Translate all code to English (#1836)

* Review the EN heading format.

* Fix pythontutor headings.

* Fix pythontutor headings.

* bug fixes

* Fix headings in **/summary.md

* Revisit the CN-to-EN translation for Python code using Claude-4.5

* Revisit the CN-to-EN translation for Java code using Claude-4.5

* Revisit the CN-to-EN translation for Cpp code using Claude-4.5.

* Fix the dictionary.

* Fix cpp code translation for the multipart strings.

* Translate Go code to English.

* Update workflows to test EN code.

* Add EN translation for C.

* Add EN translation for CSharp.

* Add EN translation for Swift.

* Trigger the CI check.

* Revert.

* Update en/hash_map.md

* Add the EN version of Dart code.

* Add the EN version of Kotlin code.

* Add missing code files.

* Add the EN version of JavaScript code.

* Add the EN version of TypeScript code.

* Fix the workflows.

* Add the EN version of Ruby code.

* Add the EN version of Rust code.

* Update the CI check for the English version  code.

* Update Python CI check.

* Fix cmakelists for en/C code.

* Fix Ruby comments
This commit is contained in:
Yudong Jin
2025-12-31 07:44:52 +08:00
committed by GitHub
parent 45e1295241
commit 2778a6f9c7
1284 changed files with 71557 additions and 3275 deletions

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@@ -6,14 +6,14 @@
#include "../utils/common.hpp"
/* Coin change: Greedy */
/* Coin change: Greedy algorithm */
int coinChangeGreedy(vector<int> &coins, int amt) {
// Assume coins list is ordered
// Assume coins list is sorted
int i = coins.size() - 1;
int count = 0;
// Loop for greedy selection until no remaining amount
// Loop to make greedy choices until no remaining amount
while (amt > 0) {
// Find the smallest coin close to and less than the remaining amount
// Find the coin that is less than and closest to the remaining amount
while (i > 0 && coins[i] > amt) {
i--;
}
@@ -27,34 +27,34 @@ int coinChangeGreedy(vector<int> &coins, int amt) {
/* Driver Code */
int main() {
// Greedy: can ensure finding a global optimal solution
// Greedy algorithm: Can guarantee finding the global optimal solution
vector<int> coins = {1, 5, 10, 20, 50, 100};
int amt = 186;
int res = coinChangeGreedy(coins, amt);
cout << "\ncoins = ";
printVector(coins);
cout << "amt = " << amt << endl;
cout << "The minimum number of coins required to make up " << amt << " is " << res << endl;
cout << "Minimum number of coins needed to make " << amt << " is " << res << endl;
// Greedy: cannot ensure finding a global optimal solution
// Greedy algorithm: Cannot guarantee finding the global optimal solution
coins = {1, 20, 50};
amt = 60;
res = coinChangeGreedy(coins, amt);
cout << "\ncoins = ";
printVector(coins);
cout << "amt = " << amt << endl;
cout << "The minimum number of coins required to make up " << amt << " is " << res << endl;
cout << "In reality, the minimum number needed is 3, i.e., 20 + 20 + 20" << endl;
cout << "Minimum number of coins needed to make " << amt << " is " << res << endl;
cout << "Actually the minimum number needed is 3, i.e., 20 + 20 + 20" << endl;
// Greedy: cannot ensure finding a global optimal solution
// Greedy algorithm: Cannot guarantee finding the global optimal solution
coins = {1, 49, 50};
amt = 98;
res = coinChangeGreedy(coins, amt);
cout << "\ncoins = ";
printVector(coins);
cout << "amt = " << amt << endl;
cout << "The minimum number of coins required to make up " << amt << " is " << res << endl;
cout << "In reality, the minimum number needed is 2, i.e., 49 + 49" << endl;
cout << "Minimum number of coins needed to make " << amt << " is " << res << endl;
cout << "Actually the minimum number needed is 2, i.e., 49 + 49" << endl;
return 0;
}

View File

@@ -16,9 +16,9 @@ class Item {
}
};
/* Fractional knapsack: Greedy */
/* Fractional knapsack: Greedy algorithm */
double fractionalKnapsack(vector<int> &wgt, vector<int> &val, int cap) {
// Create an item list, containing two properties: weight, value
// Create item list with two attributes: weight, value
vector<Item> items;
for (int i = 0; i < wgt.size(); i++) {
items.push_back(Item(wgt[i], val[i]));
@@ -29,13 +29,13 @@ double fractionalKnapsack(vector<int> &wgt, vector<int> &val, int cap) {
double res = 0;
for (auto &item : items) {
if (item.w <= cap) {
// If the remaining capacity is sufficient, put the entire item into the knapsack
// If remaining capacity is sufficient, put the entire current item into the knapsack
res += item.v;
cap -= item.w;
} else {
// If the remaining capacity is insufficient, put part of the item into the knapsack
// If remaining capacity is insufficient, put part of the current item into the knapsack
res += (double)item.v / item.w * cap;
// No remaining capacity left, thus break the loop
// No remaining capacity, so break out of the loop
break;
}
}
@@ -50,7 +50,7 @@ int main() {
// Greedy algorithm
double res = fractionalKnapsack(wgt, val, cap);
cout << "The maximum value within the bag capacity is " << res << endl;
cout << "Maximum item value not exceeding knapsack capacity is " << res << endl;
return 0;
}

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@@ -6,15 +6,15 @@
#include "../utils/common.hpp"
/* Maximum capacity: Greedy */
/* Max capacity: Greedy algorithm */
int maxCapacity(vector<int> &ht) {
// Initialize i, j, making them split the array at both ends
// Initialize i, j to be at both ends of the array
int i = 0, j = ht.size() - 1;
// Initial maximum capacity is 0
// Initial max capacity is 0
int res = 0;
// Loop for greedy selection until the two boards meet
while (i < j) {
// Update maximum capacity
// Update max capacity
int cap = min(ht[i], ht[j]) * (j - i);
res = max(res, cap);
// Move the shorter board inward
@@ -33,7 +33,7 @@ int main() {
// Greedy algorithm
int res = maxCapacity(ht);
cout << "The maximum capacity is " << res << endl;
cout << "Maximum capacity is " << res << endl;
return 0;
}

View File

@@ -6,17 +6,17 @@
#include "../utils/common.hpp"
/* Maximum product of cutting: Greedy */
/* Max product cutting: Greedy algorithm */
int maxProductCutting(int n) {
// When n <= 3, must cut out a 1
if (n <= 3) {
return 1 * (n - 1);
}
// Greedy cut out 3s, a is the number of 3s, b is the remainder
// Greedily cut out 3, a is the number of 3s, b is the remainder
int a = n / 3;
int b = n % 3;
if (b == 1) {
// When the remainder is 1, convert a pair of 1 * 3 into 2 * 2
// When the remainder is 1, convert a pair of 1 * 3 to 2 * 2
return (int)pow(3, a - 1) * 2 * 2;
}
if (b == 2) {