# Find the minimum and maximum sum of N-1 elements of the array

• Difficulty Level : Medium
• Last Updated : 21 Dec, 2022

Given an unsorted array A of size N, the task is to find the minimum and maximum values that can be calculated by adding exactly N-1 elements.

Examples:

Input: a[] = {13, 5, 11, 9, 7}
Output: 32 40
Explanation: Minimum sum is 5 + 7 + 9 + 11 = 32 and maximum sum is 7 + 9 + 11 + 13 = 40.
Input: a[] = {13, 11, 45, 32, 89, 21}
Output: 122 200
Explanation: Minimum sum is 11 + 13 + 21 + 32 + 45 = 122 and maximum sum is 13 + 21 + 32 + 45 + 89 = 200.
Input: a[] = {6, 3, 15, 27, 9}
Output: 33 57
Explanation: Minimum sum is 3 + 6 + 9 + 15 = 33 and maximum sum is 6 + 9 + 15 + 27 = 57.

Simple Approach:

1. Sort the array in ascending order.
2. Sum of the first N-1 elements in the array gives the minimum possible sum.
3. Sum of the last N-1 elements in the array gives the maximum possible sum.

Below is the implementation of the above approach:

## C++

 #includeusing namespace std; // Python Implementation of the above approachvoid minMax(vector&arr){         // Initialize the min_value    // and max_value to 0    int min_value = 0;    int max_value = 0;    int n = arr.size();         // Sort array before calculating    // min and max value    sort(arr.begin(),arr.end());    int j = n - 1;    for(int i = 0; i < n - 1; i++)    {                     // All elements except        // rightmost will be added        min_value += arr[i];                 // All elements except        // leftmost will be added        max_value += arr[j];        j -= 1;    }         // Output: min_value and max_value    cout<arr = {10, 9, 8, 7, 6, 5};    vectorarr1 = {100, 200, 300, 400, 500};     minMax(arr);    minMax(arr1); } // This code is contributed by shinjanpatra

## Java

 // Java Implementation of the above approachimport java.util.*; class GFG {     static void minMax(int[] arr)    {        // Initialize the min_value        // and max_value to 0        long min_value = 0;        long max_value = 0;        int n = arr.length;               // Sort array before calculating        // min and max value        Arrays.sort(arr);                                   for (int i = 0, j = n - 1;             i < n - 1; i++, j--)        {            // All elements except            // rightmost will be added            min_value += arr[i];                       // All elements except            // leftmost will be added            max_value += arr[j];        }                // Output: min_value and max_value        System.out.println(            min_value + " "            + max_value);    }     // Driver Code    public static void main(String[] args)    {        Scanner sc = new Scanner(System.in);         // Initialize your array elements here        int[] arr = { 10, 9, 8, 7, 6, 5 };        int[] arr1 = { 100, 200, 300, 400, 500 };        minMax(arr);        minMax(arr1);    }}

## Python3

 # Python Implementation of the above approachdef minMax(arr):         # Initialize the min_value    # and max_value to 0    min_value = 0    max_value = 0    n=len(arr)         # Sort array before calculating    # min and max value    arr.sort()    j=n-1    for i in range(n-1):                 # All elements except        # rightmost will be added        min_value += arr[i]                 # All elements except        # leftmost will be added        max_value += arr[j]        j-=1         # Output: min_value and max_value    print(min_value," ",max_value) #  Driver Codearr=[10, 9, 8, 7, 6, 5]arr1=[100, 200, 300, 400, 500] minMax(arr)minMax(arr1) #  This code is contributed by ab2127.

## C#

 using System; public class GFG{         static void minMax(int[] arr)    {        // Initialize the min_value        // and max_value to 0        long min_value = 0;        long max_value = 0;        int n = arr.Length;                // Sort array before calculating        // min and max value        Array.Sort(arr);        int j = n - 1;                          for (int i = 0 ;i < n - 1; i++)        {            // All elements except            // rightmost will be added            min_value += arr[i];                        // All elements except            // leftmost will be added            max_value += arr[j];            j--;        }                 // Output: min_value and max_value        Console.WriteLine(            min_value + " "            + max_value);    }      // Driver Code         static public void Main (){                          // Initialize your array elements here        int[] arr = { 10, 9, 8, 7, 6, 5 };        int[] arr1 = { 100, 200, 300, 400, 500 };        minMax(arr);        minMax(arr1);    }} // This code is contributed by rag2127

## Javascript



Output:

35 40
1000 1400

Time complexity: O(NlogN)
Auxiliary Space: O(1), As constant extra space is used.

Efficient Approach:

1. Find the minimum and maximum element of the array.
2. Calculate the sum of all the elements in the array.
3. Excluding maximum element from the sum gives the minimum possible sum.
4. Excluding the minimum element from the sum gives the maximum possible sum.

