update Permutations in string.cpp

4Sum.cpp

@@ -0,0 +1,80 @@
+// Question-
+
+// Suppose we have an array of numbers. It stores n integers, there are four elements a, b, c and d in the array. We have another target value, such that a + b + c + d = target. Find all unique quadruplets in the array which satisfies the situation. So if the array is like [-1,0,1,2,0,-2] and target is 0, then the result will be [[-1, 0, 0, 1],[-2, -1, 1, 2], [-2, 0, 0, 2]]
+// Input-
+// [1,0,-1,0,-2,2]
+// 0
+// Output-
+// [[1,2,-1,-2],[0,2,0,-2],[0,1,0,-1]]
+
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+
+class Solution {
+public:
+   void addAll(vector < vector <int> >& res, vector < vector <int> >& temp){
+      for(int i = 0; i < temp.size(); i++)res.push_back(temp[i]);
+   }
+   vector<vector<int>> fourSum(vector<int>& nums, int target) {
+      sort(nums.begin(), nums.end());
+      return kSum(nums, 0, 4, target);
+   }
+   vector < vector <int> > kSum(vector <int>& arr, int start, int k, int target){
+      vector < vector <int> > res;
+      if(k == 2){
+         int left = start;
+         int right = arr.size() - 1;
+         vector <int> temp(2);
+         while(left < right){
+            if(arr[left] + arr[right] == target){
+               temp[0] = arr[left];
+               temp[1] = arr[right];
+               res.push_back(temp);
+               while(left < right && arr[left] == arr[left + 1])left++;
+               while(left < right && arr[right] == arr[right - 1])right--;
+               left++;
+               right--;
+            }
+            else if(arr[left] + arr[right] > target)right--;
+            else left ++;
+         }
+      }
+      else{
+         for(int i = start; i < (int)arr.size() - k + 1; i++){
+            if(i > start && arr[i] == arr[i - 1])continue;
+            vector < vector <int> > temp = kSum(arr, i + 1, k - 1, target - arr[i]);
+            for(int j = 0; j < temp.size(); j++){
+               temp[j].push_back(arr[i]);
+            }
+            addAll(res, temp);
+         }
+      }
+      return res;
+   }
+};
+
+
+
+void print_vector(vector<vector<int> > v){
+   cout << "[";
+   for(int i = 0; i<v.size(); i++){
+      cout << "[";
+      for(int j = 0; j <v[i].size(); j++){
+         cout << v[i][j] << ", ";
+      }
+      cout << "],";
+   }
+   cout << "]"<<endl;
+}
+
+
+
+
+void main(){
+   Solution ob;
+   vector<int> v = {1,0,-1,0,-2,2};
+   print_vector(ob.fourSum(v, 0));
+}

4_Sum.cpp

@@ -0,0 +1,92 @@
+
+      /*this code is edited by batrakeshav10
+      sir please accept my pull request and guide me,
+      so that i can win hactoberfest tshirt */
+// Question-
+
+/* Suppose we have an array of numbers. It stores n integers, there are four elements a, b, c and d in the array. We have another target value, such that a  +  b  +  c  +  d = target. Find all the unique quadruplets in the array which satisfies the situation. So if the array is like [-1,0,1,2,0,-2] and target is 0, then the result will be [[-1, 0, 0, 1],[-2, -1, 1, 2], [-2, 0, 0, 2]]
+ Input-
+ [1,0,-1,0,-2,2]
+ 0
+ Output
+[[1,2,-1,-2],[0,2,0,-2],[0,1,0,-1]]   */
+
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+
+class Solution {
+public:
+   void addAll(vector < vector <int> >& res, vector < vector <int> >& temp){
+      for(int i = 0; i < temp.size(); i++)res.push_back(temp[i]);
+   }
+   vector<vector<int>> fourSum(vector<int>& nums, int target) {
+      sort(nums.begin(), nums.end());
+      return kSum(nums, 0, 4, target);
+   }
+   vector < vector <int> > kSum(vector <int>& arr, int start, int k, int target){
+      vector < vector <int> > res;
+      if(k == 2){
+         int left = start;
+         int right = arr.size() - 1;
+         vector <int> temp(2);
+         while(left < right){
+            if(arr[left] + arr[right] == target){
+               temp[0] = arr[left];
+               temp[1] = arr[right];
+               res.push_back(temp);
+               while(left < right && arr[left] == arr[left + 1])left++;
+               while(left < right && arr[right] == arr[right - 1])right--;
+               left++;
+               right--;
+            }
+
+      /*this code is edited by batrakeshav10
+      sir please accept my pull request and guide me,
+      so that i can win hactoberfest tshirt */
+            else if(arr[left] + arr[right] > target)right--;
+            else left ++;
+         }
+      }
+      else{
+         for(int i = start; i < (int)arr.size() - k + 1; i++){
+            if(i > start && arr[i] == arr[i - 1])continue;
+            vector < vector <int> > temp = kSum(arr, i + 1, k - 1, target - arr[i]);
+            for(int j = 0; j < temp.size(); j++){
+               temp[j].push_back(arr[i]);
+            }
+            addAll(res, temp);
+         }
+      }
+      return res;
+   }
+};
+
+
+
+void print_vector(vector<vector<int> > v){
+   cout << "[";
+   for(int i = 0; i<v.size(); i++){
+      cout << "[";
+      for(int j = 0; j <v[i].size(); j++){
+         cout << v[i][j] << ", ";
+      }
+      cout << "],";
+   }
+   cout << "]"<<endl;
+}
+
+
+
+
+void main(){
+   Solution ob;
+   vector<int> v = {1,0,-1,0,-2,2};
+   print_vector(ob.fourSum(v, 0));
+}                           
+
+      /*this code is edited by batrakeshav10
+      sir please accept my pull request and guide me,
+      so that i can win hactoberfest tshirt */

