# Minimize insertions in Array to make ratio of each pair as K

• Last Updated : 21 Mar, 2023

Given an array arr[] of length N the task is to find the minimum number of elements to be added in the array such that there exists every independent pair in the array whose ratio is K.

Examples:

Input: arr[] = {1, 2, 2, 2, 4, 7}, K = 2
Output: 2
Explanation: 4, 14 can be added to make the array as {1, 2, 2, 2, 4, 7, 4, 14}
Therefore if we make independent pairs as {1, 2}, {2, 4}, {2, 4}, and {7, 14}, the ratio of each is K (=2).

Input: arr[] = {5, 2, 3, 5, 15}, K = 3
Output: 3

Approach: The above problem can be solved using greedy and hashing technique, based on below observation:

Since it is given that the ratio of the pairs should be K,

If the two elements be made as pair with ratio K are a and b, then
=> a/b = K
=> a = b*K
=> The two elements will be always b and b*K

Therefore, for the ratio of a and b to be K, b and b*K should be present in the array.

So in the Array, for every element arr[i],

• we need check if arr[i] * K is present in that array.
• If it is present, make it a pair
• Else add the arr[i] * K to the array

Follow these steps to solve the above problem:

• Initialize a hashmap m to store the frequencies of the elements in the array.
• Traverse the array and store the frequencies
• Sort the array arr[].
• Initialize the cnt_pairs to store the count of the available pairs with the ration k.
• Traverse through the sorted array and check for the pairing element.
• Check if the pair with ratio k is present for arr[i], consider them as a pair and remove them from the hashmap by decreasing the frequencies.
• Keep track of available pairs in the cnt_pairs variable by incrementing it by 2.
• Print the single elements by subtracting cnt_pairs from the n.

Below is the implementation of the above approach:

## C++

 `// C++ program to minimize insertions in``// Array to make ratio of each pair as K` `#include ``using` `namespace` `std;` `// Function to minimize insertions``// in Array to make ratio``// of each pair as K``int` `find_min_additions(``    ``int` `arr[], ``int` `n, ``int` `x)``{``    ``// Initialize a hashmap``    ``// and store the frequencies``    ``// of the array elements``    ``unordered_map<``int``, ``int``> m;` `    ``// Traverse the array``    ``// and store the frequencies``    ``for` `(``int` `i = 0; i < n; i++) {``        ``m[arr[i]]++;``    ``}` `    ``// sort the array``    ``sort(arr, arr + n);` `    ``// Initialize the cnt_pairs to store the``    ``// count of the available pairs``    ``int` `cnt_pairs = 0;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Check if the pair with ratio k is``        ``// present for arr[i]``        ``if` `(m[arr[i] * x] > 0``            ``&& m[arr[i]] > 0) {` `            ``// Consider them as a pair``            ``// and remove from the hashmap``            ``m[arr[i] * x]--;``            ``m[arr[i]]--;` `            ``// Add the count of the pairs``            ``cnt_pairs += 2;``        ``}``    ``}` `    ``// Return the count of single elements``    ``// that need another element``    ``// to make ratio k``    ``return` `n - cnt_pairs;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 2, 4, 7 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 2;` `    ``cout << find_min_additions(arr, n, K);``}`

## Java

 `// Java program to minimize insertions in``// Array to make ratio of each pair as K``import` `java.util.*;` `class` `GFG{` `  ``// Function to minimize insertions``  ``// in Array to make ratio``  ``// of each pair as K``  ``static` `int` `find_min_additions(``    ``int` `[]arr, ``int` `n, ``int` `x)``  ``{` `    ``// Initialize a hashmap``    ``// and store the frequencies``    ``// of the array elements``    ``HashMap m = ``new` `HashMap<>();`  `    ``// Traverse the array``    ``// and store the frequencies``    ``for` `(``int` `i = ``0``; i < n; i++) {``      ``int` `c = ``0``;``      ``if``(m.containsKey(arr[i])) {``        ``c = m.get(arr[i]);``      ``}``      ``m.put(arr[i], c + ``1``);``    ``}` `    ``// sort the array``    ``Arrays.sort(arr);` `    ``// Initialize the cnt_pairs to store the``    ``// count of the available pairs``    ``int` `cnt_pairs = ``0``;` `    ``for` `(``int` `i = ``0``; i < n; ++i) {` `      ``// Check if the pair with ratio k is``      ``// present for arr[i]``      ``if` `(m.containsKey(arr[i] * x) && m.get(arr[i] * x) > ``0``          ``&& m.get(arr[i]) > ``0``) {` `        ``// Consider them as a pair``        ``// and remove from the hashmap``        ``m.put(arr[i] * x, m.get(arr[i] * x) - ``1``);``        ``m.put(arr[i], m.get(arr[i]) - ``1``);` `        ``// Add the count of the pairs``        ``cnt_pairs += ``2``;``      ``}``    ``}` `    ``// Return the count of single elements``    ``// that need another element``    ``// to make ratio k``    ``return` `n - cnt_pairs;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main (String[] args) {``    ``int` `arr[] = { ``1``, ``2``, ``2``, ``2``, ``4``, ``7` `};``    ``int` `n = arr.length;``    ``int` `K = ``2``;` `    ``System.out.print(find_min_additions(arr, n, K));``  ``}``}` `// This code is contributed by hrithikgarg03188.`

