]));` ` `  `        ``if` `(x ``*` `x ``=``=` `a[i]):` `            ``DP[i] ``=` `1``;` ` `  `    ``# Find the Prefix Sum` `    ``for` `i ``in` `range``(``1``, n):` `        ``DP[i] ``=` `DP[i ``-` `1``] ``+` `DP[i];` ` `  `    ``l ``=` `0` `    ``r ``=` `0` ` `  `    ``# Iterate for each query` `    ``for` `i ``in` `range``(q):` `        ``l ``=` `Query[i][``0``];` `        ``r ``=` `Query[i][``1``];` ` `  `        ``# Find the answer for each query` `        ``if` `(l ``=``=` `0``):` `            ``print``(DP[r])` `        ``else``:` `            ``print``(DP[r] ``-` `DP[l ``-` `1``])`   `# Driver code ` `if` `__name__``=``=``"__main__"``:` `    `  `    ``N ``=` `5``;` `    ``Q ``=` `3``;` ` `  `    ``# Given array arr[]` `    ``arr ``=` `[ ``2``, ``4``, ``5``, ``6``, ``9` `]` ` `  `    ``# Given Query` `    ``Query ``=` `[ [ ``0``, ``2` `], ` `              ``[ ``1``, ``3` `],` `              ``[ ``1``, ``4` `] ]` ` `  `    ``# Function call` `    ``OddDivisorsCount(N, Q, arr, Query);`   `# This code is contributed by rutvik_56`

 `// C# program for the above approach` `using` `System;`   `class` `GFG{` `class` `pair` `{ ` `    ``public` `int` `first, second; ` `    ``public` `pair(``int` `first, ``int` `second) ` `    ``{ ` `        ``this``.first = first; ` `        ``this``.second = second; ` `    ``} ` `} `   `// Function count the number of elements` `// having odd number of divisors` `static` `void` `OddDivisorsCount(``int` `n, ``int` `q,` `                             ``int` `[]a, ` `                             ``pair []Query)` `{` `    `  `    ``// Initialise []dp array` `    ``int` `[]DP = ``new` `int``[n];`   `    ``// Precomputation` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `       ``int` `x = (``int``)Math.Sqrt(a[i]);` `       ``if` `(x * x == a[i])` `           ``DP[i] = 1;` `    ``}` `    `  `    ``// Find the Prefix Sum` `    ``for``(``int` `i = 1; i < n; i++)` `    ``{` `       ``DP[i] = DP[i - 1] + DP[i];` `    ``}` `    `  `    ``int` `l, r;`   `    ``// Iterate for each query` `    ``for``(``int` `i = 0; i < q; i++)` `    ``{` `       ``l = Query[i].first;` `       ``r = Query[i].second;` `       `  `       ``// Find the answer for each query` `       ``if` `(l == 0)` `       ``{` `           ``Console.Write(DP[r] + ``"\n"``);` `       ``}` `       ``else` `       ``{` `           ``Console.Write(DP[r] - ` `                         ``DP[l - 1] + ``"\n"``);` `       ``}` `    ``}` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `N = 5;` `    ``int` `Q = 3;`   `    ``// Given array []arr` `    ``int` `[]arr = { 2, 4, 5, 6, 9 };`   `    ``// Given Query` `    ``pair []Query = { ``new` `pair(0, 2),` `                     ``new` `pair(1, 3), ` `                     ``new` `pair(1, 4) };`   `    ``// Function Call` `    ``OddDivisorsCount(N, Q, arr, Query);` `}` `}`   `// This code is contributed by gauravrajput1`

Output:
```1
1
2

```

Time complexity:

• Precomputation: O(N)
• For each query: O(1)

Auxiliary Space: O(1)

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