Minimize sum of prime numbers added to make an array non-decreasing
Last Updated :
31 Jan, 2022
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:
- Initialize a variable to store the minimum cost, say res.
- Generate and store all prime numbers up to 107 using the Sieve of Eratosthenes.
- 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.
- 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 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 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# 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
<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>
Time Complexity: O(N*log(logN))
Auxiliary Space: O(N)
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