Generate a sequence such that float division of array elements is maximized
Given an array arr[] consisting of N integers, the task is to find the expression by using parenthesis β(β and β)β and division operator β/β to maximize the value of the expression of subsequent float division of array elements.
Examples:
Input: arr[] = {1000, 100, 10, 2}
Output: β1000/(100/10/2)β
Explanation:
The value of the expression 1000/(100/10/2) can be calculated as 1000/((100/10)/2) = 200.Input: arr[] = {2, 3, 4}
Output: β2/(3/4)β
Brute Force Approach:
To solve the problem of finding the maximum and minimum value of an expression with numbers and division operators, we can divide the array into two parts, left and right. We can iterate through the array, assigning i as the dividing point where the left part is from the start to i, and the right part is from i+1 to the end.
The left and right parts can then be processed recursively, returning their maximum and minimum values along with their corresponding strings. To find the minimum value of the whole expression, we divide the minimum value of the left part by the maximum value of the right part. Similarly, to find the maximum value of the expression, we divide the maximum value of the left part by the minimum value of the right part.
To add parentheses to the expression, we need to consider the associativity of the division operator. As the associativity is from left to right, we do not need to add parentheses to the left part. However, we must add parentheses to the right part. For example, β2/(3/4)β can be formed as leftPart + β/β + β(β + rightPart + β)β, where leftPart is β2β and rightPart is β3/4β.
However, if the right part contains a single digit, we do not need to add parentheses to it. For example, in β2/3β, the left part is β2β and the right part is β3β (which contains a single digit). Therefore, we do not need to add parentheses to it, and β2/3β is a valid expression.
Algorithm:
- Create a struct/class T that contains four fields: max_val, min_val, max_str, and min_str. max_val and min_val will be used to store the maximum and minimum values of the expression, respectively. max_str and min_str will be used to store the corresponding string expressions that give rise to the maximum and minimum values, respectively.
- Create a recursive function optimal that takes in the array of integers nums, the starting index start, the ending index end, and a string res. The res string will be used to build the string expressions for the maximum and minimum values. The function should return an instance of the T struct.
- In the optimal function, if the start index is equal to the end index, then create an instance of the T struct and set all four fields to the value of nums[start]. Return this instance.
- Otherwise, create an instance of the T struct and initialize min_val to FLT_MAX and max_val to FLT_MIN.
- Loop through the array of integers from index start to index end β 1. For each index i, call the optimal function recursively with the left subarray nums[start..i] and the right subarray nums[i+1..end]. Store the returned instances of the T struct in variables left and right.
- Compute the minimum value of the expression by dividing left.min_val by right.max_val. If this value is less than the current value of min_val in the current T struct, then update min_val to this value and update min_str to be the concatenation of left.min_str, β/β, β(β, and right.max_str, followed by β)β.
- Compute the maximum value of the expression by dividing left.max_val by right.min_val. If this value is greater than the current value of max_val in the current T struct, then update max_val to this value and update max_str to be the concatenation of left.max_str, β/β, β(β, and right.min_str, followed by β)β.
- After looping through all possible dividing points, return the instance of the T struct that has the maximum value of the expression.
- Create a function optimalDivision that takes in an array of integers nums. Call the optimal function with nums, 0, nums.size() β 1, and an empty string. Return the max_str field of the returned instance of the T struct.
