Minimum and Maximum values of an expression with * and +

Last Updated : 27 Aug, 2024

Given an expression which contains numbers and two operators ‘+’ and ‘*’, we need to find maximum and minimum value which can be obtained by evaluating this expression by different parenthesization.
Examples:

Input  : expr = “1+2*3+4*5” 
Output : Minimum Value = 27, Maximum Value = 105 
Explanation:
Minimum evaluated value = 1 + (2*3) + (4*5) = 27
Maximum evaluated value = (1 + 2)*(3 + 4)*5 = 105

We can solve this problem by dynamic programming method, we can see that this problem is similar to matrix chain multiplication, here we are trying different parenthesization to maximize and minimize expression value instead of number of matrix multiplication.
In below code first we have separated the operators and numbers from given expression then two 2D arrays are taken for storing the intermediate result which are updated similar to matrix chain multiplication and different parenthesization are tried among the numbers but according to operators occurring in between them. At the end last cell of first row will store the final result in both the 2D arrays.

CPP

// C++ program to get maximum and minimum
// values of an expression
#include <bits/stdc++.h>
using namespace std;

// Utility method to check whether a character
// is operator or not
bool isOperator(char op)
{
    return (op == '+' || op == '*');
}

// method prints minimum and maximum value
// obtainable from an expression
void printMinAndMaxValueOfExp(string exp)
{
    vector<int> num;
    vector<char> opr;
    string tmp = "";

    //  store operator and numbers in different vectors
    for (int i = 0; i < exp.length(); i++)
    {
        if (isOperator(exp[i]))
        {
            opr.push_back(exp[i]);
            num.push_back(atoi(tmp.c_str()));
            tmp = "";
        }
        else
        {
            tmp += exp[i];
        }
    }
    //  storing last number in vector
    num.push_back(atoi(tmp.c_str()));

    int len = num.size();
    int minVal[len][len];
    int maxVal[len][len];

    //  initializing minval and maxval 2D array
    for (int i = 0; i < len; i++)
    {
        for (int j = 0; j < len; j++)
        {
            minVal[i][j] = INT_MAX;
            maxVal[i][j] = 0;

            //  initializing main diagonal by num values
            if (i == j)
                minVal[i][j] = maxVal[i][j] = num[i];
        }
    }

    // looping similar to matrix chain multiplication
    // and updating both 2D arrays
    for (int L = 2; L <= len; L++)
    {
        for (int i = 0; i < len - L + 1; i++)
        {
            int j = i + L - 1;
            for (int k = i; k < j; k++)
            {
                int minTmp = 0, maxTmp = 0;

                // if current operator is '+', updating tmp
                // variable by addition
                if(opr[k] == '+')
                {
                    minTmp = minVal[i][k] + minVal[k + 1][j];
                    maxTmp = maxVal[i][k] + maxVal[k + 1][j];
                }

                // if current operator is '*', updating tmp
                // variable by multiplication
                else if(opr[k] == '*')
                {
                    minTmp = minVal[i][k] * minVal[k + 1][j];
                    maxTmp = maxVal[i][k] * maxVal[k + 1][j];
                }

                //  updating array values by tmp variables
                if (minTmp < minVal[i][j])
                    minVal[i][j] = minTmp;
                if (maxTmp > maxVal[i][j])
                    maxVal[i][j] = maxTmp;
            }
        }
    }

    //  last element of first row will store the result
    cout << "Minimum value : " << minVal[0][len - 1]
         << ", Maximum value : " << maxVal[0][len - 1];
}

//  Driver code to test above methods
int main()
{
    string expression = "1+2*3+4*5";
    printMinAndMaxValueOfExp(expression);
    return 0;
}

Java

// Java program to get maximum and minimum
// values of an expression
import java.io.*;
import java.util.*;
class GFG
{

  // Utility method to check whether a character
  // is operator or not
  static boolean isOperator(char op)
  {
    return (op == '+' || op == '*');
  }

  // method prints minimum and maximum value
  // obtainable from an expression
  static void printMinAndMaxValueOfExp(String exp)
  {
    Vector<Integer> num = new Vector<Integer>();
    Vector<Character> opr = new Vector<Character>();
    String tmp = "";

    //  store operator and numbers in different vectors
    for (int i = 0; i < exp.length(); i++)
    {
      if (isOperator(exp.charAt(i)))
      {
        opr.add(exp.charAt(i));
        num.add(Integer.parseInt(tmp));
        tmp = "";
      }
      else
      {
        tmp += exp.charAt(i);
      }
    }

    //  storing last number in vector
    num.add(Integer.parseInt(tmp));

    int len = num.size();
    int[][] minVal = new int[len][len];
    int[][] maxVal = new int[len][len];

    //  initializing minval and maxval 2D array
    for (int i = 0; i < len; i++)
    {
      for (int j = 0; j < len; j++) 
      {
        minVal[i][j] = Integer.MAX_VALUE;
        maxVal[i][j] = 0;

