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Minimum removals from array to make GCD greater

  • Difficulty Level : Medium
  • Last Updated : 28 May, 2021

Given N numbers, the task is to find the minimum removal of numbers such that the GCD of the remaining numbers is greater than the initial GCD of N numbers. If it is not possible to increase the GCD, print “NO”. 

Examples: 

Input: a[] = {1, 2, 4} 
Output: 1 
Remove the first element, then the new GCD is 2, which is greater than the initial GCD i.e., 1. 

Input: a[] = {6, 9, 15, 30} 
Output: 2 
The initial gcd is 3, remove 6 and 9 to obtain a gcd of 15 which is greater than 3. You can also remove 9 and 15 to get a gcd of 6. 

The following steps are followed to solve the above problem: 



  1. Initially find the gcd of N numbers using Euclidean algorithms.
  2. Divide all numbers by the obtained gcd.
  3. Using prime factorization for multiple queries method, find the prime factorization of every number in O(log N). The method can be read here.
  4. Insert all the prime factors in the set to remove duplicates that are obtained using this method.
  5. Using a hash-map, count the frequencies of the prime factors in every i-th element.
  6. Once the factorization of numbers has been done, and the count has been stored in the frequency table, iterate in the hash-map and find out the prime factor which occurs the maximum number of times. It cannot be N, as we have already divided the array elements initially by the initial gcd of N numbers.
  7. The number of removals will always be n-(hash[prime_factor]) if there are any such factors after the division of initial gcd.

Below is the implementation of the above approach. 

C++




// C++ program to find the minimum removals
// such that gcd of remaining numbers is more
// than the initial gcd of N numbers
#include <bits/stdc++.h>
using namespace std;
 
#define MAXN 100001
 
// stores smallest prime factor for every number
int spf[MAXN];
 
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    spf[1] = 1;
    for (int i = 2; i < MAXN; i++)
 
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
 
    // separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
 
    for (int i = 3; i * i < MAXN; i++) {
 
        // checking if i is prime
        if (spf[i] == i) {
 
            // marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
 
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
vector<int> getFactorization(int x)
{
    vector<int> ret;
    while (x != 1) {
        ret.push_back(spf[x]);
        x = x / spf[x];
    }
    return ret;
}
 
// Function which returns the minimal
// removals required to make gcd
// greater than previous
int minimumRemovals(int a[], int n)
{
    int g = 0;
 
    // finding initial gcd
    for (int i = 0; i < n; i++)
        g = __gcd(a[i], g);
 
    unordered_map<int, int> mpp;
 
    // divides all number by initial gcd
    for (int i = 0; i < n; i++)
        a[i] = a[i] / g;
 
    // iterating for all numbers
    for (int i = 0; i < n; i++) {
 
        // primt factorisation to get the prime
        // factors of i-th element in the array
        vector<int> p = getFactorization(a[i]);
        set<int> s;
 
        // insert all the prime factors in
        // set to remove duplicates
        for (int j = 0; j < p.size(); j++) {
            s.insert(p[j]);
        }
 
        /// increase the count of prime
        // factor in map for every element
        for (auto it = s.begin(); it != s.end(); it++) {
            int el = *it;
            mpp[el] += 1;
        }
    }
 
    int mini = INT_MAX;
    int mini1 = INT_MAX;
 
    // iterate in map and check for every factor
    // and its count
    for (auto it = mpp.begin(); it != mpp.end(); it++) {
        int fir = it->first;
        int sec = it->second;
 
        // check for the largest appearing factor
        // which does not appears in any one or more
        if ((n - sec) <= mini) {
            mini = n - sec;
        }
    }
    if (mini != INT_MAX)
        return mini;
    else
        return -1;
}
 
// Driver code
int main()
{
    int a[] = { 6, 9, 15, 30 };
    int n = sizeof(a) / sizeof(a[0]);
    sieve();
    cout << minimumRemovals(a, n);
    return 0;
}

