Description#
Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 104.
Example 1:
Input: expression = "2-1-1"
Output: [0,2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
Input: expression = "2*3-4*5"
Output: [-34,-14,-10,-10,10]
Explanation:
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
Constraints:
1 <= expression.length <= 20expression consists of digits and the operator '+', '-', and '*'.- All the integer values in the input expression are in the range
[0, 99].
Solutions#
Solution 1#
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| class Solution:
def diffWaysToCompute(self, expression: str) -> List[int]:
@cache
def dfs(exp):
if exp.isdigit():
return [int(exp)]
ans = []
for i, c in enumerate(exp):
if c in '-+*':
left, right = dfs(exp[:i]), dfs(exp[i + 1 :])
for a in left:
for b in right:
if c == '-':
ans.append(a - b)
elif c == '+':
ans.append(a + b)
else:
ans.append(a * b)
return ans
return dfs(expression)
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| class Solution {
private static Map<String, List<Integer>> memo = new HashMap<>();
public List<Integer> diffWaysToCompute(String expression) {
return dfs(expression);
}
private List<Integer> dfs(String exp) {
if (memo.containsKey(exp)) {
return memo.get(exp);
}
List<Integer> ans = new ArrayList<>();
if (exp.length() < 3) {
ans.add(Integer.parseInt(exp));
return ans;
}
for (int i = 0; i < exp.length(); ++i) {
char c = exp.charAt(i);
if (c == '-' || c == '+' || c == '*') {
List<Integer> left = dfs(exp.substring(0, i));
List<Integer> right = dfs(exp.substring(i + 1));
for (int a : left) {
for (int b : right) {
if (c == '-') {
ans.add(a - b);
} else if (c == '+') {
ans.add(a + b);
} else {
ans.add(a * b);
}
}
}
}
}
memo.put(exp, ans);
return ans;
}
}
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| class Solution {
public:
vector<int> diffWaysToCompute(string expression) {
return dfs(expression);
}
vector<int> dfs(string exp) {
if (memo.count(exp)) return memo[exp];
if (exp.size() < 3) return {stoi(exp)};
vector<int> ans;
int n = exp.size();
for (int i = 0; i < n; ++i) {
char c = exp[i];
if (c == '-' || c == '+' || c == '*') {
vector<int> left = dfs(exp.substr(0, i));
vector<int> right = dfs(exp.substr(i + 1, n - i - 1));
for (int& a : left) {
for (int& b : right) {
if (c == '-')
ans.push_back(a - b);
else if (c == '+')
ans.push_back(a + b);
else
ans.push_back(a * b);
}
}
}
}
memo[exp] = ans;
return ans;
}
private:
unordered_map<string, vector<int>> memo;
};
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| var memo = map[string][]int{}
func diffWaysToCompute(expression string) []int {
return dfs(expression)
}
func dfs(exp string) []int {
if v, ok := memo[exp]; ok {
return v
}
if len(exp) < 3 {
v, _ := strconv.Atoi(exp)
return []int{v}
}
ans := []int{}
for i, c := range exp {
if c == '-' || c == '+' || c == '*' {
left, right := dfs(exp[:i]), dfs(exp[i+1:])
for _, a := range left {
for _, b := range right {
if c == '-' {
ans = append(ans, a-b)
} else if c == '+' {
ans = append(ans, a+b)
} else {
ans = append(ans, a*b)
}
}
}
}
}
memo[exp] = ans
return ans
}
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| using System.Collections.Generic;
public class Solution {
public IList<int> DiffWaysToCompute(string input) {
var values = new List<int>();
var operators = new List<char>();
var sum = 0;
foreach (var ch in input)
{
if (ch == '+' || ch == '-' || ch == '*')
{
values.Add(sum);
operators.Add(ch);
sum = 0;
}
else
{
sum = sum * 10 + ch - '0';
}
}
values.Add(sum);
var f = new List<int>[values.Count, values.Count];
for (var i = 0; i < values.Count; ++i)
{
f[i, i] = new List<int> { values[i] };
}
for (var diff = 1; diff < values.Count; ++diff)
{
for (var left = 0; left + diff < values.Count; ++left)
{
var right = left + diff;
f[left, right] = new List<int>();
for (var i = left; i < right; ++i)
{
foreach (var leftValue in f[left, i])
{
foreach (var rightValue in f[i + 1, right])
{
switch (operators[i])
{
case '+':
f[left, right].Add(leftValue + rightValue);
break;
case '-':
f[left, right].Add(leftValue - rightValue);
break;
case '*':
f[left, right].Add(leftValue * rightValue);
break;
}
}
}
}
}
}
return f[0, values.Count - 1];
}
}
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