Maximum and Minimum Values of an Algebraic Expression

Given an algebraic expression of the form (x₁ + x₂ + x₃ + . . . + x_n) * (y₁ + y₂ + . . . + y_m) and
(n + m) integers. Find the maximum and minimum value of the expression using the given
integers.

Consstraint :
n <= 50
m <= 50
-50 <= x1, x2, .. x_n <= 50

Examples :

Input : n = 2, m = 2
        arr[] = {1, 2, 3, 4}
Output : Maximum : 25 
         Minimum : 21
The expression is (x1 + x2) * (y1 + y2) and
the given integers are 1, 2, 3 and 4. Then
maximum value is (1 + 4) * (2 + 3) = 25 
whereas minimum value is (4 + 3) * (2 + 1) 
= 21.

Input : n = 3, m = 1
        arr[] = {1, 2, 3, 4}
Output : Maximum : 24 
         Minimum : 9

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to consider all possible combinations of n numbers and remaining m numbers and calculating their values, from which maximum value and minimum value can be derived.

Below is an efficient solution.
The idea is based on limited values of n, m, x1, x2, .. y1, y2, .. Let suppose S be the sum of all the (n + m) numbers in the expression and X be the sum of the n numbers on the left of expression. Obviously, the sum of the m numbers on the right of expression will be represented as (S – X). There can be many possible values of X from the given (n + m) numbers and hence the problem gets reduced to simply iterate through all values of X and keeping track of the minimum and maximum value of X * (S – X).
Now, the problem is equivalent to finding all possible values of X. Since the given numbers are in the range of -50 to 50 and the maximum value of (n + m) is 100, X will lie in between -2500 and 2500 which results into overall 5000 values of X. We will use dynamic programming approach to solve this problem. Consider a dp[i][j] array which can value either 1 or 0, where 1 means X can be equal to j by choosing i numbers from the (n + m) numbers and 0 otherwise. Then for each number k, if dp[i][j] is 1 then dp[i + 1][j + k] is also 1 where k belongs to given (n + m) numbers. Thus, by iterating through all k, we can determine whether a value of X is reachable by choosing a total of n numbers
Below is the implementation of the above approach.

C++

// CPP program to find the maximum 
// and minimum values of an Algebraic 
// expression of given form 
#include <bits/stdc++.h> 
using namespace std; 
  
#define INF 1e9 
#define MAX 50 
  
int minMaxValues(int arr[], int n, int m) 
{ 
    // Finding sum of array elements 
    int sum = 0; 
    for (int i = 0; i < (n + m); i++) { 
        sum += arr[i]; 
  
        // shifting the integers by 50 
        // so that they become positive 
        arr[i] += 50; 
    } 
  
// dp[i][j] represents true if sum 
// j can be reachable by choosing 
// i numbers 
bool dp[MAX+1][MAX * MAX + 1]; 
  
    // initialize the dp array to 01 
    memset(dp, 0, sizeof(dp)); 
  
    dp[0][0] = 1; 
  
    // if dp[i][j] is true, that means 
    // it is possible to select i numbers 
    // from (n + m) numbers to sum upto j 
    for (int i = 0; i < (n + m); i++) { 
  
        // k can be at max n because the 
        // left expression has n numbers 
        for (int k = min(n, i + 1); k >= 1; k--) { 
            for (int j = 0; j < MAX * MAX + 1; j++) { 
                if (dp[k - 1][j]) 
                    dp[k][j + arr[i]] = 1; 
            } 
        } 
    } 
  
    int max_value = -INF, min_value = INF; 
  
    for (int i = 0; i < MAX * MAX + 1; i++) { 
  
        // checking if a particular sum 
        // can be reachable by choosing 
        // n numbers 
        if (dp[n][i]) { 
  
            // getting the actual sum as 
            // we shifted the numbers by 
            /// 50 to avoid negative indexing 
            // in array 
            int temp = i - 50 * n; 
            max_value = max(max_value, temp * (sum - temp)); 
            min_value = min(min_value, temp * (sum - temp)); 
        } 
    } 
    cout << "Maximum Value: " << max_value 
         << "\n"
         << "Minimum Value: "
         << min_value << endl; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 2, m = 2; 
    int arr[] = { 1, 2, 3, 4 }; 
    minMaxValues(arr, n, m); 
    return 0; 
} 

Java

// Java program to find the maximum 
// and minimum values of an Algebraic 
// expression of given form 
import java.io.*; 
import java.lang.*; 
  
public class GFG { 
      
    static double INF = 1e9; 
    static int MAX = 50; 
  
    static void minMaxValues(int []arr,  
                              int n, int m) 
    { 
          
