Introduction

The Pashmak and Flowers problem on Codeforces involves distributing flowers in such a way that the total number of pairs of flowers with the same color is maximized. Given a set of flower colors represented by integers, the task is to find the minimum number of flowers that need to be replaced to achieve this goal.

Problem Statement

In this problem, we are given a list of integers representing the colors of flowers. The goal is to determine how many flowers need to be replaced so that the total number of pairs of flowers with the same color is maximized.

Example

Input:
6
1 2 2 3 3 3

Output:
4

In the example above, we have 6 flowers, and the colors are [1, 2, 2, 3, 3, 3]. The goal is to minimize the number of flowers to be replaced to maximize the number of flower pairs of the same color. In this case, the output is 4, as replacing some flowers allows us to maximize the number of pairs with matching colors.

Approach

The idea is to sort the flowers based on their colors and then check how many of each flower color we have. We will try to match flowers that have fewer occurrences to those that have more. The process is optimized by counting how many flowers need to be swapped to form pairs of identical flowers, while minimizing replacements.

Steps to Solve:

  1. Read the input values, including the number of flowers and their respective colors.
  2. Sort the array of flower colors in descending order.
  3. Check each flower color, and attempt to maximize the number of pairs by swapping colors with fewer occurrences to those with more.
  4. Calculate the total number of replacements required to achieve the optimal configuration of flower pairs.

Code Explanation

Let's break down the C++ code that solves the problem.

#include<bits/stdc++.h>

using namespace std;

void bubbleSort(int *a, int n) {
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (a[i] < a[j]) {
                swap(a[i], a[j]);
            }
        }
    }
}

int main() {
    int n, x, s1 = 0, s2 = 0, s3 = 0, cnt = 0;
    cin >> n;
    int a[n + 10];
    
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    sort(a, a + n, greater());
    
    for (int i = 0; i < n; i++) {
        if (a[i] == 4) {
            cnt++;
        }
        else if (a[i] == 3) {
            cnt++;
            s3++;
        }
        else if (a[i] == 2) {
            if (s2 > 0) {
                s2--;
            }
            else {
                s2++;
                cnt++;
            }
        }
        else if (a[i] == 1) {
            if (s3 > 0) s3--;
            else if (s2 > 0) {
                s2--;
                s3++;
            }
            else if (s1 > 0) {
                s2++;
                s1--;
            }
            else {
                s1++;
                cnt++;
            }
        }
    }
    cout << cnt << endl;
}

In this code, we begin by reading the input values and storing the number of flowers and their respective colors in an array. The array is then sorted in descending order to ensure that we try to match flowers with fewer occurrences first. The code goes through the array, and depending on the flower's color, it counts how many flowers need to be replaced in order to maximize pairs of identical flowers.

Time Complexity

The time complexity of the solution is O(n log n) due to the sorting step. Sorting the array takes O(n log n) time, where n is the number of flowers. The rest of the operations, such as iterating through the array and counting replacements, take linear time O(n).

Space Complexity

The space complexity is O(n) because we use an array of size n to store the flower colors. Other variables used for counting and tracking replacements require constant space.

Conclusion

The "Pashmak and Flowers" problem is a good exercise in sorting and counting, with the goal of minimizing replacements to maximize the number of pairs of identical flowers. By sorting the colors and making smart replacements, we can efficiently solve the problem in O(n log n) time.

Related Problems for Practice