Expression Add Operators Problem


Description

LeetCode Problem 282.

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:

1
2
3
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:

1
2
3
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:

1
2
3
4
Input: num = "105", target = 5
Output: ["1*0+5","10-5"]
Explanation: Both "1*0+5" and "10-5" evaluate to 5.
Note that "1-05" is not a valid expression because the 5 has a leading zero.

Example 4:

1
2
3
4
Input: num = "00", target = 0
Output: ["0*0","0+0","0-0"]
Explanation: "0*0", "0+0", and "0-0" all evaluate to 0.
Note that "00" is not a valid expression because the 0 has a leading zero.

Example 5:

1
2
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.
  • -2^31 <= target <= 2^31 - 1


Sample C++ Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
public:
    vector<string> ans;
    string s;
    int target;
    vector<string> addOperators(string s, int target) {
        this->s = s;
        this->target = target;
        backtrack( 0, "", 0, 0);
        return ans;
    }
    void backtrack(int i, const string& path, long resSoFar, long prevNum) {
        if (i == s.length()) {
            if (resSoFar == target) ans.push_back(path);
            return;
        }
        string numStr;
        long num = 0;
        for (int j = i; j < s.length(); j++) {
            if (j > i && s[i] == '0') break; // Skip leading zero number
            numStr += s[j];
            num = num * 10 + s[j] - '0';
            if (i == 0) {
                backtrack(j + 1, path + numStr, num, num); // First num, pick it without adding any operator!
            } else {
                backtrack(j + 1, path + "+" + numStr, resSoFar + num, num);
                backtrack(j + 1, path + "-" + numStr, resSoFar - num, -num);
                backtrack(j + 1, path + "*" + numStr, resSoFar - prevNum + prevNum * num, prevNum * num); // Can imagine with example: 1-2*3
            }
        }
    }
};




Related Posts

24 Game Problem

LeetCode 679. You are given an integer array cards of...

Matchsticks To Square Problem

LeetCode 473. You are given an integer array matchsticks where...

Increasing Subsequences Problem

LeetCode 491. Given an integer array nums, return all the...

Binary Watch Problem

LeetCode 401. A binary watch has 4 LEDs on the...

Beautiful Arrangement Problem

LeetCode 526. Suppose you have n integers labeled 1 through...

Remove Invalid Parentheses Problem

LeetCode 301. Given a string s that contains parentheses and...