Armstrongnumber.java

@@ -0,0 +1,22 @@
+//finding armstrong number in java
+public class Armstrong {
+
+    public static void main(String[] args) {
+
+        int number = 371, originalNumber, remainder, result = 0;
+
+        originalNumber = number;
+
+        while (originalNumber != 0)
+        {
+            remainder = originalNumber % 10;
+            result += Math.pow(remainder, 3);
+            originalNumber /= 10;
+        }
+
+        if(result == number)
+            System.out.println(number + " is an Armstrong number.");
+        else
+            System.out.println(number + " is not an Armstrong number.");
+    }
+}

C#/RadianDegreeConverter.cs

@@ -0,0 +1,26 @@
+using System;
+
+namespace Conversions
+{
+    public static class RadianDegreeConverter
+    {
+        /// <summary>
+        /// Converts given degree to radian
+        /// </summary>
+        /// <param name="degree"></param>
+        /// <returns></returns>
+        public static double DegreeToRadian(double degree)
+        {
+            return degree * (Math.PI / 180);
+        }
+
+        /// <summary>
+        /// Converts given radian to degree
+        /// </summary>
+        /// <returns></returns>
+        public static double RadianToDegree(double radian)
+        {
+            return radian * (180 / Math.PI);
+        }
+    }
+}

CONTRIBUTING.md

@@ -0,0 +1,2 @@
+this repository is only for algorithms in which you can upload any language's algorithms which will useful for many other people.
+happy coading 👨‍💻.

CPP/4sum.cpp

@@ -0,0 +1,143 @@
+// C++ program to find four  
+// elements with the given sum 
+#include <bits/stdc++.h>
+
+using namespace std;
+
+// Function to find 4 elements that add up to 
+// given sum 
+
+void fourSum(int X, int arr[], map<int, 
+
+             pair<int, int>> Map, int N) 
+{ 
+
+    int temp[N]; 
+
+
+    // Iterate from 0 to temp.length 
+
+    for (int i = 0; i < N; i++) 
+
+        temp[i] = 0; 
+
+
+    // Iterate from 0 to arr.length 
+
+    for (int i = 0; i < N - 1; i++) 
+
+    { 
+
+
+        // Iterate from i + 1 to arr.length 
+
+        for (int j = i + 1; j < N; j++)
+
+        { 
+
+
+            // Store curr_sum = arr[i] + arr[j] 
+
+            int curr_sum = arr[i] + arr[j]; 
+
+
+            // Check if X - curr_sum if present 
+
+            // in map 
+
+            if (Map.find(X - curr_sum) != Map.end()) 
+
+            { 
+
+
+                // Store pair having map value 
+
+                // X - curr_sum 
+
+                pair<int, int> p = Map[X - curr_sum]; 
+
+
+                if (p.first != i && p.second != i 
+
+                    && p.first != j && p.second != j 
+
+                    && temp[p.first] == 0
+
+                    && temp[p.second] == 0 && temp[i] == 0
+
+                    && temp[j] == 0)
+
+                { 
+
+
+                    // Print the output 
+
+                    cout << arr[i] << "," << arr[j] <<
+
+                      "," << arr[p.first] << "," << arr[p.second];
+
+                    temp[p.second] = 1; 
+
+                    temp[i] = 1; 
+
+                    temp[j] = 1; 
+
+                    break; 
+
+                } 
+
+            } 
+
+        } 
+
+    } 
+} 
+
+// Program for two Sum 
+
+map<int, pair<int, int>> twoSum(int nums[], int N) 
+{ 
+
+    map<int, pair<int, int>> Map; 
+
+    for (int i = 0; i < N - 1; i++)
+
+    { 
+
+        for (int j = i + 1; j < N; j++)
+
+        { 
+
+            Map[nums[i] + nums[j]].first = i;
+
+            Map[nums[i] + nums[j]].second = j;
+
+        } 
+
+    } 
+
+    return Map; 
+} 
+
+// Driver code
+
+int main()
+{
+
+    int arr[] = { 10, 20, 30, 40, 1, 2 }; 
+
+    int n = sizeof(arr) / sizeof(arr[0]); 
+
+    int X = 91; 
+
+    map<int, pair<int, int>> Map = twoSum(arr, n); 
+
+
+
+    // Function call 
+
+    fourSum(X, arr, Map, n); 
+
+
+    return 0;
+}