        //  initializing main diagonal by num values
        if (i == j)
          minVal[i][j] = maxVal[i][j]
          = num.get(i);
      }
    }

    // looping similar to matrix chain multiplication
    // and updating both 2D arrays
    for (int L = 2; L <= len; L++)
    {
      for (int i = 0; i < len - L + 1; i++)
      {
        int j = i + L - 1;
        for (int k = i; k < j; k++)
        {
          int minTmp = 0, maxTmp = 0;

          // if current operator is '+', updating
          // tmp variable by addition
          if (opr.get(k) == '+')
          {
            minTmp = minVal[i][k]
              + minVal[k + 1][j];
            maxTmp = maxVal[i][k]
              + maxVal[k + 1][j];
          }

          // if current operator is '*', updating
          // tmp variable by multiplication
          else if (opr.get(k) == '*')
          {
            minTmp = minVal[i][k]
              * minVal[k + 1][j];
            maxTmp = maxVal[i][k]
              * maxVal[k + 1][j];
          }

          //  updating array values by tmp
          //  variables
          if (minTmp < minVal[i][j])
            minVal[i][j] = minTmp;
          if (maxTmp > maxVal[i][j])
            maxVal[i][j] = maxTmp;
        }
      }
    }

    //  last element of first row will store the result
    System.out.print(
      "Minimum value : " + minVal[0][len - 1]
      + ", Maximum value : " + maxVal[0][len - 1]);
  }

  //  Driver code to test above methods
  public static void main(String[] args)
  {
    String expression = "1+2*3+4*5";
    printMinAndMaxValueOfExp(expression);
  }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 program to get maximum and minimum
# values of an expression

# Utility method to check whether a character
# is operator or not
def isOperator(op):
    return (op == '+' or op == '*')

# method prints minimum and maximum value
# obtainable from an expression
def printMinAndMaxValueOfExp(exp):
    num = []
    opr = []
    tmp = ""

    # store operator and numbers in different vectors
    for i in range(len(exp)):
        if (isOperator(exp[i])):
            opr.append(exp[i])
            num.append(int(tmp))
            tmp = ""
        else:
            tmp += exp[i]

    # storing last number in vector
    num.append(int(tmp))

    llen = len(num)
    minVal = [[ 0 for i in range(llen)] for i in range(llen)]
    maxVal = [[ 0 for i in range(llen)] for i in range(llen)]

    # initializing minval and maxval 2D array
    for i in range(llen):
        for j in range(llen):
            minVal[i][j] = 10**9
            maxVal[i][j] = 0

            # initializing main diagonal by num values
            if (i == j):
                minVal[i][j] = maxVal[i][j] = num[i]

    # looping similar to matrix chain multiplication
    # and updating both 2D arrays
    for L in range(2, llen + 1):
        for i in range(llen - L + 1):
            j = i + L - 1
            for k in range(i, j):

                minTmp = 0
                maxTmp = 0

                # if current operator is '+', updating tmp
                # variable by addition
                if(opr[k] == '+'):

                    minTmp = minVal[i][k] + minVal[k + 1][j]
                    maxTmp = maxVal[i][k] + maxVal[k + 1][j]


                # if current operator is '*', updating tmp
                # variable by multiplication
                else if(opr[k] == '*'):

                    minTmp = minVal[i][k] * minVal[k + 1][j]
                    maxTmp = maxVal[i][k] * maxVal[k + 1][j]

                # updating array values by tmp variables
                if (minTmp < minVal[i][j]):
                    minVal[i][j] = minTmp
                if (maxTmp > maxVal[i][j]):
                    maxVal[i][j] = maxTmp

    # last element of first row will store the result
    print("Minimum value : ",minVal[0][llen - 1],", \
            Maximum value : ",maxVal[0][llen - 1])

# Driver code 
expression = "1+2*3+4*5"
printMinAndMaxValueOfExp(expression)

# This code is contributed by mohit kumar 29

// C# program to get maximum and minimum
// values of an expression
using System;
using System.Collections.Generic;

public class GFG
{

  // Utility method to check whether a character
  // is operator or not
  static bool isOperator(char op)
  {
    return (op == '+' || op == '*');
  }

  // method prints minimum and maximum value
  // obtainable from an expression
  static void printMinAndMaxValueOfExp(string exp)
  {
    List<int> num = new List<int>();
    List<char> opr = new List<char>();

    string tmp = "";

    //  store operator and numbers in different vectors
    for (int i = 0; i < exp.Length; i++)
    {
      if (isOperator(exp[i]))
      {
        opr.Add(exp[i]);
        num.Add(int.Parse(tmp));
        tmp = "";
      }
      else
      {
        tmp += exp[i];
      }
    }

    //  storing last number in vector
    num.Add(int.Parse(tmp));       
    int len = num.Count;
    int[,] minVal = new int[len,len];
    int[,] maxVal = new int[len,len];

