# Smallest expression to represent a number using single digit

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 expression are +, -, *, and / . Find the minimum length expression that satisfy 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 longer time for 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
Explanation : Not possible within 10 digits.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 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 minimum level and then to find the expression that is possible in that level. So, we have two passes in this approach. Although we can get the expression in one go, but you’ll need to scratch your head for that.

 `// CPP Program to generate minimum expression containing ` `// only given digit D that evaluates to number N. ` `#include ` `#include ` `#include ` `#include ` `#include ` ` `  `// 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; ` `} `

Output:

```Expression: (4+4+4)/4+4
Expression: (((7+7)*7)*7+7+7)/7
```