Below is the implementation of the approach:
C++
// C++ program for the approach #include <bits/stdc++.h> using namespace std; // Structure to store the minimum and // maximum value of an expression along // with its string representation struct T { float max_val, min_val; string min_str, max_str; }; // Function to find the minimum and // maximum value of an expression T optimal( int nums[], int start, int end) { T t; // If only one number is left if (start == end) { t.max_val = nums[start]; t.min_val = nums[start]; t.min_str = to_string(nums[start]); t.max_str = to_string(nums[start]); return t; } t.min_val = FLT_MAX; t.max_val = FLT_MIN; t.min_str = "" ; t.max_str = "" ; // Iterate through all possible // dividing points for ( int i = start; i < end; i++) { // Find the optimal value of // the left part T left = optimal(nums, start, i); // Find the optimal value of // the right part T right = optimal(nums, i + 1, end); // Find the minimum value of // the expression if (t.min_val > left.min_val / right.max_val) { t.min_val = left.min_val / right.max_val; t.min_str = left.min_str + "/" + (i + 1 != end ? "(" : "" ) + right.max_str + (i + 1 != end ? ")" : "" ); } // Find the maximum value of // the expression if (t.max_val < left.max_val / right.min_val) { t.max_val = left.max_val / right.min_val; t.max_str = left.max_str + "/" + (i + 1 != end ? "(" : "" ) + right.min_str + (i + 1 != end ? ")" : "" ); } } // Return the value of t return t; } // Function to return the string representation // of the maximum value of the expression string optimalDivision( int nums[], int n) { T t = optimal(nums, 0, n - 1); return t.max_str; } // Driver Code int main() { int nums[] = { 1000, 100, 10, 2 }; int n = sizeof (nums) / sizeof (nums[0]); cout << optimalDivision(nums, n); return 0; } |
Java
// Java program for the approach import java.util.*; // Class to store the minimum and // maximum value of an expression along // with its string representation class T { float max_val, min_val; String min_str, max_str; } // Class for the problem class GFG { // Function to find the minimum and // maximum value of an expression public static T optimal( int nums[], int start, int end) { T t = new T(); // If only one number is left if (start == end) { t.max_val = nums[start]; t.min_val = nums[start]; t.min_str = Integer.toString(nums[start]); t.max_str = Integer.toString(nums[start]); return t; } t.min_val = Float.MAX_VALUE; t.max_val = Float.MIN_VALUE; t.min_str = "" ; t.max_str = "" ; // Iterate through all possible // dividing points for ( int i = start; i < end; i++) { // Find the optimal value of // the left part T left = optimal(nums, start, i); // Find the optimal value of // the right part T right = optimal(nums, i + 1 , end); // Find the minimum value of // the expression if (t.min_val > left.min_val / right.max_val) { t.min_val = left.min_val / right.max_val; t.min_str = left.min_str + "/" + (i + 1 != end ? "(" : "" ) + right.max_str + (i + 1 != end ? ")" : "" ); } // Find the maximum value of // the expression if (t.max_val < left.max_val / right.min_val) { t.max_val = left.max_val / right.min_val; t.max_str = left.max_str + "/" + (i + 1 != end ? "(" : "" ) + right.min_str + (i + 1 != end ? ")" : "" ); } } // Return the value of t return t; } // Function to return the string representation // of the maximum value of the expression public static String optimalDivision( int nums[], int n) { T t = optimal(nums, 0 , n - 1 ); return t.max_str; } // Driver code public static void main(String args[]) { int nums[] = { 1000 , 100 , 10 , 2 }; int n = nums.length; System.out.println(optimalDivision(nums, n)); } } |
Python3
# Python3 program for the above approach # Class to store the minimum and # maximum value of an expression along # with its string representation class T: def __init__( self ): self .max_val = float ( '-inf' ) self .min_val = float ( 'inf' ) self .min_str = "" self .max_str = "" def optimal(nums, start, end): t = T() # If only one number is left if start = = end: t.max_val = nums[start] t.min_val = nums[start] t.min_str = str (nums[start]) t.max_str = str (nums[start]) return t # Iterate through all possible dividing points for i in range (start, end): # Find the optimal value of the left part left = optimal(nums, start, i) # Find the optimal value of the right part right = optimal(nums, i + 1 , end) # Find the minimum value of the expression if t.min_val > left.min_val / right.max_val: t.min_val = left.min_val / right.max_val t.min_str = left.min_str + "/" if i + 1 ! = end: t.min_str + = "(" t.min_str + = right.max_str if i + 1 ! = end: t.min_str + = ")" # Find the maximum value of the expression if t.max_val < left.max_val / right.min_val: t.max_val = left.max_val / right.min_val t.max_str = left.max_str + "/" if i + 1 ! = end: t.max_str + = "(" t.max_str + = right.min_str if i + 1 ! = end: t.max_str + = ")" # Return the value of t return t def optimal_division(nums): n = len (nums) t = optimal(nums, 0 , n - 1 ) return t.max_str # Driver code if __name__ = = "__main__" : nums = [ 1000 , 100 , 10 , 2 ] print (optimal_division(nums)) #This code is contributed by aeroabrar_31 |
C#
