# Minimum window size containing atleast P primes in every window of given range

Given three integers X, Y and P, the task is to find the minimum window size K such that every window in the range [X, Y] of this size have atleast P prime numbers.

Examples:

Input: X = 2, Y = 8, P = 2
Output: 4
Explanation:
In the range [2, 8], window size of 4 contains atleast 2 primes in each window.
Possible Windows –
{2, 3, 4, 5} – No of Primes = 3
{3, 4, 5, 6} – No of Primes = 2
{4, 5, 6, 7} – No of Primes = 2
{5, 6, 7, 8} – No of Primes = 2

Input: X = 12, Y = 42, P = 3
Output: 14
Explanation:
In the range [12, 42], window size of 14 contains atleast 3 primes in each window.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Traverse over all the possible window sizes, for each window size traverse in range [X, Y] and check that each window contains atleast K primes. Minimum of these window size will be the desired value.

Efficient Approach: The key observation in this problem is if a window size W is the minimum window size satisfying the condition, then all window size in the range [W, Y – X + 1] will satisfy the condition. Using this we can reduce our search space at each step by half which is precisely the idea of Binary Search. Below is the illustration of the steps:

• Search Space: The search space for this problem can be the minimum length of the window size that is 1 and the maximum window size can be the difference between the ending value of the range and the starting value of the range.
```low = 1
high = Y - X + 1
```
• Next Search Space: In each step generally the idea is to check that for the given window size the primes in each window possible have P primes or not with the help of the sliding window technique. Whereas the search space for the problem can be reduced on the basis of below decision:
• Case 1: When the number of primes in each window contains at least P primes, then the size of the window can be reduced to find the window size of less than the current window.
```if (checkPPrimes(mid) == True)
high = mid - 1
```
• Case 2: When the number of primes in each window contains do not have then the window size must be greater than the current window size. Then,
```if (checkPPrimes(mid) == False)
low = mid + 1
```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// minimum window size in the range ` `// such that each window of that size ` `// contains atleast P primes ` ` `  `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to check that a number is  ` `// a prime or not in O(sqrt(N)) ` `bool` `isPrime(``int` `N) ` `{ ` `    ``if` `(N < 2) ` `        ``return` `false``; ` `    ``if` `(N < 4) ` `        ``return` `true``; ` `    ``if` `((N & 1) == 0) ` `        ``return` `false``; ` `    ``if` `(N % 3 == 0) ` `        ``return` `false``; ` `    ``int` `curr = 5, s = ``sqrt``(N); ` ` `  `    ``// Loop to check if any number ` `    ``// number is divisible by any  ` `    ``// other number or not ` `    ``while` `(curr <= s) { ` `        ``if` `(N % curr == 0) ` `            ``return` `false``; ` `        ``curr += 2; ` `        ``if` `(N % curr == 0) ` `            ``return` `false``; ` `        ``curr += 4; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check whether window ` `// size satisfies condition or not ` `bool` `check(``int` `s, ``int` `p,  ` `      ``int` `prefix_sum[], ``int` `n) ` `{ ` `    ``bool` `satisfies = ``true``; ` `     `  `    ``// Loop to check each window of  ` `    ``// size have atleast P primes ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i + s - 1 >= n) ` `            ``break``; ` `         `  `        ``// Checking condition  ` `        ``// using prefix sum ` `        ``if` `(prefix_sum[i + s - 1] -  ` `          ``(i - 1 >= 0 ?  ` `          ``prefix_sum[i - 1] : 0) < p) ` `            ``satisfies = ``false``; ` `    ``} ` `    ``return` `satisfies; ` `} ` ` `  `// Function to find the minimum  ` `// window size possible for the ` `// given range in X and Y ` `int` `minimumWindowSize(``int` `x, ``int` `y, ` `                             ``int` `p) ` `{ ` `    ``// Prefix array ` `    ``int` `prefix_sum[y - x + 1] = { 0 }; ` ` `  `    ``// Mark those numbers  ` `    ``// which are primes as 1 ` `    ``for` `(``int` `i = x; i <= y; i++) { ` `        ``if` `(isPrime(i)) ` `            ``prefix_sum[i - x] = 1; ` `    ``} ` ` `  `    ``// Convert to prefix sum ` `    ``for` `(``int` `i = 1; i < y - x + 1; i++) ` `        ``prefix_sum[i] +=  ` `              ``prefix_sum[i - 1]; ` ` `  `    ``// Applying binary search  ` `    ``// over window size ` `    ``int` `low = 1, high = y - x + 1; ` `    ``int` `mid; ` `    ``while` `(high - low > 1) { ` `        ``mid = (low + high) / 2; ` `         `  `        ``// Check whether mid satisfies  ` `        ``// the condition or not ` `        ``if` `(check(mid, p,  ` `           ``prefix_sum, y - x + 1)) { ` `             `  `            ``// If satisfies search ` `            ``// in first half ` `            ``high = mid; ` `        ``} ` `         `  `        ``// Else search in second half ` `        ``else` `            ``low = mid; ` `    ``} ` `    ``if` `(check(low, p,  ` `       ``prefix_sum, y - x + 1)) ` `        ``return` `low; ` `    ``return` `high; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `x = 12; ` `    ``int` `y = 42; ` `    ``int` `p = 3; ` ` `  `    ``cout << minimumWindowSize(x, y, p); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the ` `// minimum window size in the range ` `// such that each window of that size ` `// contains atleast P primes ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to check that a number is  ` `// a prime or not in O(Math.sqrt(N)) ` `static` `boolean` `isPrime(``int` `N) ` `{ ` `    ``if` `(N < ``2``) ` `        ``return` `false``; ` `    ``if` `(N < ``4``) ` `        ``return` `true``; ` `    ``if` `((N & ``1``) == ``0``) ` `        ``return` `false``; ` `    ``if` `(N % ``3` `== ``0``) ` `        ``return` `false``; ` `    ``int` `curr = ``5``, s = (``int``) Math.sqrt(N); ` `  `  `    ``// Loop to check if any number ` `    ``// number is divisible by any  ` `    ``// other number or not ` `    ``while` `(curr <= s) { ` `        ``if` `(N % curr == ``0``) ` `            ``return` `false``; ` `        ``curr += ``2``; ` `        ``if` `(N % curr == ``0``) ` `            ``return` `false``; ` `        ``curr += ``4``; ` `    ``} ` `    ``return` `true``; ` `} ` `  `  `// Function to check whether window ` `// size satisfies condition or not ` `static` `boolean` `check(``int` `s, ``int` `p,  ` `      ``int` `prefix_sum[], ``int` `n) ` `{ ` `    ``boolean` `satisfies = ``true``; ` `      `  `    ``// Loop to check each window of  ` `    ``// size have atleast P primes ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(i + s - ``1` `>= n) ` `            ``break``; ` `          `  `        ``// Checking condition  ` `        ``// using prefix sum ` `        ``if` `(prefix_sum[i + s - ``1``] -  ` `          ``(i - ``1` `>= ``0` `?  ` `          ``prefix_sum[i - ``1``] : ``0``) < p) ` `            ``satisfies = ``false``; ` `    ``} ` `    ``return` `satisfies; ` `} ` `  `  `// Function to find the minimum  ` `// window size possible for the ` `// given range in X and Y ` `static` `int` `minimumWindowSize(``int` `x, ``int` `y, ` `                             ``int` `p) ` `{ ` `    ``// Prefix array ` `    ``int` `[]prefix_sum = ``new` `int``[y - x + ``1``]; ` `  `  `    ``// Mark those numbers  ` `    ``// which are primes as 1 ` `    ``for` `(``int` `i = x; i <= y; i++) { ` `        ``if` `(isPrime(i)) ` `            ``prefix_sum[i - x] = ``1``; ` `    ``} ` `  `  `    ``// Convert to prefix sum ` `    ``for` `(``int` `i = ``1``; i < y - x + ``1``; i++) ` `        ``prefix_sum[i] +=  ` `              ``prefix_sum[i - ``1``]; ` `  `  `    ``// Applying binary search  ` `    ``// over window size ` `    ``int` `low = ``1``, high = y - x + ``1``; ` `    ``int` `mid; ` `    ``while` `(high - low > ``1``) { ` `        ``mid = (low + high) / ``2``; ` `          `  `        ``// Check whether mid satisfies  ` `        ``// the condition or not ` `        ``if` `(check(mid, p,  ` `           ``prefix_sum, y - x + ``1``)) { ` `              `  `            ``// If satisfies search ` `            ``// in first half ` `            ``high = mid; ` `        ``} ` `          `  `        ``// Else search in second half ` `        ``else` `            ``low = mid; ` `    ``} ` `    ``if` `(check(low, p,  ` `       ``prefix_sum, y - x + ``1``)) ` `        ``return` `low; ` `    ``return` `high; ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `x = ``12``; ` `    ``int` `y = ``42``; ` `    ``int` `p = ``3``; ` `  `  `    ``System.out.print(minimumWindowSize(x, y, p)); ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

