From 5276967a127cf3bac4a24e52e390913edc29f666 Mon Sep 17 00:00:00 2001
From: danzang100 <12danysamuel@gmail.com>
Date: Tue, 11 Mar 2025 19:11:37 +0530
Subject: [PATCH 1/2] Solution #31 - danzang100/Edited - 11.03.2025 - commit #1

---
 .../Next Permutation - Solution.cpp           | 20 +++++++++++++++++++
 .../Next Permutation.md                       |  7 +++++++
 2 files changed, 27 insertions(+)
 create mode 100644 Arrays101/#31 - Next Permutation - Medium/Next Permutation - Solution.cpp
 create mode 100644 Arrays101/#31 - Next Permutation - Medium/Next Permutation.md

diff --git a/Arrays101/#31 - Next Permutation - Medium/Next Permutation - Solution.cpp b/Arrays101/#31 - Next Permutation - Medium/Next Permutation - Solution.cpp
new file mode 100644
index 0000000..78b4adc
--- /dev/null
+++ b/Arrays101/#31 - Next Permutation - Medium/Next Permutation - Solution.cpp	
@@ -0,0 +1,20 @@
+class Solution {
+    public:
+        void nextPermutation(vector<int>& nums) {
+            int i = nums.size()-1;
+            while(i > 0 && nums[i-1] >= nums[i]){
+                i--;
+            }
+    
+            if(i==0){
+                reverse(nums.begin(), nums.end());
+                return;
+            }
+            int j = nums.size()-1;
+            while(j >= i && nums[j] <= nums[i-1]){
+                j--;
+            }
+            swap(nums[i-1], nums[j]);
+            reverse(nums.begin()+i, nums.end());
+        }
+    };
\ No newline at end of file
diff --git a/Arrays101/#31 - Next Permutation - Medium/Next Permutation.md b/Arrays101/#31 - Next Permutation - Medium/Next Permutation.md
new file mode 100644
index 0000000..9999b8c
--- /dev/null
+++ b/Arrays101/#31 - Next Permutation - Medium/Next Permutation.md	
@@ -0,0 +1,7 @@
+To solve this question,
+
+1. We find the first decreasing element from the right calling it the pivot.
+2. We check if that element does not exist, which means that the array is sorted in non-ascending order, hence we just return a reversed array.
+3. If the element exists, we find the smallest element greater than the pivot element which is to the right of it.
+4. We swap the two elements.
+5. We reverse the suffix of the pivot element to obtain the answer.

From ad2284ef524a372e7d4fcbb8ae6167b0fc084c5f Mon Sep 17 00:00:00 2001
From: danzang100 <12danysamuel@gmail.com>
Date: Wed, 12 Mar 2025 23:02:22 +0530
Subject: [PATCH 2/2] Solution #169 - Daniel/Edited - 12.03.2025

---
 .../Majority Element - Solution.cpp           | 36 +++++++++++++++++++
 .../Majority Element.md                       | 26 ++++++++++++++
 2 files changed, 62 insertions(+)
 create mode 100644 Arrays101/#169 - Majority Element - Easy/Majority Element - Solution.cpp
 create mode 100644 Arrays101/#169 - Majority Element - Easy/Majority Element.md

diff --git a/Arrays101/#169 - Majority Element - Easy/Majority Element - Solution.cpp b/Arrays101/#169 - Majority Element - Easy/Majority Element - Solution.cpp
new file mode 100644
index 0000000..33f8407
--- /dev/null
+++ b/Arrays101/#169 - Majority Element - Easy/Majority Element - Solution.cpp	
@@ -0,0 +1,36 @@
+class Solution {
+    public:
+        int majorityElement(vector<int>& nums) {
+    
+            //Sorting method
+    
+            // Hashmap method
+            unordered_map<int, int> umap;
+            for(int x: nums){
+                umap[x]++;
+            }
+            for(auto i: umap){
+                if(i.second >= ((double)nums.size()/2)){
+                    return i.first;
+                }
+            }
+            return 0;
+    
+            // Moore's voting method
+            int count = 0, candidate = nums[0];
+            for(int i=1; i<nums.size(); i++){
+                if(candidate == nums[i]){
+                    count++;
+                }
+                else{
+                    if(count == 0){
+                        candidate = nums[i];
+                    }
+                    else{
+                        count--;
+                    }
+                }
+            }
+            return candidate;
+        }
+    };
\ No newline at end of file
diff --git a/Arrays101/#169 - Majority Element - Easy/Majority Element.md b/Arrays101/#169 - Majority Element - Easy/Majority Element.md
new file mode 100644
index 0000000..f713e8c
--- /dev/null
+++ b/Arrays101/#169 - Majority Element - Easy/Majority Element.md	
@@ -0,0 +1,26 @@
+Simple question of finding an element whose frequency is greater than half of the size of the array.
+
+Three possible methods:
+
+1. Sorting:
+
+- Time Complexity: O(nlogn)
+- Space Complexity: O(1)
+
+Simply sort the array and the middle element should always be the majority element.
+
+2. Hashmap:
+
+- Time Complexity: O(n)
+- Space Complexity: O(n)
+
+Map the frequencies of the elements to a hashmap and return the element with the highest frequency.
+
+3. Moore's Voting Algorithm:
+
+- Time Complexity: O(n)
+- Space Complexity: O(1)
+
+Initialise count and candidate element. If the element is the candidate element, increment count. Else if count is 0, change the candidate element to be the current element, else decrement the count.
+
+Works on the principle that upon complete iteration of the array, the majority element should always have the highest count and therefore end up being the candidate element all the time.