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.

Constraint :
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

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

Steps to solve the problem:

1. Initialize a variable “sum” to 0.
2. Traverse the array arr[] and add each element to sum.
*Shift each element by 50 so that all integers become positive.
3. Initialize a 2D array “dp” of size (n+1)x(MAX*MAX+1) to false.
4. Set dp[0][0] to true.
5. Traverse the array arr[].
*Set k to the minimum value between n and i+1.
*Traverse the dp array for all j from 0 to MAX*MAX+1.
*If dp[k-1][j] is true, set dp[k][j+arr[i]] to true.
6. Initialize two variables, “max_value” and “min_value” to -infinity and +infinity respectively.
7. Traverse the dp array for all i from 0 to MAXMAX+1.
*If dp[n][i] is true, do the following:
*Compute the actual sum “temp” by subtracting 50n from i.
*Compute the product “temp * (sum – temp)” and update the values of max_value and min_value.
8. Print the values of max_value and min_value.

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)
?>

Javascript

<script>
 
// Javascript program to find the maximum
// and minimum values of an Algebraic
// expression of given form
 
var INF  = 1000000000
var MAX = 50
 
function minMaxValues(arr, n, m)
{
    // Finding sum of array elements
    var sum = 0;
    for (var 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
var dp = Array.from(Array(MAX+1), ()=> Array(MAX*MAX + 1).fill(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 (var i = 0; i < (n + m); i++) {
 
        // k can be at max n because the
        // left expression has n numbers
        for (var k = Math.min(n, i + 1); k >= 1; k--) {
            for (var j = 0; j < MAX * MAX + 1; j++) {
                if (dp[k - 1][j])
                    dp[k][j + arr[i]] = 1;
            }
        }
    }
 
    var max_value = -INF, min_value = INF;
 
    for (var 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
            var temp = i - 50 * n;
            max_value = Math.max(max_value, temp * (sum - temp));
            min_value = Math.min(min_value, temp * (sum - temp));
        }
    }
    document.write( "Maximum Value: " + max_value
         + "<br>"
         + "Minimum Value: "
         + min_value );
}
 
// Driver Code
var n = 2, m = 2;
var arr =[1, 2, 3, 4];
minMaxValues(arr, n, m);
 
</script>