Given an integer **N** which is the number of villagers who need to cross a river but there is only one boat on which a maximum of **2** person can travel. Each person **i** has to pay some specific price **P _{i}** to travel alone in the boat. If two person

**i, j**travel in the boat then they have to pay

**max(P**. The task is to find the minimum amount all the villagers have to pay to cross the river.

_{i}, P_{j})**Examples:**

Input:Price[] = {30, 40, 60, 70}

Output:220

P_{1}and P_{2}go together (which costs 40)

and P_{1}comes back (total cost 70 now).

Now P_{3}and P_{4}go (total cost 140) and

P_{2}comes back (total cost 180) and

finally P_{1}and P_{2}go together (total cost 220).

Input:Price[] = {892, 124}

Output:892

**Approach:** There are two ways for the two most costly person to cross the river:

- Both of them cross the river with the cheapest person turn by turn. So, the total cost will be the cost of the two costly persons + 2 * (cost of the cheapest person) (due to coming back).
- The two cheapest person cross the river and the cheapest person comes back. Now, the 2 most costly person cross the river and 2nd cheapest person comes back. So, the total cost will be the cost of the cheapest and costliest person plus 2 * cost of the second cheapest person.
- Total cost will be the minimum of the above two ways.

Let’s consider the example we used above to understand the approach:

P_{1}= 30, P_{2}= 40, P_{3}= 60, P_{4}= 70According to first method, P

_{4}goes with P_{1}and P_{1}comes back (cost is P_{4}+P_{1}).

Now, P_{3}goes with P_{1}and P_{1}comes back (cost for this ride will be P_{4}+P_{1}).

So, total cost for sending two most costly person according to method 1 is P_{4}+2*P_{1}+P_{3}= 190According to second method, P

_{2}goes with P_{1}and P_{1}comes back (cost is P_{2}+P_{1}).

Now, P_{3}goes with P_{4}and P_{2}comes back (cost for this ride will be P_{4}+P_{2}).

So, total cost for sending two most costly person according to method 2 is P_{4}+2*P_{2}+P_{1}= 180Hence, cost for sending P

_{3}and P_{4}will be minimum of 2 methods, i.e., 180.Now, we are left with P

_{1}and P_{2}whom we have to send together and cost will

be P_{2}= 40.So, total cost for travelling is 180 + 40 = 220.

Below is the implementation of the above approach:

## CPP

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `#define ll long long int ` ` ` `// Function to return the minimum cost ` `int` `minimumCost(ll price[], ` `int` `n) ` `{ ` ` ` ` ` `// Sort the price array ` ` ` `sort(price, price + n); ` ` ` `ll totalCost = 0; ` ` ` ` ` `// Calcualte minimum price ` ` ` `// of n-2 most costly person ` ` ` `for` `(` `int` `i = n - 1; i > 1; i -= 2) { ` ` ` `if` `(i == 2) { ` ` ` `totalCost += price[2] + price[0]; ` ` ` `} ` ` ` `else` `{ ` ` ` ` ` `// Both the ways as discussed above ` ` ` `ll price_first = price[i] + price[0] + 2 * price[1]; ` ` ` `ll price_second = price[i] + price[i - 1] + 2 * price[0]; ` ` ` `totalCost += min(price_first, price_second); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Calculate the minimum price ` ` ` `// of the two cheapest person ` ` ` `if` `(n == 1) { ` ` ` `totalCost += price[0]; ` ` ` `} ` ` ` `else` `{ ` ` ` `totalCost += price[1]; ` ` ` `} ` ` ` ` ` `return` `totalCost; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `ll price[] = { 30, 40, 60, 70 }; ` ` ` `int` `n = ` `sizeof` `(price) / ` `sizeof` `(price[0]); ` ` ` ` ` `cout << minimumCost(price, n); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the minimum cost ` ` ` `static` `long` `minimumCost(` `long` `price[], ` `int` `n) ` ` ` `{ ` ` ` ` ` `// Sort the price array ` ` ` `Arrays.sort(price); ` ` ` ` ` `long` `totalCost = ` `0` `; ` ` ` ` ` `// Calcualte minimum price ` ` ` `// of n-2 most costly person ` ` ` `for` `(` `int` `i = n - ` `1` `; i > ` `1` `; i -= ` `2` `) ` ` ` `{ ` ` ` `if` `(i == ` `2` `) ` ` ` `{ ` ` ` `totalCost += price[` `2` `] + price[` `0` `]; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` ` ` `// Both the ways as discussed above ` ` ` `long` `price_first = price[i] + price[` `0` `] + ` `2` `* price[` `1` `]; ` ` ` `long` `price_second = price[i] + price[i - ` `1` `] + ` `2` `* price[` `0` `]; ` ` ` `totalCost += Math.min(price_first, price_second); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Calculate the minimum price ` ` ` `// of the two cheapest person ` ` ` `if` `(n == ` `1` `) ` ` ` `{ ` ` ` `totalCost += price[` `0` `]; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `totalCost += price[` `1` `]; ` ` ` `} ` ` ` ` ` `return` `totalCost; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `long` `price[] = { ` `30` `, ` `40` `, ` `60` `, ` `70` `}; ` ` ` `int` `n = price.length; ` ` ` ` ` `System.out.println(minimumCost(price, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

