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Minimize sum of prime numbers added to make an array non-decreasing

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jyoti369
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Given an array arr[], the task is to convert it into a non-decreasing array by adding primes to array elements such that the sum of the primes added is the minimum possible.

Examples:

Input: arr[] = {2, 1, 5, 4, 3} 
Output: 7 
Explanation: 
{2, 1, 5, 4, 3} -> {2, 3, 5, 6, 6} 
By changing the 2nd element of array to 3(1 + 2), the 4th element of the array to 6(4 + 2) and the 5th element to 6( 3 + 3). 
So, the total cost is 2 + 2 + 3 = 7

Input: arr[] = {3, 3, 3, 3} 
Output: 10

Approach: The idea is to add the least prime numbers possible to the array elements to make the array non-decreasing. Below are the steps:

  1. Initialize a variable to store the minimum cost, say res.
  2. Generate and store all prime numbers up to 107 using the Sieve of Eratosthenes.
  3. Now, traverse the array and do the following steps: 
    • If the current element is less than the previous element than find the smallest prime number (say closest_prime) which can be added to make the array non-decreasing.
    • Update the current element arr[i] = arr[i] + closest_prime.
    • Add closest_prime to res.
  4. Print the final value of res obtained.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 10000000
 
// Stores if an index is a
// prime / non-prime value
bool isPrime[MAX];
 
// Stores the prime
vector<int> primes;
 
// Function to generate all
// prime numbers
void SieveOfEratosthenes()
{
    memset(isPrime, true, sizeof(isPrime));
 
    for (int p = 2; p * p <= MAX; p++) {
 
        // If current element is prime
        if (isPrime[p] == true) {
 
            // Set all its multiples non-prime
            for (int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
 
    // Store all prime numbers
    for (int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.push_back(p);
}
 
// Function to find the closest
// prime to a particular number
int prime_search(vector<int> primes,
                 int diff)
{
 
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.size() - 1;
 
    int res;
 
    while (low <= high) {
        int mid = (low + high) / 2;
 
        // If the prime added makes
        // the elements equal
        if (primes[mid] == diff) {
 
            // Return this as the
            // closest prime
            return primes[mid];
        }
 
        // If the array remains
        // non-decreasing
        else if (primes[mid] < diff) {
 
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
 
        // Otherwise
        else {
 
            res = primes[mid];
 
            // Check if a smaller prime  can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
 
    // Return closest number
    return res;
}
 
// Function to find the minimum cost
int minCost(int arr[], int n)
{
 
    // Find all primes
    SieveOfEratosthenes();
 
    // Store the result
    int res = 0;
 
    // Iterate over the array
    for (int i = 1; i < n; i++) {
 
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1]) {
            int diff = arr[i - 1] - arr[i];
 
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime
                = prime_search(primes, diff);
 
            // Add to overall cost
            res += closest_prime;
 
            // Update current element
            arr[i] += closest_prime;
        }
    }
 
    // Return the minimum cost
    return res;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 1, 5, 4, 3 };
    int n = 5;
 
    // Function Call
    cout << minCost(arr, n);
    return 0;
}

Java



// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
static final int MAX = 10000000;
 
// Stores if an index is a
// prime / non-prime value
static boolean []isPrime = new boolean[MAX + 1];
 
// Stores the prime
static Vector<Integer> primes = new Vector<Integer>();
 
// Function to generate all
// prime numbers
static void SieveOfEratosthenes()
{
    Arrays.fill(isPrime, true);
 
    for(int p = 2; p * p <= MAX; p++)
    {
         
        // If current element is prime
        if (isPrime[p] == true)
        {
             
            // Set all its multiples non-prime
            for(int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
 
    // Store all prime numbers
    for(int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.add(p);
}
 
// Function to find the closest
// prime to a particular number
static int prime_search(Vector<Integer> primes,
                        int diff)
{
 
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.size() - 1;
 
    int res = -1;
 
    while (low <= high)
    {
        int mid = (low + high) / 2;
 
