diff --git a/code_review/permutations_20190423_1832_1/readme.md b/code_review/permutations_20190423_1832_1/readme.md
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+
+# Overview 
+
+The task is to compute all the permutations for a given vector of integers (but of course the specific integer type is not relevant for the solution)
+
+The strategy is based on recursion + iterations 
+
+At each recursion, the state consists of  
+
+- the root sequence `a` which is the set of elements already placed 
+
+- the remaining elements set `b` which is the set of elements still to be placed 
+
+Inside the recursion, a loop places the `N(i)` (with `i` recursion index and) remaining elements producing the same amount of new root sequences and a new recursion is started so that `N(i+1)=N(i)-1` hence meaning the overall complexity is `O(N!)` as expected 
+
+The recursion ends when there are no more elements to place hence `b.empty()` is true 
+
+Each recursion set ends with a valid sequence hence they are all merged together in a final list of sequences 
+
+- [Article on CodeReview StackExchange](https://codereview.stackexchange.com/questions/217939/compute-all-the-permutations-for-a-given-vector-of-integers)
+
+# Solutions 
+
+- [Sol1](sol.cpp)
+
+
+
+
+
+
+
diff --git a/code_review/permutations_20190423_1832_1/sol.cpp b/code_review/permutations_20190423_1832_1/sol.cpp
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+#include <iostream>
+#include <vector>
+using namespace std;
+
+vector<int> remove_item (const vector<int>& a, const unsigned int id)
+{
+    vector<int> res; 
+    for(unsigned int i=0; i<a.size(); ++i) if(i!=id) res.push_back(a[i]); 
+    return res;
+}
+
+
+vector<int> add_item(const vector<int>& a, const int b)
+{
+    vector<int> res=a; 
+    res.push_back(b); 
+    return res;
+}
+
+vector< vector<int> > merge(const vector< vector<int> >& a, const vector< vector<int> >& b)
+{
+    vector< vector<int> > res=a; 
+    for(const auto& e : b) res.push_back(e); 
+    return res;
+}
+
+
+
+vector< vector<int> > permutations(const vector<int>& b, const vector<int>& a={})
+{
+
+    if(b.empty()) return { a }; 
+
+    vector< vector<int> > res; 
+    for(unsigned int i=0; i<b.size(); ++i) res=merge(res, permutations(remove_item(b,i), add_item(a, b[i]))); 
+    return res; 
+}
+
+int main() {
+    // your code goes here
+
+    auto res = permutations({1,2,3,4,5}); 
+    cout << "Sol Num = " << res.size() << endl; 
+    for(const auto& a : res) 
+    {
+        for(const auto& b : a) cout << to_string(b) << " "; 
+        cout << endl; 
+    }
+    return 0;
+}
+
+
+
diff --git a/code_review/readme.md b/code_review/readme.md
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+
+# Overview 
+
+This regards [Code Reviews](https://codereview.stackexchange.com/) related code submissions 
+
+# Permutations 
+
+- Compute all the possible permutations for a given vector 
+  - [Article](permutations_20190423_1832_1/readme.md)
+  - [Post on CodeReview StackExchange](https://codereview.stackexchange.com/questions/217939/compute-all-the-permutations-for-a-given-vector-of-integers)
+
+
+
+# Search in Array 
+
+- Search min element in an array which was previously sorted in ascending order and then rotated with respect to a random pivot 
+  - [Article](smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/readme.md)
+  - [Post on CodeReview StackExchange](https://codereview.stackexchange.com/questions/217897/find-the-smallest-element-in-a-sorted-and-rotated-array/217949#217949)
+
+
+
+
+
+
+
+Work in progress 
+
+
+
+
+
diff --git a/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/index_based1.cpp b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/index_based1.cpp
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+++ b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/index_based1.cpp
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+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+vector<int> a = {5,6,7,8,9,10,11, 1,2,3,4};
+
+unsigned int solve(const vector<int>& a, unsigned int l=0, unsigned int r=0)
+{
+    if(a.empty()) throw runtime_error("Empty"); 
+    if(a.size()==1) return 0; 
+    if(r==0) r=a.size()-1; ///< Overwrite the invalid initialization with the right value, unfortunately it is not possible to do this in function declaration 
+    if(a[l] < a[r]) return l; ///< Sorted in Ascending Order 
+    if(r-l==1) return r; 
+    const auto m = (r+l)/2; 
+    if(a[m] > a[l]) return solve(a, m, r); 
+    return solve(a, l, m); 
+}
+
+int main() {
+    // your code goes here
+    cout << "Min=" << a[solve(a)]; 
+    return 0;
+}
+
+
+
+
+
+
+
diff --git a/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/iterator_based1.cpp b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/iterator_based1.cpp
new file mode 100644
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+++ b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/iterator_based1.cpp
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+
+#include <iostream>
+#include <vector>
+#include <iterator>
+using namespace std;
+
+vector<int> a = {5,6,7,8,9,10,11, 1,2,3,4};
+
+vector<int>::iterator solve(const vector<int>::iterator l, const vector<int>::iterator r)
+{
+    if(l==r) return r; ///< Covers the single element array case 
+    if(*l < *r) return l; ///< Sorted in Ascending Order 
+    if(r-l==1) return r; 
+    const auto m = l + distance(l,r)/2;  
+    if(*m > *l) return solve(m, r); 
+    return solve(l, m); 
+}
+
+int main() {
+    // your code goes here
+    cout << "Min=" << *solve(a.begin(), a.end()-1); 
+    return 0;
+}
+
+
+
+
+
+
+
+
diff --git a/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/readme.md b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/readme.md
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+++ b/code_review/smallest_elem_in_sorted_and_rotated_array_20190423_1835_1/readme.md
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+
+# Overview 
+
+Here is my minimal solution about the problem 
+
+- the strategy is divide-and-conquer for the reasons already explained by @papagaga 
+
+- it is based on the idea that before the rotation the array has this structure 
+
+```
+[m .. p P .. M]
+```
+
+with 
+
+- `m` min 
+
+- `M` max 
+
+- `p` pivot
+ 
+- `P` next to the pivot so that `P>p`
+
+
+- while after the rotation it has the following structure 
+
+```
+[P .. M m .. p]
+```
+
+so the idea is to update the `l` left cursor and `r` right cursor so that `v[l] > v[r]` with a divide and conquer strategy so to have `O(LogN)` complexity and ultimately the final condition is `l` and `r` are contiguous 
+and the first identifies `M` while the second identifies `m` hence return the last one 
+
+**EDIT** Following @papagaga suggestion I provide 2 implementations 
+
+- [Article on CodeReview StackExchange](https://codereview.stackexchange.com/questions/217897/find-the-smallest-element-in-a-sorted-and-rotated-array/217949#217949)
+
+
+## 1. Index based solution 
+
+[Index based](index_based1.cpp)
+
+Comments  
+
+- Added the empty array case management using Exception 
+- An alternative could have been using Maybe Monad (a Boost Optional) to represent non meaningful results 
+
+
+## 2. Iterators based solution 
+
+[Iterators based](iterator_based1.cpp)
+
+Comments 
+
+- Working with iterators allows to represent invalid values so no need for Exception and explicit Maybe Monad as the iterator type is one 
+- The first check automatically manages the one element case 
+- The empty array case should be managed before calling this function as it expects both the iterators point to valid elements 
+
+
+
+
+
+
+