## Python3

 `# Python program to minimize insertions in``# Array to make ratio of each pair as K` `# Function to minimize insertions``# in Array to make ratio``# of each pair as K``def` `find_min_additions(arr, n,  x):` `    ``# Initialize a hashmap``    ``# and store the frequencies``    ``# of the array elements``    ``m ``=` `{}` `    ``# Traverse the array``    ``# and store the frequencies``    ``for` `i ``in` `range``(``0``, n):``        ``m[arr[i]] ``=` `m[arr[i]] ``+` `1` `if` `arr[i] ``in` `m ``else` `1` `    ``# sort the array``    ``arr.sort()` `    ``# Initialize the cnt_pairs to store the``    ``# count of the available pairs``    ``cnt_pairs ``=` `0` `    ``for` `i ``in` `range``(``0``, n):` `        ``# Check if the pair with ratio k is``        ``# present for arr[i]``        ``if` `(arr[i] ``*` `x ``in` `m ``and` `arr[i] ``in` `m ``and` `m[arr[i] ``*` `x] > ``0``                ``and` `m[arr[i]] > ``0``):` `            ``# Consider them as a pair``            ``# and remove from the hashmap``            ``m[arr[i] ``*` `x] ``-``=` `1``            ``m[arr[i]] ``-``=` `1` `            ``# Add the count of the pairs``            ``cnt_pairs ``+``=` `2` `    ``# Return the count of single elements``    ``# that need another element``    ``# to make ratio k``    ``return` `n ``-` `cnt_pairs` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``2``, ``2``, ``2``, ``4``, ``7``]``    ``n ``=` `len``(arr)``    ``K ``=` `2` `    ``print``(find_min_additions(arr, n, K))` `# This code is contributed by rakeshsahni`

## C#

 `// C# program to minimize insertions in``// Array to make ratio of each pair as K` `using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG``{` `  ``// Function to minimize insertions``  ``// in Array to make ratio``  ``// of each pair as K``  ``static` `int` `find_min_additions(``    ``int` `[]arr, ``int` `n, ``int` `x)``  ``{` `    ``// Initialize a hashmap``    ``// and store the frequencies``    ``// of the array elements``    ``Dictionary<``int``, ``int``> m = ``      ``new` `Dictionary<``int``, ``int``>();` `    ``// Traverse the array``    ``// and store the frequencies``    ``for` `(``int` `i = 0; i < n; i++) {``      ``int` `c = 0;``      ``if``(m.ContainsKey(arr[i])) {``        ``c = m[arr[i]];``      ``}``      ``m[arr[i]] = c + 1;``    ``}` `    ``// sort the array``    ``Array.Sort(arr);` `    ``// Initialize the cnt_pairs to store the``    ``// count of the available pairs``    ``int` `cnt_pairs = 0;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `      ``// Check if the pair with ratio k is``      ``// present for arr[i]``      ``if` `(m.ContainsKey(arr[i] * x) && m[arr[i] * x] > 0``          ``&& m[arr[i]] > 0) {` `        ``// Consider them as a pair``        ``// and remove from the hashmap``        ``m[arr[i] * x]--;``        ``m[arr[i]]--;` `        ``// Add the count of the pairs``        ``cnt_pairs += 2;``      ``}``    ``}` `    ``// Return the count of single elements``    ``// that need another element``    ``// to make ratio k``    ``return` `n - cnt_pairs;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main()``  ``{``    ``int` `[]arr = { 1, 2, 2, 2, 4, 7 };``    ``int` `n = arr.Length;``    ``int` `K = 2;` `    ``Console.Write(find_min_additions(arr, n, K));``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

`2`

Time Complexity: O(N*log N) where N is the size of the array.
Space Complexity: O(N)

Another Approach:

1. Initialize an unordered_map m and a set s.                                                                                                                                                                           (a) m will store the frequencies of each element in the array.
(b) s will store the unique elements in the array.
(c) Traverse the array and store the frequencies of each element in the map m.
2. Traverse the set s, and for each element i, check if its pair with ratio K*i is present in the map m.
3. If the pair exists, decrement the frequencies of both elements in the map.
(a) Increment the cnt_pairs by 2, as two elements make a pair.
4. Return the count of elements that were not part of a pair, which is equal to the total number of elements minus the cnt_pairs.In the main function:
5. Initialize an array and its size.
Call the function find_min_additions() with the array, size, and a given value of K as arguments.
Print the output.