Java




// Java program to find the minimum removals
// such that gcd of remaining numbers is more
// than the initial gcd of N numbers
import java.util.*;
 
class GFG{
 
static final int MAXN = 100001;
 
// stores smallest prime factor for every number
static int []spf = new int[MAXN];
 
// Calculating SPF (Smallest Prime Factor)
// for every number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
    spf[1] = 1;
    for(int i = 2; i < MAXN; i++)
     
        // Marking smallest prime factor
        // for every number to be itself
        spf[i] = i;
 
    // Separately marking spf for every even
    // number as 2
    for(int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
 
    for(int i = 3; i * i < MAXN; i++)
    {
         
        // Checking if i is prime
        if (spf[i] == i)
        {
             
            // Marking SPF for all numbers
            // divisible by i
            for(int j = i * i; j < MAXN; j += i)
             
                // Marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
static Vector<Integer> getFactorization(int x)
{
    Vector<Integer> ret = new Vector<>();
    while (x != 1)
    {
        ret.add(spf[x]);
        x = x / spf[x];
    }
    return ret;
}
 
// Recursive function to return gcd of a and b 
static int __gcd(int a, int b) 
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Function which returns the minimal
// removals required to make gcd
// greater than previous
static int minimumRemovals(int a[], int n)
{
    int g = 0;
 
    // Finding initial gcd
    for(int i = 0; i < n; i++)
        g = __gcd(a[i], g);
 
    HashMap<Integer, Integer> mpp = new HashMap<>();
 
    // Divides all number by initial gcd
    for(int i = 0; i < n; i++)
        a[i] = a[i] / g;
 
    // Iterating for all numbers
    for(int i = 0; i < n; i++)
    {
         
        // Prime factorisation to get the prime
        // factors of i-th element in the array
        Vector<Integer> p = getFactorization(a[i]);
        HashSet<Integer> s = new HashSet<>();
 
        // Insert all the prime factors in
        // set to remove duplicates
        for(int j = 0; j < p.size(); j++)
        {
            s.add(p.get(j));
        }
 
        // Increase the count of prime
        // factor in map for every element
        for(int it: s)
        {
            int el = it;
            if (mpp.containsKey(el))
            {
                mpp.put(el, mpp.get(el) + 1);
            }
            else
            {
                mpp.put(el, 1);
            }
        }
    }
 
    int mini = Integer.MAX_VALUE;
    int mini1 = Integer.MAX_VALUE;
 
    // Iterate in map and check for
    // every factor and its count
    for(Map.Entry<Integer, Integer> it : mpp.entrySet())
    {
        int fir = it.getKey();
        int sec = it.getValue();
 
        // Check for the largest appearing factor
        // which does not appears in any one or more
        if ((n - sec) <= mini)
        {
            mini = n - sec;
        }
    }
    if (mini != Integer.MAX_VALUE)
        return mini;
    else
        return -1;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 6, 9, 15, 30 };
    int n = a.length;
     
    sieve();
     
    System.out.print(minimumRemovals(a, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program to find the minimum removals
# such that gcd of remaining numbers is more
# than the initial gcd of N numbers
from math import gcd as __gcd
 
MAXN = 100001
 
# stores smallest prime factor for every number
spf = [i for i in range(MAXN)]
 
# Calculating SPF (Smallest Prime Factor) for every
# number till MAXN.
# Time Complexity : O(nloglogn)
def sieve():
 
    # separately marking spf for every even
    # number as 2
    for i in range(4, MAXN, 2):
        spf[i] = 2
 
    for i in range(3, MAXN):
 
        if i * i > MAXN:
            break
 
        # checking if i is prime
        if (spf[i] == i):
 
            # marking SPF for all numbers divisible by i
            for j in range(2 * i, MAXN, i):
 
                # marking spf[j] if it is not
                # previously marked
                if (spf[j] == j):
                    spf[j] = i
 
# A O(log n) function returning primefactorization
# by dividing by smallest prime factor at every step
def getFactorization(x):
    ret = []
    while (x != 1):
        ret.append(spf[x])
        x = x // spf[x]
    return ret
 
