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# Expression Add Operators LeetCode Solution

## Problem – Expression Add Operators LeetCode Solution

Given a string `num` that contains only digits and an integer `target`, return all possibilities to insert the binary operators `'+'``'-'`, and/or `'*'` between the digits of `num` so that the resultant expression evaluates to the `target` value.

Note that operands in the returned expressions should not contain leading zeros.

Example 1:

``````Input: num = "123", target = 6
Output: ["1*2*3","1+2+3"]
Explanation: Both "1*2*3" and "1+2+3" evaluate to 6.
``````

Example 2:

``````Input: num = "232", target = 8
Output: ["2*3+2","2+3*2"]
Explanation: Both "2*3+2" and "2+3*2" evaluate to 8.
``````

Example 3:

``````Input: num = "3456237490", target = 9191
Output: []
Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191.
``````

Constraints:

• `1 <= num.length <= 10`
• `num` consists of only digits.
• `-231 <= target <= 231 - 1`

## Expression Add Operators LeetCode Solution in Java

``````public class Solution {
public List<String> addOperators(String num, int target) {
List<String> rst = new ArrayList<String>();
if(num == null || num.length() == 0) return rst;
helper(rst, "", num, target, 0, 0, 0);
return rst;
}
public void helper(List<String> rst, String path, String num, int target, int pos, long eval, long multed){
if(pos == num.length()){
if(target == eval)
return;
}
for(int i = pos; i < num.length(); i++){
if(i != pos && num.charAt(pos) == '0') break;
long cur = Long.parseLong(num.substring(pos, i + 1));
if(pos == 0){
helper(rst, path + cur, num, target, i + 1, cur, cur);
}
else{
helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur);

helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur);

helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur );
}
}
}
}
``````

## Expression Add Operators LeetCode Solution in Python

``````def addOperators(self, num, target):
res, self.target = [], target
for i in range(1,len(num)+1):
if i == 1 or (i > 1 and num != "0"): # prevent "00*" as a number
self.dfs(num[i:], num[:i], int(num[:i]), int(num[:i]), res) # this step put first number in the string
return res

def dfs(self, num, temp, cur, last, res):
if not num:
if cur == self.target:
res.append(temp)
return
for i in range(1, len(num)+1):
val = num[:i]
if i == 1 or (i > 1 and num != "0"): # prevent "00*" as a number
self.dfs(num[i:], temp + "+" + val, cur+int(val), int(val), res)
self.dfs(num[i:], temp + "-" + val, cur-int(val), -int(val), res)
``````

## Expression Add Operators LeetCode Solution in C++

``````class Solution {
private:
// cur: {string} expression generated so far.
// pos: {int}    current visiting position of num.
// cv:  {long}   cumulative value so far.
// pv:  {long}   previous operand value.
// op:  {char}   previous operator used.
void dfs(std::vector<string>& res, const string& num, const int target, string cur, int pos, const long cv, const long pv, const char op) {
if (pos == num.size() && cv == target) {
res.push_back(cur);
} else {
for (int i=pos+1; i<=num.size(); i++) {
string t = num.substr(pos, i-pos);
long now = stol(t);
if (to_string(now).size() != t.size()) continue;
dfs(res, num, target, cur+'+'+t, i, cv+now, now, '+');
dfs(res, num, target, cur+'-'+t, i, cv-now, now, '-');
dfs(res, num, target, cur+'*'+t, i, (op == '-') ? cv+pv - pv*now : ((op == '+') ? cv-pv + pv*now : pv*now), pv*now, op);
}
}
}

public:
vector<string> addOperators(string num, int target) {
vector<string> res;
if (num.empty()) return res;
for (int i=1; i<=num.size(); i++) {
string s = num.substr(0, i);
long cur = stol(s);
if (to_string(cur).size() != s.size()) continue;
dfs(res, num, target, s, i, cur, cur, '#');         // no operator defined.
}

return res;
}
};
``````
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