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 = 6
Output: 15
Explanation: One will need to pick all the socks to get at least 6 pairs of matching socks.Input: arr[] = {4, 5, 6}, K = 3
Output: 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 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
// Java program for the above approachimport 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 approachusing 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 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. |
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!