# Function which returns the minimal
# removals required to make gcd
# greater than previous
def minimumRemovals(a, n):
    g = 0
 
    # finding initial gcd
    for i in range(n):
        g = __gcd(a[i], g)
 
    mpp = dict()
 
    # divides all number by initial gcd
    for i in range(n):
        a[i] = a[i] // g
 
    # iterating for all numbers
    for i in range(n):
 
        # primt factorisation to get the prime
        # factors of i-th element in the array
        p = getFactorization(a[i])
        s = dict()
 
        # insert all the prime factors in
        # set to remove duplicates
        for j in range(len(p)):
            s[p[j]] = 1
 
        # increase the count of prime
        # factor in map for every element
        for i in s:
            mpp[i] = mpp.get(i, 0) + 1
 
    mini = 10**9
    mini1 = 10**9
 
    # iterate in map and check for every factor
    # and its count
    for i in mpp:
        fir = i
        sec = mpp[i]
 
        # check for the largest appearing factor
        # which does not appears in any one or more
        if ((n - sec) <= mini):
            mini = n - sec
 
    if (mini != 10**9):
        return mini
    else:
        return -1
 
# Driver code
a = [6, 9, 15, 30]
n = len(a)
sieve()
print(minimumRemovals(a, n))
 
# This code is contributed by mohit kumar 29

C#




// C# program to find the minimum
// removals such that gcd of remaining
// numbers is more than the initial
// gcd of N numbers
using System;
using System.Collections.Generic;
class GFG{
 
static readonly int MAXN = 100001;
 
// stores smallest prime
// factor for every number
static int []spf =
       new int[MAXN];
 
// Calculating SPF (Smallest
// Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
  spf[1] = 1;
  for(int i = 2; i < MAXN; i++)
 
    // Marking smallest prime factor
    // for every number to be itself
    spf[i] = i;
 
  // Separately marking spf for
  // every even number as 2
  for(int i = 4; i < MAXN; i += 2)
    spf[i] = 2;
 
  for(int i = 3; i * i < MAXN; i++)
  {
    // Checking if i is prime
    if (spf[i] == i)
    {
      // Marking SPF for all numbers
      // divisible by i
      for(int j = i * i;
              j < MAXN; j += i)
 
        // Marking spf[j] if it is
        // not previously marked
        if (spf[j] == j)
          spf[j] = i;
    }
  }
}
 
// A O(log n) function returning
// primefactorization by dividing
// by smallest prime factor at
// every step
static List<int> getFactorization(int x)
{
  List<int> ret = new List<int>();
   
  while (x != 1)
  {
    ret.Add(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
 
// Recursive function to
// return gcd of a and b 
static int __gcd(int a,
                 int b) 
  return b == 0 ? a :
         __gcd(b, a % b);    
}
 
// Function which returns the
// minimal removals required
// to make gcd greater than
// previous
static int minimumRemovals(int []a,
                           int n)
{
  int g = 0;
 
  // Finding initial gcd
  for(int i = 0; i < n; i++)
    g = __gcd(a[i], g);
 
  Dictionary<int,
             int> mpp = new Dictionary<int,
                                       int>();
 
  // Divides all number by
  // initial gcd
  for(int i = 0; i < n; i++)
    a[i] = a[i] / g;
 
  // Iterating for all numbers
  for(int i = 0; i < n; i++)
  {
    // Prime factorisation to get the prime
    // factors of i-th element in the array
    List<int> p = getFactorization(a[i]);
    HashSet<int> s = new HashSet<int>();
 
    // Insert all the prime factors in
    // set to remove duplicates
    for(int j = 0; j < p.Count; j++)
    {
      s.Add(p[j]);
    }
 
    // Increase the count of prime
    // factor in map for every
    // element
    foreach(int it in s)
    {
      int el = it;
      if (mpp.ContainsKey(el))
      {
        mpp[el]= mpp[el] + 1;
      }
      else
      {
        mpp.Add(el, 1);
      }
    }
  }
 
  int mini = int.MaxValue;
  int mini1 = int.MaxValue;
 