## Python3

 `# Python3 implementation to find the ` `# minimum window size in the range ` `# such that each window of that size ` `# contains atleast P primes ` ` `  `from` `math ``import` `sqrt ` ` `  `# Function to check that a number is  ` `# a prime or not in O(sqrt(N)) ` `def` `isPrime(N): ` `    ``if` `(N < ``2``): ` `        ``return` `False` `    ``if` `(N < ``4``): ` `        ``return` `True` `    ``if` `((N & ``1``) ``=``=` `0``): ` `        ``return` `False` `    ``if` `(N ``%` `3` `=``=` `0``): ` `        ``return` `False` `     `  `    ``curr ``=` `5` `    ``s ``=` `sqrt(N) ` `     `  `    ``# Loop to check if any number ` `    ``# number is divisible by any  ` `    ``# other number or not ` `    ``while` `(curr <``=` `s): ` `        ``if` `(N ``%` `curr ``=``=` `0``): ` `            ``return` `False` `        ``curr ``+``=` `2` `        ``if` `(N ``%` `curr ``=``=` `0``): ` `            ``return` `False` `         `  `        ``curr ``+``=` `4` `     `  `    ``return` `True` ` `  `# Function to check whether window ` `# size satisfies condition or not ` `def` `check(s, p, prefix_sum, n): ` `     `  `    ``satisfies ``=` `True` `    ``# Loop to check each window of  ` `    ``# size have atleast P primes ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``+` `s ``-` `1` `>``=` `n): ` `            ``break` `        ``# Checking condition  ` `        ``# using prefix sum ` `        ``if` `(i ``-` `1` `>``=` `0``): ` `            ``x ``=` `prefix_sum[i ``-` `1``] ` `        ``else``: ` `            ``x ``=` `0` `        ``if` `(prefix_sum[i ``+` `s ``-` `1``] ``-` `x < p): ` `            ``satisfies ``=` `False` `         `  `    ``return` `satisfies ` ` `  `# Function to find the minimum  ` `# window size possible for the ` `# given range in X and Y ` `def` `minimumWindowSize(x, y, p): ` `     `  `    ``# Prefix array ` `    ``prefix_sum ``=` `[``0``]``*``(y ``-` `x ``+` `1``) ` `     `  `    ``# Mark those numbers  ` `    ``# which are primes as 1     ` `    ``for` `i ``in` `range``(x ,y``+``1``): ` `        ``if` `(isPrime(i)): ` `            ``prefix_sum[i ``-` `x] ``=` `1` `     `  `    ``# Convert to prefix sum ` `    ``for` `i ``in` `range``(``1` `,y ``-` `x ``+` `1``): ` `        ``prefix_sum[i] ``+``=` `prefix_sum[i ``-` `1``] ` `         `  `    ``# Applying binary search  ` `    ``# over window size ` `    ``low ``=` `1` `    ``high ``=` `y ``-` `x ``+` `1` `     `  `    ``while` `(high ``-` `low > ``1``): ` `        ``mid ``=` `(low ``+` `high) ``/``/` `2` `         `  `        ``# Check whether mid satisfies  ` `        ``# the condition or not ` `        ``if` `(check(mid, p ,prefix_sum, y ``-` `x ``+` `1``)): ` `             `  `            ``# If satisfies search ` `            ``# in first half ` `            ``high ``=` `mid ` `         `  `        ``# Else search in second half ` `        ``else``: ` `            ``low ``=` `mid ` `    ``if` `(check(low, p, prefix_sum, y ``-` `x ``+` `1``)): ` `        ``return` `low ` `    ``return` `high ` ` `  `# Driver Code ` `x ``=` `12` `y ``=` `42` `p ``=` `3` ` `  `print``(minimumWindowSize(x, y, p)) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation to find the ` `// minimum window size in the range ` `// such that each window of that size ` `// contains atleast P primes ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to check that a number is  ` `// a prime or not in O(Math.Sqrt(N)) ` `static` `bool` `isPrime(``int` `N) ` `{ ` `    ``if` `(N < 2) ` `        ``return` `false``; ` `    ``if` `(N < 4) ` `        ``return` `true``; ` `    ``if` `((N & 1) == 0) ` `        ``return` `false``; ` `    ``if` `(N % 3 == 0) ` `        ``return` `false``; ` `    ``int` `curr = 5, s = (``int``) Math.Sqrt(N); ` `   `  `    ``// Loop to check if any number ` `    ``// number is divisible by any  ` `    ``// other number or not ` `    ``while` `(curr <= s) { ` `        ``if` `(N % curr == 0) ` `            ``return` `false``; ` `        ``curr += 2; ` `        ``if` `(N % curr == 0) ` `            ``return` `false``; ` `        ``curr += 4; ` `    ``} ` `    ``return` `true``; ` `} ` `   `  `// Function to check whether window ` `// size satisfies condition or not ` `static` `bool` `check(``int` `s, ``int` `p,  ` `      ``int` `[]prefix_sum, ``int` `n) ` `{ ` `    ``bool` `satisfies = ``true``; ` `       `  `    ``// Loop to check each window of  ` `    ``// size have atleast P primes ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i + s - 1 >= n) ` `            ``break``; ` `           `  `        ``// Checking condition  ` `        ``// using prefix sum ` `        ``if` `(prefix_sum[i + s - 1] -  ` `          ``(i - 1 >= 0 ?  ` `          ``prefix_sum[i - 1] : 0) < p) ` `            ``satisfies = ``false``; ` `    ``} ` `    ``return` `satisfies; ` `} ` `   `  `// Function to find the minimum  ` `// window size possible for the ` `// given range in X and Y ` `static` `int` `minimumWindowSize(``int` `x, ``int` `y, ` `                             ``int` `p) ` `{ ` `    ``// Prefix array ` `    ``int` `[]prefix_sum = ``new` `int``[y - x + 1]; ` `   `  `    ``// Mark those numbers  ` `    ``// which are primes as 1 ` `    ``for` `(``int` `i = x; i <= y; i++) { ` `        ``if` `(isPrime(i)) ` `            ``prefix_sum[i - x] = 1; ` `    ``} ` `   `  `    ``// Convert to prefix sum ` `    ``for` `(``int` `i = 1; i < y - x + 1; i++) ` `        ``prefix_sum[i] +=  ` `              ``prefix_sum[i - 1]; ` `   `  `    ``// Applying binary search  ` `    ``// over window size ` `    ``int` `low = 1, high = y - x + 1; ` `    ``int` `mid; ` `    ``while` `(high - low > 1) { ` `        ``mid = (low + high) / 2; ` `           `  `        ``// Check whether mid satisfies  ` `        ``// the condition or not ` `        ``if` `(check(mid, p,  ` `           ``prefix_sum, y - x + 1)) { ` `               `  `            ``// If satisfies search ` `            ``// in first half ` `            ``high = mid; ` `        ``} ` `           `  `        ``// Else search in second half ` `        ``else` `            ``low = mid; ` `    ``} ` `    ``if` `(check(low, p,  ` `       ``prefix_sum, y - x + 1)) ` `        ``return` `low; ` `    ``return` `high; ` `} ` `   `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `x = 12; ` `    ``int` `y = 42; ` `    ``int` `p = 3; ` `   `  `    ``Console.Write(minimumWindowSize(x, y, p)); ` `} ` `} ` `  `  `// This code is contributed by 29AjayKumar `

Output:

```14
```

Time complexity: O(N*log(N))
Auxillary Space: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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