Given a number N and a digit D, we have to form an expression or equation that contains only D and that expression evaluates to N. Allowed operators in an expression are +, -, *, and / . Find the minimum length expression that satisfies the condition above and D can only appear in the expression at most 10(limit) times. Hence limit the values of N (Although the value of limit depends upon how far you want to go. But a large value of limit can take a longer time for the below approach).
Remember, there can be more than one minimum expression of D that evaluates to N but the length of that expression will be minimum.
Examples:
Input : N = 7, D = 3 Output : 3/3+ 3 + 3 Explanation : 3/3 = 1, and 1+3+3 = 7 This is the minimum expression. Input : N = 7, D = 4 Output : (4+4+4)/4 + 4 Explanation : (4+4+4) = 12, and 12/4 = 3 and 3+4 = 7 Also this is the minimum expression. Although you may find another expression but that expression can have only five 4's Input : N = 200, D = 9 Output : Expression not found! Explanation : Not possible within 10 digits.
The approach we use is Backtracking. We start with the given Digit D and start multiplying, adding, subtracting, and dividing if possible. This process is done until we find the total as N or we reach the end and we backtrack to start another path. To find the minimum expression, we find the minimum level of the recursive tree. And then apply our backtracking algorithm.
Let’s say N = 7, D = 3
The above approach is exponential. At every level, we recurse 4 more ways (at-most). So, we can say the time complexity of the method is 
where n is the number of levels in the recursive tree (or we can say the number of times we want D to appear at-most in the expression which in our case is 10).
Note: We use the above approach two times. First to find the minimum level and then to find the expression that is possible at that level. So, we have two passes in this approach. We can get the expression in one go, but you’ll need to scratch your head for that.
C++
// CPP Program to generate minimum expression containing // only given digit D that evaluates to number N. #include <climits> #include <iostream> #include <map> #include <sstream> #include <stack> // limit of Digits in the expression #define LIMIT 10 using namespace std; // map that store if a number is seen or not map<int, int> seen; // stack for storing operators stack<char> operators; int minimum = LIMIT; // function to find minimum levels in the recursive tree void minLevel(int total, int N, int D, int level) { // if total is equal to given N if (total == N) { // store if level is minimum minimum = min(minimum, level); return; } // if the last level is reached if (level == minimum) return; // if total can be divided by D. // recurse by dividing the total by D if (total % D == 0) minLevel(total / D, N, D, level + 1); // recurse for total + D minLevel(total + D, N, D, level + 1); // if total - D is greater than 0 if (total - D > 0) // recurse for total - D minLevel(total - D, N, D, level + 1); // recurse for total multiply D minLevel(total * D, N, D, level + 1); } // function to generate the minimum expression bool generate(int total, int N, int D, int level) { // if total is equal to N if (total == N) return true; // if the last level is reached if (level == minimum) return false; // if total is seen at level greater than current level // or if we haven't seen total before. Mark the total // as seen at current level if (seen.find(total) == seen.end() || seen.find(total)->second >= level) { seen[total] = level; int divide = INT_MAX; // if total is divisible by D if (total % D == 0) { divide = total / D; // if divide isn't seen before // mark it as seen if (seen.find(divide) == seen.end()) seen[divide] = level + 1; } int addition = total + D; // if addition isn't seen before // mark it as seen if (seen.find(addition) == seen.end()) seen[addition] = level + 1; int subtraction = INT_MAX; // if D can be subtracted from total if (total - D > 0) { subtraction = total - D; // if subtraction isn't seen before // mark it as seen if (seen.find(subtraction) == seen.end()) seen[subtraction] = level + 1; } int multiply = total * D; // if multiply isn't seen before // mark it as seen if (seen.find(multiply) == seen.end()) seen[multiply] = level + 1; // recurse by dividing the total if possible if (divide != INT_MAX) if (generate(divide, N, D, level + 1)) { // store the operator. operators.push('/'); return true; } // recurse by adding D to total if (generate(addition, N, D, level + 1)) { // store the operator. operators.push('+'); return true; } // recurse by subtracting D from total if (subtraction != INT_MAX) if (generate(subtraction, N, D, level + 1)) { // store the operator. operators.push('-'); return true; } // recurse by multiplying D by total if (generate(multiply, N, D, level + 1)) { // store the operator. operators.push('*'); return true; } } // expression is not found yet return false; } // function to print the expression void printExpression(int N, int D) { // find minimum level minLevel(D, N, D, 1); // generate expression if possible