chore: add C++ permutation and add Go contains duplicate (#441)

Co-authored-by: Ming Tsai <37890026+ming-tsai@users.noreply.github.com>

algorithms/C/README.md

@@ -9,6 +9,7 @@
 - [Multiply](bit-manipulation/multiply-bitwise.c)
 - [Divide bitwise](bit-manipulation/divide-bitwise.c)
 
+
 ## Graphs
 - [Prim's Algorithm](graphs/Prim's-algorithm.c)
 

algorithms/CPlusPlus/README.md

@@ -114,3 +114,4 @@
 
 1. [Tower of Hanoi](Recursion/towerofHanoi.cpp)
 2. [Factorial](Recursion/factorial.cpp)
+3. [Permutation](Recursion/permutation.cpp)

algorithms/CPlusPlus/Recursion/permutation.cpp

@@ -0,0 +1,48 @@
+/*
+Given an array of distinct numbers, nums
+find all the possible permutations. You can return the answer in any order.
+
+Example:
+    Input: nums = [1,2,3]
+    Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
+*/
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+vector<vector<int>> permute(vector<int> &nums);
+void helper(vector<int> newNum, vector<int> current, vector<vector<int>> &result);
+
+//Input will be similar to the example above, free feel to test with any kind of input
+int main(){
+    vector<int> nums = {1, 2, 3};
+    vector<vector<int>> permutations = permute(nums);
+    for(int i  = 0 ; i < permutations.size(); i++){
+        for(int j = 0; j < permutations[i].size();j++){
+            cout<< permutations[i][j]<<" ";
+        }
+        cout<<endl;
+    }
+}
+
+vector<vector<int>> permute(vector<int>& nums) {
+    vector<vector<int>> result;
+    helper(nums, {}, result);
+    return result;
+}
+
+void helper(vector<int> newNum, vector<int> current, vector<vector<int>> &result){
+    if(newNum.size() == 0 && current.size() > 0){
+        result.push_back(current);
+    }else{
+        for(int i = 0; i < newNum.size(); i++){
+            vector<int> newArray;
+            newArray.insert(newArray.end(), newNum.begin(), newNum.begin() + i);
+            newArray.insert(newArray.end(), newNum.begin()+i+1, newNum.end());
+            vector<int> newPerms = current;
+            newPerms.push_back(newNum[i]);
+            helper(newArray, newPerms, result);
+        }
+    }
+}

algorithms/Go/README.md

@@ -3,6 +3,8 @@
 ## Arrays
 1. [Maximum subarray sum (Kadane's Algorithm)](arrays/maximum-subarray-sum.go)
 2. [Two Sum](arrays/two-sum.go)
+3. [Majority Element](arrays/majority_element.go)
+4. [Contains Duplicate](arrays/contains_duplicate.go)
 
 ## Scheduling
 1. [Interval Scheduling](scheduling/interval-scheduling.go)
@@ -10,6 +12,7 @@
 ## Searching
 1. [Binary Search](searching/binary-search.go)
 2. [Linear Search](searching/linear-search.go)
+3. [Find Minimum in Rotated Sorted Array](searching/rotated-array-search.go)
 
 ## Sorting
 1. [Bubble Sort](sorting/bubble-sort.go)
@@ -17,5 +20,4 @@
 3. [Quicksort](sorting/quicksort.go)
 
 ## Recursion
-1. [Fibonacci](recursion/fibonacci.go)
+1. [Fibonacci](recursion/fibonacci.go)
-

algorithms/Go/arrays/contains_duplicate.go

@@ -0,0 +1,33 @@
+package arrays
+import("fmt")
+/*
+Given an integer array nums, 
+return true if any value appears at least twice in the array, and return false if every element is distinct.
+
+Example:
+	Input: nums = [1, 2, 3, 1]
+	Output: true
+
+Time: O(n)
+Space: O(n)
+
+*/
+
+func containsDuplicate(nums []int) bool {
+    dictionary := make(map[int]int)
+    
+    for _, num := range(nums){
+    if _, found := dictionary[num]; found{
+			return true;
+		}else{
+			dictionary[num] = 1
+		}
+    }
+    return false;
+}
+
+func runContainDuplicate(){
+	nums := []int{1,2,3,1}
+	hasDuplicate := containsDuplicate(nums)
+	fmt.Printf("Contain Duplicate %t\n", hasDuplicate)
+}

