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 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 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
<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.
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