if (generate(D, N, D, 1)) { // stringstream for converting to D to string ostringstream num; num << D; string expression; // if stack is not empty if (!operators.empty()) { // concatenate D and operator at top of stack expression = num.str() + operators.top(); operators.pop(); } // until stack is empty // concatenate the operator with parenthesis for precedence while (!operators.empty()) { if (operators.top() == '/' || operators.top() == '*') expression = "(" + expression + num.str() + ")" + operators.top(); else expression = expression + num.str() + operators.top(); operators.pop(); } expression = expression + num.str(); // print the expression cout << "Expression: " << expression << endl; } // not possible within 10 digits. else cout << "Expression not found!" << endl; } // Driver's Code int main() { int N = 7, D = 4; // print the Expression if possible printExpression(N, D); // print expression for N =100, D =7 minimum = LIMIT; printExpression(100, 7); // print expression for N =200, D =9 minimum = LIMIT; printExpression(200, 9); return 0; } |
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Python3
# Python3 program to generate minimum expression containing # only the given digit D that evaluates to number N. LIMIT = 10 # to restrict the maximum number of times # the number D can be used to obtain N minimum = LIMIT # to store the value of intermediate D # and the D of operands used to get that intermediate D, ie # seen[intermediateNumber] = numberOfOperandsUsed seen = {} # stack to store the operators used to print the expression operators = [] # function to obtain minimum D of operands in recursive tree def minLevel(total, N, D, level): global minimum # if total is equal to given N if total == N: # store if D of operands is minimum minimum = min(minimum, level) return # if the last level (limit) is reached if level == minimum: return # if total can be divided by D, recurse # by dividing the total by D if total % D == 0: minLevel(total / D, N, D, level + 1) # recurse for total + D minLevel(total + D, N, D, level + 1) # if total - D is greater than 0, recurse for total - D if total - D > 0: minLevel(total - D, N, D, level + 1) # recurse for total * D minLevel(total * D, N, D, level + 1) # function to generate the minimum expression def generate(total, N, D, level): # if total is equal to N if total == N: return True # if the last level is reached if level == minimum: return False # if the total is not already seen or if # the total is seen with more level # then mark total as seen with current level if seen.get(total) is None or seen.get(total) >= level: seen[total] = level divide = -1 # if total is divisible by D if total % D == 0: divide = total / D # if the number (divide) is not seen, mark as seen if seen.get(divide) is None: seen[divide] = level + 1 addition = total + D # if the number (addition) is not seen, mark as seen if seen.get(addition) is None: seen[addition] = level + 1 subtraction = -1 # if D can be subtracted from total if total - D > 0: subtraction = total - D # if the number (subtraction) is not seen, mark as seen if seen.get(subtraction) is None: seen[subtraction] = level + 1 multiplication = total * D # if the number (multiplication) is not seen, mark as seen if seen.get(multiplication) is None: seen[multiplication] = level + 1 # recurse by dividing the total if possible and store the operator if divide != -1 and generate(divide, N, D, level + 1): operators.append('/') return True # recurse by adding D to total and store the operator if generate(addition, N, D, level + 1): operators.append('+') return True # recurse by subtracting D from total # if possible and store the operator if subtraction != -1 and generate(subtraction, N, D, level + 1): operators.append('-') return True # recurse by multiplying D by total and store the operator if generate(multiplication, N, D, level + 1): operators.append('*') return True # expression is not found yet return False # function to print the expression def printExpression(N, D): # find the minimum number of operands minLevel(D, N, D, 1) # generate expression if possible if generate(D, N, D, 1): expression = '' # if stack is not empty, concatenate D and # operator popped from stack if len(operators) > 0: expression = str(D) + operators.pop() # until stack is empty, concatenate the # operator popped with parenthesis for precedence while len(operators) > 0: popped = operators.pop() if popped == '/' or popped == '*': expression = '(' + expression + str(D) + ')' + popped else: expression = expression + str(D) + popped expression = expression + str(D) print("Expression: " + expression) # not possible within 10 digits. else: print("Expression not found!") # Driver's code if __name__ == '__main__': # N = 7, D = 4 minimum = LIMIT printExpression(7, 4) minimum = LIMIT printExpression(100, 7) minimum = LIMIT printExpression(200, 9) # This code is contributed by thecodingpanda |
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Output:
Expression: (4+4+4)/4+4 Expression: (((7+7)*7)*7+7+7)/7 Expression not found!
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