        // Finding sum of array elements 
        int sum = 0; 
        for (int i = 0; i < (n + m); i++) 
        { 
            sum += arr[i]; 
      
            // shifting the integers by 50 
            // so that they become positive 
            arr[i] += 50; 
        } 
      
        // dp[i][j] represents true if sum 
        // j can be reachable by choosing 
        // i numbers 
        boolean dp[][] =  
             new boolean[MAX+1][MAX * MAX + 1]; 
      
        dp[0][0] = true; 
      
        // if dp[i][j] is true, that means 
        // it is possible to select i numbers 
        // from (n + m) numbers to sum upto j 
        for (int i = 0; i < (n + m); i++) { 
      
            // k can be at max n because the 
            // left expression has n numbers 
            for (int k = Math.min(n, i + 1); k >= 1; k--)  
            { 
                for (int j = 0; j < MAX * MAX + 1; j++) 
                { 
                    if (dp[k - 1][j]) 
                        dp[k][j + arr[i]] = true; 
                } 
            } 
        } 
      
        double max_value = -1 * INF, min_value = INF; 
      
        for (int i = 0; i < MAX * MAX + 1; i++) 
        { 
      
            // checking if a particular sum 
            // can be reachable by choosing 
            // n numbers 
            if (dp[n][i]) { 
      
                // getting the actual sum as 
                // we shifted the numbers by 
                /// 50 to avoid negative indexing 
                // in array 
                int temp = i - 50 * n; 
                max_value = Math.max(max_value, temp * 
                                        (sum - temp)); 
                                              
                min_value = Math.min(min_value, temp *  
                                        (sum - temp)); 
            } 
        } 
          
        System.out.print("Maximum Value: " +  
                     (int)max_value + "\n" +  
          "Minimum Value: " + (int)min_value + "\n"); 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        int n = 2, m = 2; 
        int []arr = { 1, 2, 3, 4 }; 
        minMaxValues(arr, n, m); 
    } 
} 
  
// This code is contributed by Manish Shaw 
// (manishshaw1) 

Python3

# Python3 program to find the  
# maximum and minimum values  
# of an Algebraic expression  
# of given form 
def minMaxValues(arr, n, m) :      
    # Finding sum of 
    # array elements 
    sum = 0
    INF = 1000000000
    MAX = 50
    for i in range(0, (n + m)) : 
        sum += arr[i] 
  
        # shifting the integers by 50 
        # so that they become positive 
        arr[i] += 50
  
    # dp[i][j] represents true  
    # if sum j can be reachable 
    # by choosing i numbers 
    dp = [[0 for x in range(MAX * MAX + 1)] 
                  for y in range( MAX + 1)] 
      
    dp[0][0] = 1
  
    # if dp[i][j] is true, that  
    # means it is possible to  
    # select i numbers from (n + m) 
    # numbers to sum upto j 
    for i in range(0, (n + m)) :  
          
        # k can be at max n because the 
        # left expression has n numbers 
        for k in range(min(n, i + 1), 0, -1) : 
            for j in range(0, MAX * MAX + 1) : 
                if (dp[k - 1][j]) : 
                    dp[k][j + arr[i]] = 1
  
    max_value = -1 * INF  
    min_value = INF 
  
    for i in range(0, MAX * MAX + 1) : 
  
        # checking if a particular  
        # sum can be reachable by  
        # choosing n numbers 
        if (dp[n][i]) : 
  
            # getting the actual sum  
            # as we shifted the numbers  
            # by 50 to avoid negative  
            # indexing in array 
            temp = i - 50 * n 
            max_value = max(max_value,  
                         temp * (sum - temp)) 
                                          
            min_value = min(min_value, 
                         temp * (sum - temp)) 
      
    print ("Maximum Value: {}\nMinimum Value: {}"
                 .format(max_value, min_value)) 
  
# Driver Code 
n = 2
m = 2
arr = [ 1, 2, 3, 4 ] 
  
minMaxValues(arr, n, m)  
  
# This code is contributed by  
# Manish Shaw(manishshaw1) 

C#

// C# program to find the maximum 
// and minimum values of an Algebraic 
// expression of given form 
using System; 
using System.Collections.Generic; 
  
class GFG { 
      
    static double INF = 1e9; 
    static int MAX = 50; 
  
    static void minMaxValues(int []arr, int n, int m) 
    { 
          
        // Finding sum of array elements 
        int sum = 0; 
        for (int i = 0; i < (n + m); i++) 
        { 
            sum += arr[i]; 
      
            // shifting the integers by 50 
            // so that they become positive 
            arr[i] += 50; 
        } 
      