Below is the implementation of above approach:

## C++

 `// C++ program to minimize insertions in``// Array to make ratio of each pair as K` `#include ``using` `namespace` `std;` `// Function to minimize insertions``// in Array to make ratio``// of each pair as K``int` `find_min_additions(``int` `arr[], ``int` `n, ``int` `x)``{``// Initialize a hashmap``// and store the frequencies``// of the array elements``unordered_map<``int``, ``int``> m;``// Initialize a set to store``// the unique elements in the array``set<``int``> s;` `// Traverse the array``// and store the frequencies``for` `(``int` `i = 0; i < n; i++) {``    ``m[arr[i]]++;``    ``s.insert(arr[i]);``}` `// Initialize the cnt_pairs to store the``// count of the available pairs``int` `cnt_pairs = 0;` `// Traverse through the set and check``// for the pairing element``for` `(``auto` `i : s) {` `    ``// Check if the pair with ratio k is``    ``// present for i``    ``if` `(m[i * x] > 0 && m[i] > 0) {` `        ``// Consider them as a pair``        ``// and remove from the hashmap``        ``m[i * x]--;``        ``m[i]--;` `        ``// Add the count of the pairs``        ``cnt_pairs += 2;``    ``}``}` `// Return the count of single elements``// that need another element``// to make ratio k``return` `n - cnt_pairs;``}` `// Driver code``int` `main()``{``int` `arr[] = { 1, 2, 2, 2, 4, 7 };``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``int` `K = 2;``cout << find_min_additions(arr, n, K);``}`

## Javascript

 `// Javascript program to minimize insertions in``// Array to make ratio of each pair as K` `// Function to minimize insertions``// in Array to make ratio``// of each pair as K``function` `find_min_additions(arr, n, x) {``  ``// Initialize a hashmap``  ``// and store the frequencies``  ``// of the array elements``  ``const m = ``new` `Map();``  ``// Initialize a set to store``  ``// the unique elements in the array``  ``const s = ``new` `Set();` `  ``// Traverse the array``  ``// and store the frequencies``  ``for` `(let i = 0; i < n; i++) {``    ``const num = arr[i];``    ``m.set(num, (m.get(num) || 0) + 1);``    ``s.add(num);``  ``}` `  ``// Initialize the cnt_pairs to store the``  ``// count of the available pairs``  ``let cnt_pairs = 0;` `  ``// Traverse through the set and check``  ``// for the pairing element``  ``for` `(let num of s) {` `    ``// Check if the pair with ratio k is``    ``// present for i``    ``if` `(m.get(num * x) > 0 && m.get(num) > 0) {` `      ``// Consider them as a pair``      ``// and remove from the hashmap``      ``m.set(num * x, m.get(num * x) - 1);``      ``m.set(num, m.get(num) - 1);` `      ``// Add the count of the pairs``      ``cnt_pairs += 2;``    ``}``  ``}` `  ``// Return the count of single elements``  ``// that need another element``  ``// to make ratio k``  ``return` `n - cnt_pairs;``}` `// Driver code``const arr = [1, 2, 2, 2, 4, 7];``const n = arr.length;``const K = 2;``console.log(find_min_additions(arr, n, K));` `// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL`

## C#

 `// C# program to minimize insertions in``// Array to make ratio of each pair as K` `using` `System;``using` `System.Collections.Generic;` `class` `GFG {``    ``// Function to minimize insertions``    ``// in Array to make ratio``    ``// of each pair as K``    ``static` `int` `FindMinAdditions(``int``[] arr, ``int` `n, ``int` `x)``    ``{``        ``// Initialize a dictionary``        ``// and store the frequencies``        ``// of the array elements``        ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();` `        ``// Initialize a set to store``        ``// the unique elements in the array``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();` `        ``// Traverse the array``        ``// and store the frequencies``        ``for` `(``int` `i = 0; i < n; i++) {``            ``// Check if the pair with ratio k is``            ``// present for i``            ``if` `(m.ContainsKey(arr[i]))``                ``m[arr[i]]++;``            ``else``                ``m[arr[i]] = 1;` `            ``s.Add(arr[i]);``        ``}` `        ``// Initialize the cnt_pairs to store the``        ``// count of the available pairs``        ``int` `cnt_pairs = 0;` `        ``// Traverse through the set and check``        ``// for the pairing element``        ``foreach``(``int` `i ``in` `s)``        ``{``            ``if` `(m.ContainsKey(i * x) && m[i * x] > 0``                ``&& m[i] > 0) {``                ``m[i * x]--;``                ``m[i]--;``                ``cnt_pairs += 2;``            ``}``        ``}` `        ``return` `n - cnt_pairs;``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``int``[] arr = { 1, 2, 2, 2, 4, 7 };``        ``int` `n = arr.Length;``        ``int` `K = 2;` `        ``// Function Call``        ``Console.WriteLine(FindMinAdditions(arr, n, K));``    ``}``}`

