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++

`// C++ program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to count the minimum` `// number of socks to be picked` `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;` `}` `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

`// 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#

`// 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.` |

## Python3

`# Python program for the above approach` `# Function to count the minimum` `# number of socks to be picked` `def` `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 ` `=` `3` `print` `(findMin(arr, ` `3` `, k));` ` ` `# This code is contributed by SoumikMondal.` |

## Javascript

`<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>` |

**Output:**

8

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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