    //  initializing minval and maxval 2D array
    for (int i = 0; i < len; i++)
    {
      for (int j = 0; j < len; j++) 
      {
        minVal[i, j] = Int32.MaxValue;
        maxVal[i, j] = 0;

        //  initializing main diagonal by num values
        if (i == j)
        {
          minVal[i, j] = maxVal[i, j] = num[i];
        }
      }
    }

    // looping similar to matrix chain multiplication
    // and updating both 2D arrays
    for (int L = 2; L <= len; L++)
    {
      for (int i = 0; i < len - L + 1; i++)
      {
        int j = i + L - 1;
        for (int k = i; k < j; k++)
        {
          int minTmp = 0, maxTmp = 0;

          // if current operator is '+', updating
          // tmp variable by addition
          if (opr[k] == '+')
          {
            minTmp = minVal[i, k] + minVal[k + 1, j];
            maxTmp = maxVal[i, k] + maxVal[k + 1, j];
          }

          // if current operator is '*', updating
          // tmp variable by multiplication
          else if (opr[k] == '*')
          {
            minTmp = minVal[i, k] * minVal[k + 1, j];
            maxTmp = maxVal[i, k] * maxVal[k + 1, j];
          }

          //  updating array values by tmp
          //  variables
          if (minTmp < minVal[i, j])
            minVal[i, j] = minTmp;
          if (maxTmp > maxVal[i, j])
            maxVal[i, j] = maxTmp;
        }
      }
    }

    //  last element of first row will store the result
    Console.Write("Minimum value : " + 
                  minVal[0, len - 1] + 
                  ", Maximum value : " +
                  maxVal[0,len - 1]);

  }

  //  Driver code to test above methods
  static public void Main ()
  {
    string expression = "1+2*3+4*5";
    printMinAndMaxValueOfExp(expression);
  }
}

// This code is contributed by avanitrachhadiya2155

JavaScript

<script>

// JavaScript program to get maximum and minimum
// values of an expression

// Utility method to check whether a character
// is operator or not
function isOperator(op)
{
    return (op == '+' || op == '*');
}

 // method prints minimum and maximum value
  // obtainable from an expression
function printMinAndMaxValueOfExp(exp)
{
    let num = [];
    let opr = [];
    let tmp = "";
 
    //  store operator and numbers in different vectors
    for (let i = 0; i < exp.length; i++)
    {
      if (isOperator(exp[i]))
      {
        opr.push(exp[i]);
        num.push(parseInt(tmp));
        tmp = "";
      }
      else
      {
        tmp += exp[i];
      }
    }
 
    //  storing last number in vector
    num.push(parseInt(tmp));
 
    let len = num.length;
    let minVal = new Array(len);
    let maxVal = new Array(len);
 
    //  initializing minval and maxval 2D array
    for (let i = 0; i < len; i++)
    {
        minVal[i]=new Array(len);
        maxVal[i]=new Array(len);
      for (let j = 0; j < len; j++)
      {
        minVal[i][j] = Number.MAX_VALUE;
        maxVal[i][j] = 0;
 
        //  initializing main diagonal by num values
        if (i == j)
          minVal[i][j] = maxVal[i][j]
          = num[i];
      }
    }
 
    // looping similar to matrix chain multiplication
    // and updating both 2D arrays
    for (let L = 2; L <= len; L++)
    {
      for (let i = 0; i < len - L + 1; i++)
      {
        let j = i + L - 1;
        for (let k = i; k < j; k++)
        {
          let minTmp = 0, maxTmp = 0;
 
          // if current operator is '+', updating
          // tmp variable by addition
          if (opr[k] == '+')
          {
            minTmp = minVal[i][k]
              + minVal[k + 1][j];
            maxTmp = maxVal[i][k]
              + maxVal[k + 1][j];
          }
 
          // if current operator is '*', updating
          // tmp variable by multiplication
          else if (opr[k] == '*')
          {
            minTmp = minVal[i][k]
              * minVal[k + 1][j];
            maxTmp = maxVal[i][k]
              * maxVal[k + 1][j];
          }
 
          //  updating array values by tmp
          //  variables
          if (minTmp < minVal[i][j])
            minVal[i][j] = minTmp;
          if (maxTmp > maxVal[i][j])
            maxVal[i][j] = maxTmp;
        }
      }
    }
 
    //  last element of first row will store the result
    document.write(
      "Minimum value : " + minVal[0][len - 1]
      + ", Maximum value : " + maxVal[0][len - 1]);
}

//  Driver code to test above methods
let expression = "1+2*3+4*5";
printMinAndMaxValueOfExp(expression);

// This code is contributed by ab2127

</script>

Output:

Minimum value : 27, Maximum value : 105

Time Complexity: O(n^3), as we have three nested loops in the function.

Space Complexity: O(n^2), as Two 2D arrays (minVal and maxVal) are created, each with dimensions n x n.

Analysis of Algorithms

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Minimum and Maximum values of an expression with * and +

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