        // If the prime added makes
        // the elements equal
        if (primes.get(mid) == diff)
        {
             
            // Return this as the
            // closest prime
            return primes.get(mid);
        }
 
        // If the array remains
        // non-decreasing
        else if (primes.get(mid) < diff)
        {
             
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
 
        // Otherwise
        else
        {
            res = primes.get(mid);
 
            // Check if a smaller prime can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
 
    // Return closest number
    return res;
}
 
// Function to find the minimum cost
static int minCost(int arr[], int n)
{
     
    // Find all primes
    SieveOfEratosthenes();
 
    // Store the result
    int res = 0;
 
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
         
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1])
        {
            int diff = arr[i - 1] - arr[i];
 
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime = prime_search(primes,
                                             diff);
 
            // Add to overall cost
            res += closest_prime;
 
            // Update current element
            arr[i] += closest_prime;
        }
    }
 
    // Return the minimum cost
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 2, 1, 5, 4, 3 };
    int n = 5;
 
    // Function call
    System.out.print(minCost(arr, n));
}
}
 
// This code is contributed by Amit Katiyar
Python3



# Python3 Program to implement
# the above approach
MAX = 10000000
 
# Stores if an index is a
# prime / non-prime value
isPrime = [True] * (MAX + 1)
 
# Stores the prime
primes = []
 
# Function to generate all
# prime numbers
def SieveOfEratosthenes():
 
    global isPrime
 
    p = 2
    while p * p <= MAX:
 
        # If current element is prime
        if (isPrime[p] == True):
 
            # Set all its multiples non-prime
            for i in range (p * p, MAX + 1, p):
                isPrime[i] = False
               
        p += 1
       
    # Store all prime numbers
    for p in range (2, MAX + 1):
        if (isPrime[p]):
            primes.append(p)
   
# Function to find the closest
# prime to a particular number
def prime_search(primes, diff):
 
    # Applying binary search
    # on primes vector
    low = 0
    high = len(primes) - 1
    
    while (low <= high):
        mid = (low + high) // 2
 
        # If the prime added makes
        # the elements equal
        if (primes[mid] == diff):
 
            # Return this as the
            # closest prime
            return primes[mid]
        
        # If the array remains
        # non-decreasing
        elif (primes[mid] < diff):
 
            # Search for a bigger
            # prime number
            low = mid + 1
      
        # Otherwise
        else:
            res = primes[mid]
 
            # Check if a smaller prime  can
            # make array non-decreasing or not
            high = mid - 1
        
    # Return closest number
    return res
   
# Function to find the minimum cost
def minCost(arr, n):
 
    # Find all primes
    SieveOfEratosthenes()
 
    # Store the result
    res = 0
 
    # Iterate over the array
    for i in range (1, n):
 
        # Current element is less
        # than the previous element
        if (arr[i] < arr[i - 1]):
            diff = arr[i - 1] - arr[i]
 
            # Find the closest prime which
            # makes the array non decreasing
            closest_prime = prime_search(primes, diff)
 
            # Add to overall cost
            res += closest_prime
 
            # Update current element
            arr[i] += closest_prime
       
    # Return the minimum cost
    return res
 
# Driver Code
if __name__ == "__main__":
   
    # Given array
    arr = [2, 1, 5, 4, 3]
    n = 5
 
    #Function Call
    print (minCost(arr, n))
    
# This code is contributed by Chitranayal
C#



// C# program to implement 
// the above approach 
using System;
using System.Collections;
using System.Collections.Generic; 
  
class GFG{
    
static int MAX = 10000000;
   
// Stores if an index is a
// prime / non-prime value
static bool []isPrime = new bool[MAX + 1];
   
// Stores the prime
static ArrayList primes = new ArrayList();
   
// Function to generate all
// prime numbers
static void SieveOfEratosthenes()
{
    Array.Fill(isPrime, true);
   
    for(int p = 2; p * p <= MAX; p++)
    {
           
        // If current element is prime
        if (isPrime[p] == true)
        {
               
            // Set all its multiples non-prime
            for(int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
   
