Minimum number that can be obtained by applying '+' and '*' operations on array elements
Last Updated : 04 Aug, 2021
Given an array arr[] consisting of N positive integers and a string S of length (N - 1), containing characters '+' and '*', the task is to find the smallest number that can be obtained after applying the arithmetic operations mentioned in the string S on the array elements in any order.
Examples:
Input: A[] ={2, 2, 2, 2}, S = "**+"
Output: 8
Explanation:
Operation 1: Perform the multiplication operation on the first two elements operation i.e., arr[0]*arr[1] = 2*2 = 4. Now the array modifies to {4, 2, 2}.
Operation 2: Perform the multiplication operation on the remaining two elements i.e., arr[1]*arr[2] = 2*2 = 4. Now the array modifies to {4, 4}.
Operation 3: Perform the addition operation on the remaining elements arr[0] + arr[1] = 4 + 4 = 8. Now the array modifies to {8}.
Therefore, the result of the operation performed is 8, which is minimum.
Input: A[] = {1, 2, 3, 4}, S = "*++"
Output: 9
Approach: The given problem can be solved using bit-masking. Follow the steps below to solve the problem:
- Store the count of the multiplication and addition operation in the string S in variables, say mul and add respectively.
- Now, the operation applied on the elements of the array can be encoded in a mask(binary string) such that if the ith bit of mask is set then it is equal to '1', and multiplication operation has to be performed. Otherwise, perform addition.
- Initialize a variable, say ans as INT_MAX that stores the minimum value of the result.
- So, create all masks in the range [0, 2(N - 1)] and find the number of set bits in the mask and perform the following steps:
- If the number of set bits in the mask is equal to mul, then apply this mask on the array A[] i.e., if the ith bit of mask is '1' then perform multiplication operation on the ith element and (i + 1)th element.
- Otherwise, perform addition.
- Update the value of ans to the minimum of the ans and the calculated value in the above step.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the smallest number // that can be obtained after applying // the arithmetic operations mentioned // in the string S int minimumSum(int A[], int N, string S) { // Stores the count of multiplication // operator in the string int mul = 0; for (int i = 0; i < (int)S.size(); i++) { if (S[i] == '*') mul += 1; } // Store the required result int ans = 1000000; // Iterate in the range to // create the mask for (int i = 0; i < (1 << (N - 1)); i++) { int cnt = 0; vector<char> v; // Checking the number of bits // that are set in the mask for (int j = 0; j < N - 1; j++) { if ((1 << j) & (i)) { cnt += 1; v.push_back('*'); } else { v.push_back('+'); } } // Check if the number of bits // that are set in the mask is // multiplication operation if (cnt == mul) { // Storing the elements // that is to be added deque<int> d; d.push_back(A[0]); // Apply the multiplications // operation first for (int j = 0; j < N - 1; j++) { // If sign is '*', then // multiply last element // of deque with arr[i] if (v[j] == '*') { int x = d.back(); d.pop_back(); x = x * A[j + 1]; // Push last multiplied // element in the deque d.push_back(x); } else { // If the element is to // be added, then add // it to the deque d.push_back(A[j + 1]); } } int sum = 0; // Add all the element of // the deque while (d.size() > 0) { int x = d.front(); sum += x; d.pop_front(); } // Minimize the answer with // the given sum ans = min(ans, sum); } } return ans; } // Driver Code int main() { int A[] = { 2, 2, 2, 2 }; string S = "**+"; int N = sizeof(A) / sizeof(A[0]); cout << minimumSum(A, N, S); return 0; } // Java program for the above approach import java.util.