Minimum number of socks required to picked to have at least K pairs of the same color

05 May, 2021
Given an array arr[] consisting of N integers such that arr[i] representing the number of socks of the color i and an integer K, the task is to find the minimum number of socks required to be picked to get at least K pairs of socks of the same color.
Examples:
Input: arr[] = {3, 4, 5, 3}, K = 6Output: 15Explanation: One will need to pick all the socks to get at least 6 pairs of matching socks.
Input: arr[] = {4, 5, 6}, K = 3Output: 8
Approach: The given problem can be solved based on the following observations: 
According to Pigeonhole’s Principle i.e., in the worst-case scenario if N socks of different colors have been picked then the next pick will form a matching pair of socks.
Suppose one has picked N socks of different colors then, for each (K – 1) pairs one will need to pick two socks, one for forming a pair and another for maintaining N socks of all different colors, and for the last pair, there is only need to pick a single sock of any color available.
Therefore, the idea is to find the total number of pairs that can be formed by the same colors and if the total count is at most K then print (2*K + N – 1) as the minimum count of pairs to be picked. Otherwise, print “-1” as there are not enough socks to formed K pairs.
Below is the implementation of the above approach:
C++
Java
C#
Python3
Javascript
// C++ program for the above approach #include <iostream>using namespace std; // Function to count the minimum// number of socks to be pickedint findMin(int arr[], int N, int k){    // Stores the total count    // of pairs of socks    int pairs = 0;     // Find the total count of pairs    for (int i = 0; i < N; i++) {        pairs += arr[i] / 2;    }     // If K is greater than pairs    if (k > pairs)        return -1;     // Otherwise    else        return 2 * k + N - 1;} int main(){    int arr[3] = { 4, 5, 6 };    int K = 3;    cout << findMin(arr, 3, K);    return 0;} // This code is contributed by RohitOberoi.
// Java program for the above approach import java.io.*; class GFG {     // Function to count the minimum    // number of socks to be picked    public static int findMin(        int[] arr, int N, int k)    {        // Stores the total count        // of pairs of socks        int pairs = 0;         // Find the total count of pairs        for (int i = 0; i < N; i++) {            pairs += arr[i] / 2;        }         // If K is greater than pairs        if (k > pairs)            return -1;         // Otherwise        else            return 2 * k + N - 1;    }     // Driver Code    public static void main(String[] args)    {        int[] arr = { 4, 5, 6 };        int K = 3;        int N = arr.length;        System.out.println(findMin(arr, N, K));    }}
// C# program for the above approach using System; class GFG {     // Function to count the minimum    // number of socks to be picked    public static int findMin(int[] arr, int N, int k)    {        // Stores the total count        // of pairs of socks        int pairs = 0;         // Find the total count of pairs        for (int i = 0; i < N; i++) {            pairs += arr[i] / 2;        }         // If K is greater than pairs        if (k > pairs)            return -1;         // Otherwise        else            return 2 * k + N - 1;    }     // Driver Code    public static void Main(string[] args)    {        int[] arr = { 4, 5, 6 };        int K = 3;        int N = arr.Length;        Console.WriteLine(findMin(arr, N, K));    }} // This code is contributed by ukasp.
# Python program for the above approach # Function to count the minimum# number of socks to be pickeddef findMin(arr, N, k):       # Stores the total count    # of pairs of socks    pairs = 0         # Find the total count of pairs    for i in range(N):        pairs += arr[i] / 2         # If K is greater than pairs    if (k > pairs):        return -1     # Otherwise    else:        return 2 * k + N - 1         arr = [4, 5, 6]k = 3print(findMin(arr, 3, k));     # This code is contributed by SoumikMondal.
<script> // JavaScript program to implement// the above approach     // Function to count the minimum    // number of socks to be picked    function findMin(        arr, N, k)    {        // Stores the total count        // of pairs of socks        let pairs = 0;          // Find the total count of pairs        for (let i = 0; i < N; i++) {            pairs += arr[i] / 2;        }          // If K is greater than pairs        if (k > pairs)            return -1;          // Otherwise        else            return 2 * k + N - 1;    } // Driver code         let arr = [ 4, 5, 6 ];        let K = 3;        let N = arr.length;        document.write(findMin(arr, N, K));            </script>
8
 
Time Complexity: O(N)Auxiliary Space: O(1)
RohitOberoi
ukasp
SoumikMondal
souravghosh0416
Pigeonhole Principle
Arrays
Greedy
Mathematical
Arrays
Greedy
Mathematical
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