    // dp[i][j] represents true if sum 
    // j can be reachable by choosing 
    // i numbers 
        bool[,] dp = new bool[MAX+1, MAX * MAX + 1]; 
      
        dp[0,0] = true; 
      
        // if dp[i][j] is true, that means 
        // it is possible to select i numbers 
        // from (n + m) numbers to sum upto j 
        for (int i = 0; i < (n + m); i++) { 
      
            // k can be at max n because the 
            // left expression has n numbers 
            for (int k = Math.Min(n, i + 1); k >= 1; k--)  
            { 
                for (int j = 0; j < MAX * MAX + 1; j++) 
                { 
                    if (dp[k - 1,j]) 
                        dp[k,j + arr[i]] = true; 
                } 
            } 
        } 
      
        double max_value = -1 * INF, min_value = INF; 
      
        for (int i = 0; i < MAX * MAX + 1; i++) 
        { 
      
            // checking if a particular sum 
            // can be reachable by choosing 
            // n numbers 
            if (dp[n,i]) { 
      
                // getting the actual sum as 
                // we shifted the numbers by 
                /// 50 to avoid negative indexing 
                // in array 
                int temp = i - 50 * n; 
                max_value = Math.Max(max_value, temp * 
                                          (sum - temp)); 
                                            
                min_value = Math.Min(min_value, temp *  
                                          (sum - temp)); 
            } 
        } 
          
        Console.WriteLine("Maximum Value: " + max_value 
         + "\n" + "Minimum Value: " + min_value + "\n"); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int n = 2, m = 2; 
        int []arr = { 1, 2, 3, 4 }; 
        minMaxValues(arr, n, m); 
    } 
} 
  
// This code is contributed by Manish Shaw 
// (manishshaw1) 

PHP

<?php 
// PHP program to find the  
// maximum and minimum values  
// of an Algebraic expression  
// of given form 
function minMaxValues($arr, $n, $m) 
{          
    // Finding sum of 
    // array elements 
    $sum = 0; 
    $INF = 1000000000; 
    $MAX = 50; 
    for ($i = 0; $i < ($n + $m); $i++) 
    { 
        $sum += $arr[$i]; 
  
        // shifting the integers by 50 
        // so that they become positive 
        $arr[$i] += 50; 
    } 
  
    // dp[i][j] represents true  
    // if sum j can be reachable 
    // by choosing i numbers 
    $dp = array(); 
      
    // new bool[MAX+1, MAX * MAX + 1]; 
    for($i = 0; $i < $MAX + 1; $i++) 
    { 
        for($j = 0; $j < $MAX * $MAX + 1; $j++) 
            $dp[$i][$j] = 0;  
    } 
      
    $dp[0][0] = 1; 
  
    // if dp[i][j] is true, that  
    // means it is possible to  
    // select i numbers from (n + m) 
    // numbers to sum upto j 
    for ($i = 0; $i < ($n + $m); $i++)  
    { 
  
        // k can be at max n because the 
        // left expression has n numbers 
        for ($k = min($n, $i + 1);  
                      $k >= 1; $k--)  
        { 
            for ($j = 0; $j < $MAX *  
                              $MAX + 1; $j++) 
            { 
                if ($dp[$k - 1][$j]) 
                    $dp[$k][$j + $arr[$i]] = 1; 
            } 
        } 
    } 
  
    $max_value = -1 * $INF;  
    $min_value = $INF; 
  
    for ($i = 0; $i < $MAX * $MAX + 1; $i++) 
    { 
  
        // checking if a particular  
        // sum can be reachable by  
        // choosing n numbers 
        if ($dp[$n][$i])  
        { 
  
            // getting the actual sum  
            // as we shifted the numbers  
            // by 50 to avoid negative  
            // indexing in array 
            $temp = $i - 50 * $n; 
            $max_value = max($max_value, $temp * 
                                ($sum - $temp)); 
                                          
            $min_value = min($min_value, $temp *  
                                ($sum - $temp)); 
        } 
    } 
      
    echo ("Maximum Value: ". $max_value. "\n".  
          "Minimum Value: ". $min_value. "\n"); 
} 
  
// Driver Code 
$n = 2; 
$m = 2; 
$arr = [ 1, 2, 3, 4 ]; 
  
minMaxValues($arr, $n, $m);  
  
// This code is contributed by  
// Manish Shaw(manishshaw1) 
?> 

Output :

Maximum Value: 25
Minimum Value: 21

This approach will have a runtime complexity of O(MAX * MAX * (n+m)²).

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My Personal Notes

Jul	AUG	Sep
	14
2019	2020	2021

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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