using System; // Class to store the minimum and // maximum value of an expression along // with its string representation class T { public float max_val, min_val; public string min_str, max_str; } // Class for the problem class GFG { // Function to find the minimum and // maximum value of an expression public static T Optimal( int [] nums, int start, int end) { T t = new T(); // If only one number is left if (start == end) { t.max_val = nums[start]; t.min_val = nums[start]; t.min_str = nums[start].ToString(); t.max_str = nums[start].ToString(); return t; } t.min_val = float .MaxValue; t.max_val = float .MinValue; t.min_str = "" ; t.max_str = "" ; // Iterate through all possible // dividing points for ( int i = start; i < end; i++) { // Find the optimal value of // the left part T left = Optimal(nums, start, i); // Find the optimal value of // the right part T right = Optimal(nums, i + 1, end); // Find the minimum value of // the expression if (t.min_val > left.min_val / right.max_val) { t.min_val = left.min_val / right.max_val; t.min_str = left.min_str + "/" + (i + 1 != end ? "(" : "" ) + right.max_str + (i + 1 != end ? ")" : "" ); } // Find the maximum value of // the expression if (t.max_val < left.max_val / right.min_val) { t.max_val = left.max_val / right.min_val; t.max_str = left.max_str + "/" + (i + 1 != end ? "(" : "" ) + right.min_str + (i + 1 != end ? ")" : "" ); } } // Return the value of t return t; } // Function to return the string representation // of the maximum value of the expression public static string OptimalDivision( int [] nums, int n) { T t = Optimal(nums, 0, n - 1); return t.max_str; } // Driver code public static void Main( string [] args) { int [] nums = { 1000, 100, 10, 2 }; int n = nums.Length; Console.WriteLine(OptimalDivision(nums, n)); } } |
Javascript
// Class to store the minimum and maximum value of an expression along // with its string representation class T { constructor() { this .max_val = 0; this .min_val = 0; this .min_str = '' ; this .max_str = '' ; } } // Function to find the minimum and maximum value of an expression function optimal(nums, start, end) { const t = new T(); // If only one number is left if (start === end) { t.max_val = nums[start]; t.min_val = nums[start]; t.min_str = nums[start].toString(); t.max_str = nums[start].toString(); return t; } t.min_val = Number.MAX_VALUE; t.max_val = Number.MIN_VALUE; t.min_str = '' ; t.max_str = '' ; // Iterate through all possible dividing points for (let i = start; i < end; i++) { // Find the optimal value of the left part const left = optimal(nums, start, i); // Find the optimal value of the right part const right = optimal(nums, i + 1, end); // Find the minimum value of the expression if (t.min_val > left.min_val / right.max_val) { t.min_val = left.min_val / right.max_val; t.min_str = left.min_str + '/' + (i + 1 !== end ? '(' : '' ) + right.max_str + (i + 1 !== end ? ')' : '' ); } // Find the maximum value of the expression if (t.max_val < left.max_val / right.min_val) { t.max_val = left.max_val / right.min_val; t.max_str = left.max_str + '/' + (i + 1 !== end ? '(' : '' ) + right.min_str + (i + 1 !== end ? ')' : '' ); } } // Return the value of t return t; } // Function to return the string representation // of the maximum value of the expression function optimalDivision(nums) { const n = nums.length; const t = optimal(nums, 0, n - 1); return t.max_str; } // Driver code const nums = [1000, 100, 10, 2]; console.log(optimalDivision(nums)); //This code is contributed by aeroabrar_31 |
1000/(100/10/2)
Time Complexity: O(n!). The number of permutations of expression after applying brackets will be in O(n!)where n is the number of items in the list.
Space Complexity: O(n2). The depth of the recursion tree will be O(n) and each node contains a string of maximum length O(n).
Approach: The idea is based on the observation that for every division, the result is maximum only when the denominator is minimum. Therefore, the task reduces to placing the parentheses and operators in such a way that the denominator is minimum. Consider the following example to solve the problem:
Consider an expression 1000 / 100 / 10 / 2.
To make its value maximum, denominator needs to be minimized. Therefore, the denominators need to be in the sequence 100, 10, 2.
Now, consider the following cases:
- 100 / (10 / 2) = (100 Γ 2) / 10 = 20
- (100 / 10) / 2 = 10 / 2 = 5
Therefore, the minimized value for the expression is obtained for the second case. Therefore, 1000 / (100 / 10 / 2) is the required sequence.
Therefore, from the above example, it can be concluded that the parenthesis needs to be placed to the sequence after the first integer that makes the whole sequence from the second integer reduced to the minimum value possible.
Follow the steps below to solve the problem:
- Initialize a string S as ββ, to store the final expression.
- If N is equal to 1, print the integer in string form.
- Otherwise, append arr[0] and β/(β in S, and then append all the remaining integers of arr[] separated by β/β.
