Find minimum chocolates left after following the given rules

Geek has N packs of chocolates and the amount of chocolates in each pack is given in an array arr[]. His sister wants to have a pack of chocolate. Geeks being greedy will pick the packs with more number of chocolates but he can pick from either first or last. Find the number of chocolates his sister will get.

Examples:

Input: arr[ ] = {5, 3, 1, 6, 9}
Output: 1
Explanation: Geek will have the packs with chocolates
5, 3, 6, 9.

Input: arr[ ] = {5, 9, 2, 6}
Output:  2

Approach: The problem can be solved using the following observation:

As Geek is picking chocolate packs greedily, so he can always make sure that the pack with the minimum number of chocolates is left at the end.

Follow the below steps to implement the idea:

• Declare a variable (say ans) and initialize ans = arr[0] to store the minimum number of chocolates.
• Iterate from i = 1 to N-1:
  • If the value of arr[i] is less than ans, update ans = arr[i].
• The final value stored in ans will be the minimum value. So return ans as the required answer.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find tastiness level
// of the square that his sister gets
int chocolates(int arr[], int n)
{
    int ans = arr[0];

    // Loop to find the minimum element
    for (int i = 1; i < n; i++) {
        if (arr[i] < ans) {
            ans = arr[i];
        }
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    int arr[] = { 5, 3, 1, 6, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << chocolates(arr, N) << endl;
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
public class GFG {

  // Function to find tastiness level
  // of the square that his sister gets
  public static int chocolates(int[] arr, int n)
  {
    int ans = arr[0];

    // Loop to find the minimum element
    for (int i = 1; i < n; i++) {
      if (arr[i] < ans) {
        ans = arr[i];
      }
    }

    // Return the answer
    return ans;
  }

// driver code
public static void main(String[] args)
{
    int[] arr = { 5, 3, 1, 6, 9 };
    int N = arr.length;

    // Function Call
    System.out.print(chocolates(arr, N));
}
}

// This code is contributed by code_hunt.

C#

using System;

public class GFG {

  // Function to find tastiness level
  // of the square that his sister gets
  public static int chocolates(int[] arr, int n)
  {
    int ans = arr[0];

    // Loop to find the minimum element
    for (int i = 1; i < n; i++) {
      if (arr[i] < ans) {
        ans = arr[i];
      }
    }

    // Return the answer
    return ans;
  }

  static public void Main()
  {

    int[] arr = { 5, 3, 1, 6, 9 };
    int N = arr.Length;

    // Function Call
    Console.WriteLine(chocolates(arr, N));
  }
}

// This code is contributed by akashish__

Python3

# Python code to implement the approach

# Function to find tastiness level
# of the square that his sister gets
def chocolates(arr, n):
    ans = arr[0]

    # Loop to find the minimum element
    for i in range(1, n):
        if (arr[i] < ans):
            ans = arr[i]
        # Return the answer
    return ans

# Driver Code
if __name__ == '__main__':
    arr = [5, 3, 1, 6, 9]
    N = len(arr)

    # Function call
    print(chocolates(arr, N))

    # This code is contributed by arohirai2616.

Javascript

<script>
// Javascript code for the above approach

// Function to find tastiness level
// of the square that his sister gets
function chocolates(arr,n)
  {
    let ans = arr[0];

    // Loop to find the minimum element
    for (let i = 1; i < n; i++) {
      if (arr[i] < ans) {
        ans = arr[i];
      }
    }

    // Return the answer
    return ans;
  }
  // driver code

    let arr = [ 5, 3, 1, 6, 9 ];
    let N = arr.length;

    // Function Call
   console.log(chocolates(arr, N));

// This code is contributed by arohirai2616.
</script>

PHP

// PHP Code
<?php
    // Function to find tastiness level
    // of the square that his sister gets
    function chocolates($arr, $n) {
        $ans = $arr[0];

        // Loop to find the minimum element
        for ($i = 1; $i < $n; $i++) {
            if ($arr[$i] < $ans) {
                $ans = $arr[$i];
            }
        }

        // Return the answer
        return $ans;
    }

    // Driver Code
    $arr = [5, 3, 1, 6, 9];
    $n = sizeof($arr)/sizeof($arr[0]);

    // Function Call
    echo chocolates($arr, $n);
 // This code is contributed by Kanishka Gupta
?>

Output

1

Time Complexity: O(N) As we are traversing the array once
Auxiliary Space: O(1) As we are not using any extra space.