Below is the implementation of the above approach:

## C++

 // C++ program to find the minimum and maximum// sum from an array.#include using namespace std; // Function to calculate minimum and maximum sumstatic void miniMaxSum(int arr[], int n){     // Initialize the minElement, maxElement    // and sum by 0.    int minElement = 0, maxElement = 0, sum = 0;     // Assigning maxElement, minElement    // and sum as the first array element    minElement = arr[0];    maxElement = minElement;    sum = minElement;     // Traverse the entire array    for(int i = 1; i < n; i++)    {                 // Calculate the sum of        // array elements        sum += arr[i];         // Keep updating the        // minimum element        if (arr[i] < minElement)        {            minElement = arr[i];        }         // Keep updating the        // maximum element        if (arr[i] > maxElement)        {            maxElement = arr[i];        }    }     // print the minimum and maximum sum    cout << (sum - maxElement) << " "         << (sum - minElement) << endl;} // Driver Codeint main(){         // Test Case 1:    int a1[] = { 13, 5, 11, 9, 7 };    int n = sizeof(a1) / sizeof(a1[0]);         // Call miniMaxSum()    miniMaxSum(a1, n);     // Test Case 2:    int a2[] = { 13, 11, 45, 32, 89, 21 };    n = sizeof(a2) / sizeof(a2[0]);    miniMaxSum(a2, n);     // Test Case 3:    int a3[] = { 6, 3, 15, 27, 9 };    n = sizeof(a3) / sizeof(a3[0]);    miniMaxSum(a3, n);} // This code is contributed by chitranayal

## Java

 // Java program to find the minimum and maximum// sum from an array.class GFG {     // Function to calculate minimum and maximum sum    static void miniMaxSum(int[] arr)    {         // Initialize the minElement, maxElement        // and sum by 0.        int minElement = 0, maxElement = 0, sum = 0;         // Assigning maxElement, minElement        // and sum as the first array element        minElement = arr[0];        maxElement = minElement;        sum = minElement;         // Traverse the entire array        for (int i = 1; i < arr.length; i++) {             // calculate the sum of            // array elements            sum += arr[i];             // Keep updating the            // minimum element            if (arr[i] < minElement) {                minElement = arr[i];            }             // Keep updating the            // maximum element            if (arr[i] > maxElement) {                maxElement = arr[i];            }        }         // print the minimum and maximum sum        System.out.println((sum - maxElement) + " "                        + (sum - minElement));    }     // Driver Code    public static void main(String args[])    {         // Test Case 1:        int a1[] = { 13, 5, 11, 9, 7 };        // Call miniMaxSum()        miniMaxSum(a1);         // Test Case 2:        int a2[] = { 13, 11, 45, 32, 89, 21 };        miniMaxSum(a2);         // Test Case 3:        int a3[] = { 6, 3, 15, 27, 9 };        miniMaxSum(a3);    }}

## Python3

 # Python3 program to find the minimum and# maximum sum from a list. # Function to calculate minimum and maximum sumdef miniMaxSum(arr, n):     # Initialize the minElement, maxElement    # and sum by 0.    minElement = 0    maxElement = 0    sum = 0     # Assigning maxElement, minElement    # and sum as the first list element    minElement = arr[0]    maxElement = minElement    sum = minElement     # Traverse the entire list    for i in range(1, n):         # Calculate the sum of        # list elements        sum += arr[i]         # Keep updating the        # minimum element        if (arr[i] < minElement):            minElement = arr[i]         # Keep updating the        # maximum element        if (arr[i] > maxElement):            maxElement = arr[i]     # Print the minimum and maximum sum    print(sum - maxElement,          sum - minElement) # Driver Code # Test Case 1:a1 = [ 13, 5, 11, 9, 7 ]n = len(a1) # Call miniMaxSum()miniMaxSum(a1, n) # Test Case 2:a2 = [ 13, 11, 45, 32, 89, 21 ]n = len(a2)miniMaxSum(a2, n) # Test Case 3:a3 = [ 6, 3, 15, 27, 9 ]n = len(a3)miniMaxSum(a3, n) # This code is contributed by vishu2908

## C#

 // C# program to find the minimum and maximum// sum from an array.using System; class GFG{ // Function to calculate minimum and maximum sumstatic void miniMaxSum(int[] arr){         // Initialize the minElement, maxElement    // and sum by 0.    int minElement = 0, maxElement = 0, sum = 0;     // Assigning maxElement, minElement    // and sum as the first array element    minElement = arr[0];    maxElement = minElement;    sum = minElement;     // Traverse the entire array    for(int i = 1; i < arr.Length; i++)    {                 // Calculate the sum of        // array elements        sum += arr[i];         // Keep updating the        // minimum element        if (arr[i] < minElement)        {            minElement = arr[i];        }         // Keep updating the        // maximum element        if (arr[i] > maxElement)        {            maxElement = arr[i];        }    }     // Print the minimum and maximum sum    Console.WriteLine((sum - maxElement) + " " +                      (sum - minElement));} // Driver Codepublic static void Main(){         // Test Case 1:    int[] a1 = new int[]{ 13, 5, 11, 9, 7 };         // Call miniMaxSum()    miniMaxSum(a1);     // Test Case 2:    int[] a2 = new int[]{ 13, 11, 45, 32, 89, 21 };    miniMaxSum(a2);     // Test Case 3:    int[] a3 = new int[]{ 6, 3, 15, 27, 9 };    miniMaxSum(a3);}} // This code is contributed by sanjoy_62

## Javascript



Output

32 40
122 200
33 57

Time complexity: O(N)
Auxiliary Space: O(1), As constant extra space is used.

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