hacktoberfest-2022

3SumClosest.cpp

@@ -0,0 +1,39 @@
+// Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target.
+
+// Return the sum of the three integers.
+
+// You may assume that each input would have exactly one solution.
+
+// https://leetcode.com/problems/3sum-closest/
+
+// Example 1:
+
+// Input: nums = [-1,2,1,-4], target = 1
+// Output: 2
+// Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
+
+
+
+int threeSumClosest(vector<int>& nums, int target) {
+    if(nums.size() < 3) return 0;
+    int closest = nums[0]+nums[1]+nums[2];
+    sort(nums.begin(), nums.end());
+    for(int first = 0 ; first < nums.size()-2 ; ++first) {
+        if(first > 0 && nums[first] == nums[first-1]) continue;
+        int second = first+1;
+        int third = nums.size()-1;            
+        while(second < third) {
+            int curSum = nums[first]+nums[second]+nums[third];
+            if(curSum == target) return curSum;
+            if(abs(target-curSum)<abs(target-closest)) {
+                closest = curSum;
+            }
+            if(curSum > target) {
+                --third;
+            } else {
+                ++second;
+            }
+        }
+    }
+    return closest;
+}

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.");
+    }
+}

Array_Rotate.c

@@ -0,0 +1,34 @@
+#include <stdio.h>
+
+void rotate(int arr[], int n)
+{
+	int x = arr[n - 1], i;
+	for (i = n - 1; i > 0; i--)
+		arr[i] = arr[i - 1];
+	arr[0] = x;
+}
+
+int main()
+{
+	int size, i;
+	printf("Enter Size : ");
+	scanf("%d", &size);
+	int arr[size];
+	printf("Enter Array Elements :  ");
+	for (i = 0; i < size; i++)
+	{
+		scanf("%d", &arr[i]);
+	}
+
+	printf("Given array is\n");
+	for (i = 0; i < size; i++)
+		printf("%d ", arr[i]);
+
+	rotate(arr, size);
+
+	printf("\nRotated array is\n");
+	for (i = 0; i < size; i++)
+		printf("%d ", arr[i]);
+
+	return 0;
+}