    // Store all prime numbers
    for(int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.Add(p);
}
   
// Function to find the closest
// prime to a particular number
static int prime_search(ArrayList primes,
                        int diff)
{
   
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.Count - 1;
   
    int res = -1;
   
    while (low <= high)
    {
        int mid = (low + high) / 2;
   
        // If the prime added makes
        // the elements equal
        if ((int)primes[mid] == diff)
        {
               
            // Return this as the
            // closest prime
            return (int)primes[mid];
        }
   
        // If the array remains
        // non-decreasing
        else if ((int)primes[mid] < diff)
        {
               
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
   
        // Otherwise
        else
        {
            res = (int)primes[mid];
   
            // Check if a smaller prime can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
   
    // Return closest number
    return res;
}
   
// Function to find the minimum cost
static int minCost(int []arr, int n)
{
       
    // Find all primes
    SieveOfEratosthenes();
   
    // Store the result
    int res = 0;
   
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
           
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1])
        {
            int diff = arr[i - 1] - arr[i];
   
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime = prime_search(primes,
                                             diff);
   
            // Add to overall cost
            res += closest_prime;
   
            // Update current element
            arr[i] += closest_prime;
        }
    }
   
    // Return the minimum cost
    return res;
}
  
// Driver Code
public static void Main(string[] args)
{
     
    // Given array
    int []arr = { 2, 1, 5, 4, 3 };
    int n = 5;
   
    // Function call
    Console.Write(minCost(arr, n));
}
}
 
// This code is contributed by rutvik_56
Javascript



<script>
 
// Javascript Program to implement
// the above approach
 
let MAX = 10000000;
 
// Stores if an index is a
// prime / non-prime value
let isPrime = new Array(MAX);
 
// Stores the prime
let primes = new Array();
 
// Function to generate all
// prime numbers
function SieveOfEratosthenes()
{
   isPrime.fill(true);
  
    for (let p = 2; p * p <= MAX; p++) {
 
        // If current element is prime
        if (isPrime[p] == true) {
 
            // Set all its multiples non-prime
            for (let i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
 
    // Store all prime numbers
    for (let p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.push(p);
}
 
// Function to find the closest
// prime to a particular number
function prime_search(primes, diff)
{
 
    // Applying binary search
    // on primes vector
    let low = 0;
    let high = primes.length - 1;
 
    let res = 0;
 
    while (low <= high) {
        let mid = Math.floor((low + high) / 2);
 
        // If the prime added makes
        // the elements equal
        if (primes[mid] == diff) {
 
            // Return this as the
            // closest prime
 
            return primes[mid];
        }
 
        // If the array remains
        // non-decreasing
        else if (primes[mid] < diff) {
 
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
 
        // Otherwise
        else {
 
            res = primes[mid];
 
            // Check if a smaller prime  can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
 
    // Return closest number
    return res;
}
 
// Function to find the minimum cost
function minCost(arr, n)
{
 
    // Find all primes
    SieveOfEratosthenes();
 
    // Store the result
    let res = 0;
 
    // Iterate over the array
    for (let i = 1; i < n; i++) {
 
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1]) {
            let diff = arr[i - 1] - arr[i];
 
            // Find the closest prime which
            // makes the array non decreasing
            let closest_prime
                = prime_search(primes, diff);
 
            // Add to overall cost
            res += closest_prime;
 
            // Update current element
            arr[i] += closest_prime;
        }
    }
 
    // Return the minimum cost
    return res;
}
 
// Driver Code
    // Given array
    let arr = [ 2, 1, 5, 4, 3 ];
    let n = 5;
 
    // Function Call
    document.write(minCost(arr, n))
 
 
 
// This code is contributed by gfgking
 
</script>
Output: 
7
Time Complexity: O(N*log(logN)) 
Auxiliary Space: O(N)

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Last Updated :
31 Jan, 2022
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