*; class GFG{ // Function to find the smallest number // that can be obtained after applying // the arithmetic operations mentioned // in the String S static int minimumSum(int A[], int N, String S) { // Stores the count of multiplication // operator in the String int mul = 0; for(int i = 0; i < (int)S.length(); i++) { if (S.charAt(i) == '*') mul += 1; } // Store the required result int ans = 1000000; // Iterate in the range to // create the mask for(int i = 0; i < (1 << (N - 1)); i++) { int cnt = 0; Vector<Character> v = new Vector<Character>(); // Checking the number of bits // that are set in the mask for(int j = 0; j < N - 1; j++) { if (((1 << j) & (i)) > 0) { cnt += 1; v.add('*'); } else { v.add('+'); } } // Check if the number of bits // that are set in the mask is // multiplication operation if (cnt == mul) { // Storing the elements // that is to be added LinkedList<Integer> d = new LinkedList<Integer>(); d.add(A[0]); // Apply the multiplications // operation first for(int j = 0; j < N - 1; j++) { // If sign is '*', then // multiply last element // of deque with arr[i] if (v.get(j) == '*') { int x = d.getLast(); d.removeLast(); x = x * A[j + 1]; // Push last multiplied // element in the deque d.add(x); } else { // If the element is to // be added, then add // it to the deque d.add(A[j + 1]); } } int sum = 0; // Add all the element of // the deque while (d.size() > 0) { int x = d.peek(); sum += x; d.removeFirst(); } // Minimize the answer with // the given sum ans = Math.min(ans, sum); } } return ans; } // Driver Code public static void main(String[] args) { int A[] = { 2, 2, 2, 2 }; String S = "**+"; int N = A.length; System.out.print(minimumSum(A, N, S)); } } // This code is contributed by shikhasingrajput # Python3 program for the above approach # Function to find the smallest number # that can be obtained after applying # the arithmetic operations mentioned # in the string S def minimumSum(A, N, S): # Stores the count of multiplication # operator in the string mul = 0 for i in range(len(S)): if (S[i] == "*"): mul += 1 # Store the required result ans = 1000000 # Iterate in the range to # create the mask for i in range(1 << (N - 1)): cnt = 0 v = [] # Checking the number of bits # that are set in the mask for j in range(N - 1): if ((1 << j) & i): cnt += 1 v.append("*") else: v.append("+") # Check if the number of bits # that are set in the mask is # multiplication operation if (cnt == mul): # Storing the elements # that is to be added d = [] d.append(A[0]) # Apply the multiplications # operation first for j in range(N - 1): # If sign is '*', then # multiply last element # of deque with arr[i] if (v[j] == "*"): x = d[len(d) - 1] d.pop() x = x * A[j + 1] # append last multiplied # element in the deque d.append(x) else: # If the element is to # be added, then add # it to the deque d.append(A[j + 1]) sum = 0 # Add all the element of # the deque while (len(d) > 0): x = d[0] sum += x d.pop(0) # Minimize the answer with # the given sum ans = min(ans, sum) return ans # Driver Code A = [ 2, 2, 2, 2 ] S = "**+" N = len(A) print(minimumSum(A, N, S)) # This code is contributed by gfgking // C# program for the above approach using System; using System.Collections.Generic; public class GFG{ // Function to find the smallest number // that can be obtained after applying // the arithmetic operations mentioned // in the String S static int minimumSum(int []A, int N, String S) { // Stores the count of multiplication // operator in the String int mul = 0; for(int i = 0; i < (int)S.Length; i++) { if (S[i] == '*') mul += 1; } // Store the required result int ans = 1000000; // Iterate in the range to // create the mask for(int i = 0; i < (1 << (N - 1)); i++) { int cnt = 0; List<char> v = new List<char>(); // Checking the number of bits // that are set in the mask for(int j = 0; j < N - 1; j++) { if (((1 << j) & (i)) > 0) { cnt += 1; v.Add('*'); } else { v.Add('+'); } } // Check if the number of bits // that are set in the mask is // multiplication operation if (cnt == mul) { // Storing the elements // that is to be added List<int> d = new List<int>(); d.Add(A[0]); // Apply the multiplications // operation first for(int j = 0; j < N - 1; j++) { // If sign is '*', then // multiply last element // of deque with arr[i] if (v[j] == '*') { int x = d[d.Count-1]; d.RemoveAt(d.Count-1); x = x * A[j + 1]; // Push last multiplied // element in the deque d.Add(x); } else { // If the element is to // be added, then