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Leetcode Expression Add Operators problem solution

YASH PAL, 31 July 202421 January 2026

In this Leetcode Expression Add Operators problem solution, we have given a string num that contains only digits and an integer target, return all possibilities to add the binary operators ‘+’, ‘-‘, or ‘*’ between the digits of num so that the resultant expression evaluates to the target value.

Leetcode Expression Add Operators problem solution

Leetcode Expression Add Operators problem solution in Python.

class Solution(object):
    def addOperators(self, num, target):
        """
        :type num: str
        :type target: int
        :rtype: List[str]
        """
        if not num:
            return []
        ans = [num[0]]
        ams = [[int(num[0])]]
        for n in range(1,len(num)):
            temp = []
            tenp = []
            for i,a in enumerate(ans):
                temp.append(a+"+"+num[n]);
                tenp.append(ams[i]+[int(num[n])])
                temp.append(a+"-"+num[n]);
                tenp.append(ams[i]+[-int(num[n])])
                
                temp.append(a+"*"+num[n]);
                tamsa = [b for b in ams[i]]
                tamsa[-1]*= int(num[n])
                tenp.append(tamsa)
                
                if a[-1]!= "0" or (len(a)>1 and a[-2].isalnum()):
                    temp.append(a+""+num[n]);
                    tams = [b for b in ams[i]]
                    tams[-1]= tams[-1]*10+int(num[n])
                    tenp.append(tams)
            ans = temp
            ams = tenp
            # print(ans, ams)
        aans = []
        for i,a in enumerate(ans):
            if sum(ams[i])==target:
                aans.append(a)
        return aans

Expression Add Operators problem solution in Java.

class Solution {
    int n;
    String num;
    List<String> list;
    public List<String> addOperators(String num, int target) {
        n = num.length();
        this.num = num;
        list = new ArrayList<>();
        if(n>0){
            f(0,target,1,1,num.charAt(0)-'0',"" + num.charAt(0));
        }
        return list;
    }
    
    void f(int i,long target,int c,long x,long y,String expr){
        if(i==n-1){
            if(target == c*x*y){
                list.add(expr);
            }
            return;
        }
        int p = num.charAt(i+1) - '0';
        long prd = y*x;
        f(i+1,target             ,c,prd,p,expr + "*" + num.charAt(i+1));
        f(i+1,target - prd*c,1,    1,p,expr + "+" + num.charAt(i+1));
        f(i+1,target - prd*c,-1,   1,p,expr + "-" + num.charAt(i+1));
        if(y!=0){
            f(i+1,target, c,x, 10*y + p,expr + num.charAt(i+1));
        }
    }
}

Problem solution in C++.

class Solution 
{
public:    
    vector<string> addOperators(string num, long long target) 
    {
        _number = std::move(num);
        
        if (!_number.empty())
        {
            _buffer.resize(_number.size());
            tryAddSymbol(0, 0, -target, 1, _number[0] - '0');
        }
        
        return move(_result);
    }
private:
    
    void addResultString()
    {
        string str;
        for (size_t i = 0; i < _buffer.size(); ++i)
        {
            if (_buffer[i])
            {
                str += _buffer[i];
            }
            str += _number[i];
        }
        _result.push_back(std::move(str));
    }
    
    void tryAddSymbol(char symbol, size_t index, long long value1, long long value2, long long value3)
    {
        _buffer[index++] = symbol;
        if (index != _number.size())
        {
            const auto digit = _number[index] - '0';
            tryAddSymbol('+', index, value1 + value2 * value3, 1, digit);
            tryAddSymbol('-', index, value1 + value2 * value3, -1, digit);
            tryAddSymbol('*', index, value1, value2 * value3, digit);
            if (value3)
            {
                tryAddSymbol('', index, value1, value2, 10 * value3 + digit);
            }
        }
        else if (value1 + value2 * value3 == 0)
        {
            addResultString();
        }
    }
    
private:
    
    std::string _number;
    std::vector<char> _buffer;
    vector<string> _result;
};

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