Minimize total time taken by two persons to visit N cities such that none of them meet

Given an array arr[] of size N, where arr[i] is the time required to visit i^th city, the task is to find the minimum total time required to visit all N cities by two persons such that none of them meet in any of the cities.

Examples:

Input: arr[] = {2, 8, 3}
Output: 16
Explanation:
Visiting cities in below given order will take minimum time:
First person: 2nd city ? 1st city ? 3rd city
Second person: 1st city ? 3rd city ? 2nd city.

Input: arr[]={1, 10, 6, 7, 5}
Output: 29

Approach: The given problem can be solved based on the following observations:

• Suppose i^th city takes the longest time T to visit and the total time to visit all cities by one person is the sum of all array elements, say sum.
• If the 1^st person visits the i^th city, then in T time, the second person will visit other cities in that time, if possible.
• If the value T is at most (sum - T), then both people can visit the place individually in sum time.
• Otherwise, the 2^nd person will have to wait to visit the i^th city. Then, the total time required will be 2 * T as the 2^nd person will be able to visit the i^th city only if the first person comes out.
• Therefore, from the above observations, the answer will be the maximum of 2 * T and sum.

Therefore, from the above observations, find the sum of the array elements/a> and find the maximum element present in the array and print the maximum among twice the maximum element and the sum.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum time
// to visit all the cities such that
// both the person never meets
void minimumTime(int* arr, int n)
{
    // Initialize sum as 0
    int sum = 0;

    // Find the maximum element
    int T = *max_element(arr, arr + n);

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // Increment sum by arr[i]
        sum += arr[i];
    }

    // Print maximum of 2*T and sum
    cout << max(2 * T, sum);
}

// Driver Code
int main()
{
    int arr[] = { 2, 8, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    minimumTime(arr, N);

    return 0;
}

// Java program for the above approach 
import java.util.*;
class GFG
{
      
// Function to find the minimum time
// to visit all the cities such that
// both the person never meets
static void minimumTime(int[] arr, int n)
{
  
    // Initialize sum as 0
    int sum = 0;

    // Find the maximum element
    int T = Arrays.stream(arr).max().getAsInt(); 

    // Traverse the array
    for (int i = 0; i < n; i++) 
    {

        // Increment sum by arr[i]
        sum += arr[i];
    }

    // Print maximum of 2*T and sum
    System.out.println(Math.max(2 * T, sum));
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 8, 3 };
    int N = arr.length;

    // Function Call
    minimumTime(arr, N);
}
}

// This code is contributed by sanjoy_62

# Python3 program for the above approach

# Function to find the minimum time
# to visit all the cities such that
# both the person never meets
def minimumTime(arr, n):
  
  # Initialize sum as 0
    sum = 0

    # Find the maximum element
    T = max(arr)

    # Traverse the array
    for i in range(n):
      
        # Increment sum by arr[i]
        sum += arr[i]

    # Print maximum of 2*T and sum
    print(max(2 * T, sum))

# Driver Code
if __name__ == '__main__':
    arr = [2, 8, 3]
    N = len(arr)

    # Function Call
    minimumTime(arr, N)

    # This code is contributed by mohit kumar 29

// C# program for the above approach 
using System;
using System.Linq;
class GFG
{
      
// Function to find the minimum time
// to visit all the cities such that
// both the person never meets
static void minimumTime(int[] arr, int n)
{
  
    // Initialize sum as 0
    int sum = 0;

    // Find the maximum element
    int T = arr.Min(); 

    // Traverse the array
    for (int i = 0; i < n; i++) 
    {

        // Increment sum by arr[i]
        sum += arr[i];
    }

    // Print maximum of 2*T and sum
    Console.WriteLine(Math.Max(2 * T, sum));
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 8, 3 };
    int N = arr.Length;

    // Function Call
    minimumTime(arr, N);
}
}

// This code is contributed by 29AjayKumar

<script>
// javascript program for the above approach 

    // Function to find the minimum time
    // to visit all the cities such that
    // both the person never meets
    function minimumTime(arr , n) {

        // Initialize sum as 0
        var sum = 0;

        // Find the maximum element
        var T =Math.max(...arr);

        // Traverse the array
        for (i = 0; i < n; i++) {

            // Increment sum by arr[i]
            sum += arr[i];
        }

        // Print maximum of 2*T and sum
        document.write(Math.max(2 * T, sum));
    }

    // Driver code
    var arr = [ 2, 8, 3 ];
    var N = arr.length;

    // Function Call
    minimumTime(arr, N);

// This code is contributed by todaysgaurav
</script>

16

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