algorithms/Go/arrays/majority_element.go

@@ -0,0 +1,33 @@
+package arrays
+import ("fmt")
+/*
+Given an array nums of size n, return the majority element.
+The majority element appears more than n/2 times
+
+Time: O(n)
+Space: O(1)
+*/
+
+func majorityElement(nums []int) int {
+    count := 1
+    candidate := nums[0]
+    
+    for i:= 1; i < len(nums); i++{
+        num := nums[i]
+        if count == 0{
+            candidate = nums[i]
+            count = 1
+        }else if num == candidate{
+            count++
+        }else if num != candidate{
+            count--
+        }
+    }
+    return candidate
+}
+
+func runmajorityElement(){
+	nums1 := []int{2,2,1,1,1,2,2}
+    target := majorityElement(nums1)
+    fmt.Printf("Majority Element is %d\n", target)
+}

algorithms/Go/recursion/fibonacci.go

@@ -0,0 +1,43 @@
+package recursion
+import "fmt"
+/*
+Time: O(2^n) 
+Space: O(n) - call stack
+*/
+func fibonacci_using_recursion(n int) int {
+	if n == 2{
+		return 1
+	}
+	if n == 1{
+		return 0;
+	}
+	return fibonacci_using_recursion(n-1) + fibonacci_using_recursion(n-2)
+}
+
+/*
+Time: O(N)
+Space: O(1)
+*/
+func fibonacci_using_constant_space(n int) int{
+	first:= 0
+	second:= 1
+	
+	for i := 3; i<= n; i++{
+		next:= first + second
+		first = second
+		second = next
+	}
+	
+	if n > 1{
+		return second
+	}
+	return first	// this handles n == 1 or n == 0 case
+}
+
+func run_fibonacci(){
+	n := 10
+	recursionValue := fibonacci_using_recursion(n)
+	nonrecursionValue := fibonacci_using_constant_space(n)
+	fmt.Printf("Recursion value: %d\n", recursionValue)
+	fmt.Printf("Non-recursion value: %d\n", nonrecursionValue)
+}

algorithms/Go/searching/rotated-array-search.go

@@ -0,0 +1,45 @@
+package searching
+
+import "fmt"
+
+/*
+
+Find the minimum element in the rotated sorted array(ascending order).
+An array [a[0], a[1], a[2], ..., a[n-1]] after 1 rotation will be [a[n-1], a[0], a[1], a[2], ..., a[n-2]]
+
+All elements are unique
+
+*/
+func findMin(nums []int) int {
+    if len(nums) == 1 || nums[0] < nums[len(nums)-1]{
+        return nums[0]
+    }
+    
+    front, back := 0, len(nums) -1 
+    for front <= back {
+        mid := (front+back) / 2
+        if nums[mid] > nums[mid+1]{
+            return nums[mid+1]
+        }else if nums[mid] < nums[mid-1]{
+            return nums[mid]
+        }
+
+        if nums[mid] > nums[0]{
+            front = mid + 1
+        }else if nums[mid] < nums[0]{
+            back = mid - 1
+        }
+    }
+    return 0
+}
+
+//I will hard-code the input for now but feel free to modify
+func runFindMin(){
+	nums1 := []int{4,5,6,7,8,0,1,2}
+	result1 := findMin(nums1)
+	fmt.Println(result1) 	// should print 0
+
+	nums2 := []int{1,2,3,4,5,6,7,8}
+	result2 := findMin(nums2)
+	fmt.Println(result2) 	// should print 1
+}