added all arrays assignment programs

Author: thatsabhishek

README.md

@@ -50,6 +50,12 @@ This repo includes:
     - [Linear Search](./milestone2/arrays/linearsearch.cpp)
     - [Arrange Numbers in Array](./milestone2/arrays/ArrangeNumbersinArray.cpp)
     - [Swap Alternate](./milestone2/arrays/SwapAlternate.cpp)
+    - [Find Unique](./milestone2/arrays/FindUnique.cpp)
+    - [Find Duplicate](./milestone2/arrays/FindDuplicate.cpp)
+    - [Intersection of Two Arrays II](./milestone2/arrays/IntersectionofTwoArraysII.cpp)
+    - [Pair Sum](./milestone2/arrays/PairSum.cpp)
+    - [Triplet Sum](./milestone2/arrays/TripletSum.cpp)
+    - [Sort 0 1](./milestone2/arrays/Sort01.cpp)
 
 # Coding Ninjas
 ## Rate my Repo ⭐!!!

milestone2/arrays/FindDuplicate.cpp

@@ -0,0 +1,64 @@
+// You have been given an integer array/list(ARR) of size N which contains numbers from 0 to (N - 2). 
+// Each number is present at least once. That is, if N = 5, the array/list constitutes values ranging from 0 to 3 and among these, there is a single integer value that is present twice. 
+// You need to find and return that duplicate number present in the array.
+// Note : Duplicate number is always present in the given array/list.
+
+// My Code
+int duplicateNumber(int *arr, int size)
+{
+    int count, res;
+    for(int i=0; i<size; i++){
+        int j= arr[i];
+        count = 0;
+        for(int k=0; k<size; k++){
+            if (arr[k] == j){
+                count++;
+            }
+        }
+        if(count > 1){
+            res = arr[i];
+        }
+    }
+    return res;
+}
+
+// Main Code
+#include <iostream>
+using namespace std;
+
+#include "solution.h"
+
+int main()
+{
+
+	int t;
+	cin >> t;
+	
+	while (t--)
+	{
+		int size;
+		cin >> size;
+		int *input = new int[size];
+
+		for (int i = 0; i < size; i++)
+		{
+			cin >> input[i];
+		}
+
+		cout << duplicateNumber(input, size) << endl;
+	}
+
+	return 0;
+}
+
+// CN Code
+int duplicateNumber(int *arr, int size)
+{
+    for(int i=0; i<size; i++){
+        for(int j=i+1; j<size; j++){
+            if(arr[i] == arr[j]){
+                return arr[i];
+            }
+        }
+    }
+}

milestone2/arrays/FindUnique.cpp

@@ -0,0 +1,67 @@
+// You have been given an integer array/list(ARR) of size N. Where N is equal to [2M + 1].
+// Now, in the given array/list, 'M' numbers are present twice and one number is present only once.
+// You need to find and return that number which is unique in the array/list.
+//  Note: Unique element is always present in the array/list according to the given condition.
+
+//My Code
+int findUnique(int *arr, int size)
+{
+    int count, res;
+    for(int i=0; i<size; i++){
+        int j= arr[i];
+        count = 0;
+        for(int k=0; k<size; k++){
+            if (arr[k] == j){
+                count++;
+            }
+        }
+        if(count == 1){
+            res = arr[i];
+        }
+    }
+    return res;
+}
+
+// Main Code
+#include <iostream>
+#include "solution.h"
+using namespace std;
+
+int main()
+{
+
+	int t;
+	cin >> t;
+
+	while (t--)
+	{
+		int size;
+		cin >> size;
+		int *input = new int[size];
+
+		for (int i = 0; i < size; ++i)
+		{
+			cin >> input[i];
+		}
+
+		cout << findUnique(input, size) << endl;
+	}
+
+	return 0;
+}
+
+// CN Code
+int findUnique(int *arr, int size)
+{
+    for(int i=0; i<size; i++){
+        int j=0;
+        for(; j<size; j++){
+            if(i != j && arr[i] == arr[j]){
+                break;
+            }
+        }
+        if(j == size){
+            return arr[i];
+        }
+    }
+}

milestone2/arrays/IntersectionofTwoArraysII.cpp

@@ -0,0 +1,61 @@
+// You have been given two integer arrays/list(ARR1 and ARR2) of size N and M, respectively. 
+// You need to print their intersection; An intersection for this problem can be defined when both the arrays/lists contain a particular value or to put it in other words, 
+// when there is a common value that exists in both the arrays/lists.
+// Note : Input arrays/lists can contain duplicate elements.
+// The intersection elements printed would be in the order they appear in the first array/list(ARR1)
+
+// My Code
+void intersection(int *input1, int *input2, int size1, int size2)
+{
+   for(int i=0; i<size1; i++){
+       for (int j=0; j<size2; j++){
+           if(input1[i] == input2[j]){
+               cout<< input1[i] << " ";
+               input2[j] = -1;
+               break;
+           }
+       }
+   }
+}
+
+// Main Code
+#include <iostream>
+#include <algorithm>
+using namespace std;
+#include "solution.h"
+
+int main()
+{
+	int t;
+	cin >> t;
+	while (t--)
+	{
+
+		int size1, size2;
+
+		cin >> size1;
+		int *input1 = new int[size1];
+
+		for (int i = 0; i < size1; i++)
+		{
+			cin >> input1[i];
+		}
+
+		cin >> size2;
+		int *input2 = new int[size2];
+
+		for (int i = 0; i < size2; i++)
+		{
+			cin >> input2[i];
+		}
+
+		intersection(input1, input2, size1, size2);
+		
+		delete[] input1;
+		delete[] input2;
+		
+		cout << endl;
+	}
+
+	return 0;
+}

milestone2/arrays/Sort01.cpp

@@ -0,0 +1,57 @@
+// You have been given an integer array/list(ARR) of size N that contains only integers, 0 and 1. 
+// Write a function to sort this array/list. Think of a solution which scans the array/list only once and don't require use of an extra array/list.
+// Note: You need to change in the given array/list itself. 
+//       Hence, no need to return or print anything. 
+
+// My Code 
+void sortZeroesAndOne(int *input, int size)
+{
+   int zero = 0;
+   for (int i=0; i<size; i++){
+       if (input[i] == 0){
+           int temp = input[zero];
+           input[zero] = input[i];
+           input[i] = temp;
+           zero++;
+       }
+   }
+}
+
+// Main Code
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+#include "solution.h"
+
+int main()
+{
+
+	int t;
+	cin >> t;
+
+	while (t--)
+	{
+		int size;
+
+		cin >> size;
+		int *input = new int[size];
+
+		for (int i = 0; i < size; ++i)
+		{
+			cin >> input[i];
+		}
+
+		sortZeroesAndOne(input, size);
+
+		for (int i = 0; i < size; ++i)
+		{
+			cout << input[i] << " ";
+		}
+
+		cout << endl;
+		delete[] input;
+	}
+
+	return 0;
+}