- At last, append β)β in the string S and print the string S as the result.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to place the parenthesis // such that the result is maximized void generateSequence( int arr[], int n) { // Store the required string string ans; // Add the first integer to string ans = to_string(arr[0]); // If the size of array is 1 if (n == 1) cout << ans; // If the size of array is 2, print // the 1st integer followed by / operator // followed by the next integer else if (n == 2) { cout << ans + "/" << to_string(arr[1]); } // If size of array is exceeds two, // print 1st integer concatenated // with operators '/', '(' and next // integers with the operator '/' else { ans += "/(" + to_string(arr[1]); for ( int i = 2; i < n; i++) { ans += "/" + to_string(arr[i]); } // Add parenthesis at the end ans += ")" ; // Print the final expression cout << ans; } } // Driver Code int main() { int arr[] = { 1000, 100, 10, 2 }; int N = sizeof (arr) / sizeof (arr[0]); generateSequence(arr, N); return 0; } |
Java
// Java program for the above approach import java.io.*; class GFG { // Function to place the parenthesis // such that the result is maximized static void generateSequence( int arr[], int n) { // Store the required string String ans; // Add the first integer to string ans = Integer.toString(arr[ 0 ]); // If the size of array is 1 if (n == 1 ) System.out.println(ans); // If the size of array is 2, print // the 1st integer followed by / operator // followed by the next integer else if (n == 2 ) { System.out.println(ans + "/" + Integer.toString(arr[ 1 ])); } // If size of array is exceeds two, // print 1st integer concatenated // with operators '/', '(' and next // integers with the operator '/' else { ans += "/(" + Integer.toString(arr[ 1 ]); for ( int i = 2 ; i < n; i++) { ans += "/" + Integer.toString(arr[i]); } // Add parenthesis at the end ans += ")" ; // Print the final expression System.out.println(ans); } } // Driver Code public static void main(String[] args) { int arr[] = { 1000 , 100 , 10 , 2 }; int N = arr.length; generateSequence(arr, N); } } // This code is contributed by code_hunt. |
Python3
# Python3 program for the above approach # Function to place the parenthesis # such that the result is maximized def generateSequence(arr, n): # Store the required string ans = "" # Add the first integer to string ans = str (arr[ 0 ]) # If the size of array is 1 if (n = = 1 ): print (ans) # If the size of array is 2, print # the 1st integer followed by / operator # followed by the next integer elif (n = = 2 ): print (ans + "/" + str (arr[ 1 ])) # If size of array is exceeds two, # pr1st integer concatenated # with operators '/', '(' and next # integers with the operator '/' else : ans + = "/(" + str (arr[ 1 ]) for i in range ( 2 , n): ans + = "/" + str (arr[i]) # Add parenthesis at the end ans + = ")" # Print final expression print (ans) # Driver Code if __name__ = = '__main__' : arr = [ 1000 , 100 , 10 , 2 ] N = len (arr) generateSequence(arr, N) # This code is contributed by mohit kumar 29. |
C#
// C# program for the above approach using System; class GFG { // Function to place the parenthesis // such that the result is maximized static void generateSequence( int []arr, int n) { // Store the required string string ans= "" ; // Add the first integer to string ans = arr[0].ToString(); // If the size of array is 1 if (n == 1) Console.WriteLine(ans); // If the size of array is 2, print // the 1st integer followed by / operator // followed by the next integer else if (n == 2) { Console.WriteLine(ans + "/" + arr[1].ToString()); } // If size of array is exceeds two, // print 1st integer concatenated // with operators '/', '(' and next // integers with the operator '/' else { ans += "/(" + arr[1].ToString(); for ( int i = 2; i < n; i++) { ans += "/" + arr[i].ToString(); } // Add parenthesis at the end ans += ")" ; // Print the final expression Console.WriteLine(ans); } } // Driver Code public static void Main( string [] args) { int []arr = { 1000, 100, 10, 2 }; int N = arr.Length; generateSequence(arr, N); } } // This code is contributed by chitranayal. |
Javascript
<script> // Javascript program for the above approach // Function to place the parenthesis // such that the result is maximized function generateSequence(arr, n) { // Store the required string var ans; // Add the first integer to string ans = (arr[0].toString()); // If the size of array is 1 if (n == 1) document.write( ans); // If the size of array is 2, print // the 1st integer followed by / operator // followed by the next integer else if (n == 2) { document.write( ans + "/" + (arr[1].toString())); } // If size of array is exceeds two, // print 1st integer concatenated // with operators '/', '(' and next // integers with the operator '/' else { ans += "/(" + (arr[1].toString()); for ( var i = 2; i < n; i++) { ans += "/" + (arr[i].toString()); } // Add parenthesis at the end ans += ")" ; // Print the final expression document.write( ans); } } // Driver Code var arr = [1000, 100, 10, 2]; var N = arr.length; generateSequence(arr, N); // This code is contributed by noob2000. </script> |
1000/(100/10/2)
Time Complexity: O(N)
Auxiliary Space: O(N)
Contact Us