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feat: add solutions to lc problem: No.2232
No.2232.Minimize Result by Adding Parentheses to Expression
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Diff for: ‎solution/2200-2299/2232.Minimize Result by Adding Parentheses to Expression/README.md

+40-2
Original file line numberDiff line numberDiff line change
@@ -55,22 +55,60 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举左右括号的插入位置**
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minimizeResult(self, expression: str) -> str:
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l, r = expression.split("+")
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m, n = len(l), len(r)
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mi = inf
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ans = None
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for i in range(m):
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for j in range(n):
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c = int(l[i:]) + int(r[: j + 1])
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a = 1 if i == 0 else int(l[:i])
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b = 1 if j == n - 1 else int(r[j + 1 :])
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if (t := a * b * c) < mi:
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mi = t
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ans = f"{l[:i]}({l[i:]}+{r[: j + 1]}){r[j + 1:]}"
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String minimizeResult(String expression) {
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int idx = expression.indexOf('+');
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String l = expression.substring(0, idx);
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String r = expression.substring(idx + 1);
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int m = l.length(), n = r.length();
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int mi = Integer.MAX_VALUE;
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String ans = "";
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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int c = Integer.parseInt(l.substring(i)) + Integer.parseInt(r.substring(0, j + 1));
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int a = i == 0 ? 1 : Integer.parseInt(l.substring(0, i));
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int b = j == n - 1 ? 1 : Integer.parseInt(r.substring(j + 1));
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int t = a * b * c;
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if (t < mi) {
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mi = t;
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ans = String.format("%s(%s+%s)%s", l.substring(0, i), l.substring(i), r.substring(0, j + 1), r.substring(j + 1));
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}
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}
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}
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return ans;
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}
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}
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```
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### **TypeScript**

Diff for: ‎solution/2200-2299/2232.Minimize Result by Adding Parentheses to Expression/README_EN.md

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@@ -57,13 +57,49 @@ It can be shown that 170 is the smallest possible value.
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### **Python3**
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```python
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class Solution:
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def minimizeResult(self, expression: str) -> str:
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l, r = expression.split("+")
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m, n = len(l), len(r)
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mi = inf
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ans = None
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for i in range(m):
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for j in range(n):
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c = int(l[i:]) + int(r[: j + 1])
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a = 1 if i == 0 else int(l[:i])
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b = 1 if j == n - 1 else int(r[j + 1 :])
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if (t := a * b * c) < mi:
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mi = t
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ans = f"{l[:i]}({l[i:]}+{r[: j + 1]}){r[j + 1:]}"
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return ans
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```
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### **Java**
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```java
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class Solution {
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public String minimizeResult(String expression) {
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int idx = expression.indexOf('+');
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String l = expression.substring(0, idx);
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String r = expression.substring(idx + 1);
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int m = l.length(), n = r.length();
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int mi = Integer.MAX_VALUE;
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String ans = "";
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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int c = Integer.parseInt(l.substring(i)) + Integer.parseInt(r.substring(0, j + 1));
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int a = i == 0 ? 1 : Integer.parseInt(l.substring(0, i));
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int b = j == n - 1 ? 1 : Integer.parseInt(r.substring(j + 1));
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int t = a * b * c;
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if (t < mi) {
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mi = t;
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ans = String.format("%s(%s+%s)%s", l.substring(0, i), l.substring(i), r.substring(0, j + 1), r.substring(j + 1));
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}
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}
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}
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return ans;
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}
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}
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```
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### **TypeScript**

Diff for: ‎solution/2200-2299/2232.Minimize Result by Adding Parentheses to Expression/Solution.java

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@@ -0,0 +1,23 @@
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class Solution {
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public String minimizeResult(String expression) {
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int idx = expression.indexOf('+');
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String l = expression.substring(0, idx);
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String r = expression.substring(idx + 1);
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int m = l.length(), n = r.length();
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int mi = Integer.MAX_VALUE;
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String ans = "";
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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int c = Integer.parseInt(l.substring(i)) + Integer.parseInt(r.substring(0, j + 1));
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int a = i == 0 ? 1 : Integer.parseInt(l.substring(0, i));
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int b = j == n - 1 ? 1 : Integer.parseInt(r.substring(j + 1));
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int t = a * b * c;
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if (t < mi) {
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mi = t;
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ans = String.format("%s(%s+%s)%s", l.substring(0, i), l.substring(i), r.substring(0, j + 1), r.substring(j + 1));
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}
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}
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}
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return ans;
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}
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}

Diff for: ‎solution/2200-2299/2232.Minimize Result by Adding Parentheses to Expression/Solution.py

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class Solution:
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def minimizeResult(self, expression: str) -> str:
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l, r = expression.split("+")
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m, n = len(l), len(r)
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mi = inf
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ans = None
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for i in range(m):
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for j in range(n):
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c = int(l[i:]) + int(r[: j + 1])
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a = 1 if i == 0 else int(l[:i])
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b = 1 if j == n - 1 else int(r[j + 1 :])
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if (t := a * b * c) < mi:
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mi = t
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ans = f"{l[:i]}({l[i:]}+{r[: j + 1]}){r[j + 1:]}"
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return ans

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