Given an array arr[] of size N, where arr[i] is the time required to visit ith 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 ith 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 1st person visits the ith 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 2nd person will have to wait to visit the ith city. Then, the total time required will be 2 * T as the 2nd person will be able to visit the ith 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++
// 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 meetsvoid 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 Codeint main(){ int arr[] = { 2, 8, 3 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call minimumTime(arr, N); return 0;} |
Java
// 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 meetsstatic 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 codepublic 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
# Python3 program for the above approach# Function to find the minimum time# to visit all the cities such that# both the person never meetsdef 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 Codeif __name__ == '__main__': arr = [2, 8, 3] N = len(arr) # Function Call minimumTime(arr, N) # This code is contributed by mohit kumar 29 |
C#
// 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 meetsstatic 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 codepublic 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 |
Javascript
<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> |
Output:
16
Time Complexity: O(N)
Auxiliary Space: O(1)
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