  // Iterate in map and check for
  // every factor and its count
  foreach(KeyValuePair<int,
                       int> it in mpp)
  {
    int fir = it.Key;
    int sec = it.Value;
 
    // Check for the largest appearing
    // factor which does not appears
    // in any one or more
    if ((n - sec) <= mini)
    {
      mini = n - sec;
    }
  }
  if (mini != int.MaxValue)
    return mini;
  else
    return -1;
}
 
// Driver code
public static void Main(String[] args)
{
  int []a = {6, 9, 15, 30};
  int n = a.Length;
  sieve();
  Console.Write(minimumRemovals(a, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// Javascript program to find the minimum removals
// such that gcd of remaining numbers is more
// than the initial gcd of N numbers
 
let  MAXN = 100001;
// stores smallest prime factor for every number
let spf = new Array(MAXN);
 
// Calculating SPF (Smallest Prime Factor)
// for every number till MAXN.
// Time Complexity : O(nloglogn)
function sieve()
{
    spf[1] = 1;
    for(let i = 2; i < MAXN; i++)
      
        // Marking smallest prime factor
        // for every number to be itself
        spf[i] = i;
  
    // Separately marking spf for every even
    // number as 2
    for(let i = 4; i < MAXN; i += 2)
        spf[i] = 2;
  
    for(let i = 3; i * i < MAXN; i++)
    {
          
        // Checking if i is prime
        if (spf[i] == i)
        {
              
            // Marking SPF for all numbers
            // divisible by i
            for(let j = i * i; j < MAXN; j += i)
              
                // Marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
function getFactorization(x)
{
    let ret = [];
    while (x != 1)
    {
        ret.push(spf[x]);
        x = x / spf[x];
    }
    return ret;
}
 
// Recursive function to return gcd of a and b    
function __gcd(a,b)
{
    return b == 0 ? a : __gcd(b, a % b);
}
 
// Function which returns the minimal
// removals required to make gcd
// greater than previous
function minimumRemovals(a,n)
{
    let g = 0;
  
    // Finding initial gcd
    for(let i = 0; i < n; i++)
        g = __gcd(a[i], g);
  
    let mpp = new Map();
  
    // Divides all number by initial gcd
    for(let i = 0; i < n; i++)
        a[i] = a[i] / g;
  
    // Iterating for all numbers
    for(let i = 0; i < n; i++)
    {
          
        // Prime factorisation to get the prime
        // factors of i-th element in the array
        let p = getFactorization(a[i]);
        let s = new Set();
  
        // Insert all the prime factors in
        // set to remove duplicates
        for(let j = 0; j < p.length; j++)
        {
            s.add(p[j]);
        }
  
        // Increase the count of prime
        // factor in map for every element
        for(let it of s.values())
        {
            let el = it;
            if (mpp.has(el))
            {
                mpp.set(el, mpp.get(el) + 1);
            }
            else
            {
                mpp.set(el, 1);
            }
        }
    }
  
    let mini = Number.MAX_VALUE;
    let mini1 = Number.MAX_VALUE;
  
    // Iterate in map and check for
    // every factor and its count
    for(let [key, value] of mpp.entries())
    {
        let fir = key;
        let sec = value;
  
        // Check for the largest appearing factor
        // which does not appears in any one or more
        if ((n - sec) <= mini)
        {
            mini = n - sec;
        }
    }
    if (mini != Number.MAX_VALUE)
        return mini;
    else
        return -1;
}
 
// Driver code
let a = [6, 9, 15, 30];
let n = a.length;
sieve();
document.write(minimumRemovals(a, n));
    
// This code is contributed by unknown2108
</script>
Output: 
2

 

Time Complexity: O(log log N) for precalculation of Sieve, and O(N * log N) for calculation. 
Auxiliary Space: O(N)

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