add // it to the deque d.Add(A[j + 1]); } } int sum = 0; // Add all the element of // the deque while (d.Count > 0) { int x = d[0]; sum += x; d.RemoveAt(0); } // Minimize the answer with // the given sum ans = Math.Min(ans, sum); } } return ans; } // Driver Code public static void Main(String[] args) { int []A = { 2, 2, 2, 2 }; String S = "**+"; int N = A.Length; Console.Write(minimumSum(A, N, S)); } } // This code contributed by shikhasingrajput <script> // Javascript program for the above approach // Function to find the smallest number // that can be obtained after applying // the arithmetic operations mentioned // in the string S function minimumSum(A, N, S) { // Stores the count of multiplication // operator in the string let mul = 0; for(let i = 0; i < S.length; i++) { if (S[i] == "*") mul += 1; } // Store the required result let ans = 1000000; // Iterate in the range to // create the mask for(let i = 0; i < 1 << (N - 1); i++) { let cnt = 0; let v = []; // Checking the number of bits // that are set in the mask for(let j = 0; j < N - 1; j++) { if ((1 << j) & i) { cnt += 1; v.push("*"); } else { v.push("+"); } } // Check if the number of bits // that are set in the mask is // multiplication operation if (cnt == mul) { // Storing the elements // that is to be added let d = []; d.push(A[0]); // Apply the multiplications // operation first for(let j = 0; j < N - 1; j++) { // If sign is '*', then // multiply last element // of deque with arr[i] if (v[j] == "*") { let x = d[d.length - 1]; d.pop(); x = x * A[j + 1]; // Push last multiplied // element in the deque d.push(x); } else { // If the element is to // be added, then add // it to the deque d.push(A[j + 1]); } } let sum = 0; // Add all the element of // the deque while (d.length > 0) { let x = d[0]; sum += x; d.shift(); } // Minimize the answer with // the given sum ans = Math.min(ans, sum); } } return ans; } // Driver Code let A = [ 2, 2, 2, 2 ]; let S = "**+"; let N = A.length; document.write(minimumSum(A, N, S)); // This code is contributed by gfgking </script> Time Complexity: O(2(N - 1)*N)
Auxiliary Space: O(N)
Similar Reads
Minimum number of steps required to obtain the given Array by the given operations Given an array arr[] of N positive integers, the task is to find the minimum number of operations required of the following types to obtain the array arr[] from an array of zeroes only. Select any index i and increment all the elements at the indices [i, N - 1] by 1.Select any index i and decrease a
12 min read
Minimum number of increment/decrement operations such that array contains all elements from 1 to N Given an array of N elements, the task is to convert it into a permutation (Each number from 1 to N occurs exactly once) by using the following operations a minimum number of times: Increment any number.Decrement any number. Examples: Input: arr[] = {1, 1, 4} Output: 2 The array can be converted int
4 min read
Check if K can be obtained by performing arithmetic operations on any permutation of an Array Given an array arr[] of N integers and an integer K, the task is to check if the expression formed for any permutation of the given array after assigning arithmetic operators(+, -, /, *) gives the value K or not. If found to be true, then print the order of operations performed. Otherwise, print "-1
15 min read
Minimum operations required to make all elements in an array of first N odd numbers equal Given an array consisting of first N odd numbers, the task is to find the minimum number of operations required to make all the array elements equal by repeatedly selecting a pair and incrementing one element and decrementing the other element in the pair by 1. Examples: Input: N = 3Output: 2Explana
10 min read
Minimize operations to convert Array elements to 0s Given an array nums[] of length N containing non-negative integers, the task is to convert all elements from index 0 to N-2 to zero, after doing the below operations minimum number of times. In one operation select two indices i and j such that i < j and all the elements between i and j has to be
6 min read