## Java

 `// Java program to minimize insertions in``// Array to make ratio of each pair as K` `import` `java.util.*;` `public` `class` `GFG {``    ``// Function to minimize insertions``    ``// in Array to make ratio``    ``// of each pair as K``    ``public` `static` `int` `findMinAdditions(``int``[] arr, ``int` `n, ``int` `x) {``        ``// Initialize a hashmap``        ``// and store the frequencies``        ``// of the array elements``        ``HashMap m = ``new` `HashMap();``        ``// Initialize a set to store``        ``// the unique elements in the array``        ``HashSet s = ``new` `HashSet();` `        ``// Traverse the array``        ``// and store the frequencies``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(m.containsKey(arr[i])) {``                ``m.put(arr[i], m.get(arr[i]) + ``1``);``            ``} ``else` `{``                ``m.put(arr[i], ``1``);``            ``}``            ``s.add(arr[i]);``        ``}` `        ``// Initialize the cnt_pairs to store the``        ``// count of the available pairs``        ``int` `cntPairs = ``0``;` `        ``// Traverse through the set and check``        ``// for the pairing element``        ``for` `(``int` `i : s) {``            ``// Check if the pair with ratio k is``            ``// present for i``            ``if` `(m.containsKey(i * x) && m.containsKey(i) && m.get(i * x) > ``0` `&& m.get(i) > ``0``) {``                ``// Consider them as a pair``                ``// and remove from the hashmap``                ``m.put(i * x, m.get(i * x) - ``1``);``                ``m.put(i, m.get(i) - ``1``);` `                ``// Add the count of the pairs``                ``cntPairs += ``2``;``            ``}``        ``}` `        ``// Return the count of single elements``        ``// that need another element``        ``// to make ratio k``        ``return` `n - cntPairs;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = { ``1``, ``2``, ``2``, ``2``, ``4``, ``7` `};``        ``int` `n = arr.length;``        ``int` `k = ``2``;``        ``System.out.println(findMinAdditions(arr, n, k));``    ``}``}`

## Python3

 `# Python3 program to minimize insertions in``# Array to make ratio of each pair as K` `# Python program to minimize insertions in``# Array to make ratio of each pair as K` `from` `collections ``import` `defaultdict` `# Function to minimize insertions``# in Array to make ratio``# of each pair as K``def` `find_min_additions(arr, n, x):``    ``# Initialize a dictionary``    ``# and store the frequencies``    ``# of the array elements``    ``m ``=` `defaultdict(``int``)``    ``# Initialize a set to store``    ``# the unique elements in the array``    ``s ``=` `set``()` `    ``# Traverse the array``    ``# and store the frequencies``    ``for` `i ``in` `range``(n):``        ``m[arr[i]] ``+``=` `1``        ``s.add(arr[i])` `    ``# Initialize the cnt_pairs to store the``    ``# count of the available pairs``    ``cnt_pairs ``=` `0` `    ``# Traverse through the set and check``    ``# for the pairing element``    ``for` `i ``in` `s:``        ``# Check if the pair with ratio k is``        ``# present for i``        ``if` `m[i ``*` `x] > ``0` `and` `m[i] > ``0``:``            ``# Consider them as a pair``            ``# and remove from the dictionary``            ``m[i ``*` `x] ``-``=` `1``            ``m[i] ``-``=` `1` `            ``# Add the count of the pairs``            ``cnt_pairs ``+``=` `2` `    ``# Return the count of single elements``    ``# that need another element``    ``# to make ratio k``    ``return` `n ``-` `cnt_pairs` `# Driver code``arr ``=` `[``1``, ``2``, ``2``, ``2``, ``4``, ``7``]``n ``=` `len``(arr)``K ``=` `2``print``(find_min_additions(arr, n, K))`

Output

`2`

Time Complexity: O(N*logN)

Auxiliary Space: O(N)

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