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

  • Difficulty Level : Hard
  • Last Updated : 03 May, 2021

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:
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. 
 

 

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 <bits/stdc++.h>
 
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

Javascript




<script>
 
// javascript implementation to find the
// minimum window size in the range
// such that each window of that size
// contains atleast P primes
 
// Function to check that a number is
// a prime or not in O(Math.sqrt(N))
function isPrime(N)
{
    if (N < 2)
        return false;
    if (N < 4)
        return true;
    if ((N & 1) == 0)
        return false;
    if (N % 3 == 0)
        return false;
    let curr = 5, s = Math.floor(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
function check(s, p,
      prefix_sum, n)
{
    let satisfies = true;
      
    // Loop to check each window of
    // size have atleast P primes
    for (let 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
function minimumWindowSize(x, y, p)
{
    // Prefix array
    let prefix_sum = new Array(y - x + 1).fill(0);
  
    // Mark those numbers
    // which are primes as 1
    for (let i = x; i <= y; i++) {
        if (isPrime(i))
            prefix_sum[i - x] = 1;
    }
  
    // Convert to prefix sum
    for (let i = 1; i < y - x + 1; i++)
        prefix_sum[i] +=
              prefix_sum[i - 1];
  
    // Applying binary search
    // over window size
    let low = 1, high = y - x + 1;
    let mid;
    while (high - low > 1) {
        mid = Math.floor((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
     
    let x = 12;
    let y = 42;
    let p = 3;
  
    document.write(minimumWindowSize(x, y, p));
     
</script>
Output: 
14

 

Time complexity: O(N*log(N)) 
Auxiliary Space: O(N)
 

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