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## Python

`# Python3 implementation of the approach ` ` ` `# Function to return the minimum cost ` `def` `minimumCost(price, n): ` ` ` ` ` `# Sort the price array ` ` ` `price ` `=` `sorted` `(price) ` ` ` `totalCost ` `=` `0` ` ` ` ` `# Calcualte minimum price ` ` ` `# of n-2 most costly person ` ` ` `for` `i ` `in` `range` `(n ` `-` `1` `, ` `1` `, ` `-` `2` `): ` ` ` `if` `(i ` `=` `=` `2` `): ` ` ` `totalCost ` `+` `=` `price[` `2` `] ` `+` `price[` `0` `] ` ` ` ` ` `else` `: ` ` ` ` ` `# Both the ways as discussed above ` ` ` `price_first ` `=` `price[i] ` `+` `price[` `0` `] ` `+` `2` `*` `price[` `1` `] ` ` ` `price_second ` `=` `price[i] ` `+` `price[i ` `-` `1` `] ` `+` `2` `*` `price[` `0` `] ` ` ` `totalCost ` `+` `=` `min` `(price_first, price_second) ` ` ` ` ` `# Calculate the minimum price ` ` ` `# of the two cheapest person ` ` ` `if` `(n ` `=` `=` `1` `): ` ` ` `totalCost ` `+` `=` `price[` `0` `] ` ` ` ` ` `else` `: ` ` ` `totalCost ` `+` `=` `price[` `1` `] ` ` ` ` ` `return` `totalCost ` ` ` `# Driver code ` ` ` `price ` `=` `[` `30` `, ` `40` `, ` `60` `, ` `70` `] ` `n ` `=` `len` `(price) ` ` ` `print` `(minimumCost(price, n)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the minimum cost ` ` ` `static` `long` `minimumCost(` `long` `[]price, ` `int` `n) ` ` ` `{ ` ` ` ` ` `// Sort the price array ` ` ` `Array.Sort(price); ` ` ` ` ` `long` `totalCost = 0; ` ` ` ` ` `// Calcualte minimum price ` ` ` `// of n-2 most costly person ` ` ` `for` `(` `int` `i = n - 1; i > 1; i -= 2) ` ` ` `{ ` ` ` `if` `(i == 2) ` ` ` `{ ` ` ` `totalCost += price[2] + price[0]; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` ` ` `// Both the ways as discussed above ` ` ` `long` `price_first = price[i] + price[0] + 2 * price[1]; ` ` ` `long` `price_second = price[i] + price[i - 1] + 2 * price[0]; ` ` ` `totalCost += Math.Min(price_first, price_second); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Calculate the minimum price ` ` ` `// of the two cheapest person ` ` ` `if` `(n == 1) ` ` ` `{ ` ` ` `totalCost += price[0]; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `totalCost += price[1]; ` ` ` `} ` ` ` ` ` `return` `totalCost; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `long` `[]price = { 30, 40, 60, 70 }; ` ` ` `int` `n = price.Length; ` ` ` ` ` `Console.WriteLine(minimumCost(price, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

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**Output:**

220

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