solved unionIntersection

Author: kotAPI

Arrays and Strings/C++/unionAndIntersection.cpp

@@ -0,0 +1,132 @@
+
+/*******************************************************************************************************
+/* ___ _ _ _ _ 
+/* / _ \ | | (_)| | | | 
+/* / /_\ \__ __ ___ | | __ _ ___ _ __ _ | |_ | |__ _ __ ___ ___ 
+/* | _ |\ \ /\ / / / _ \| | / _` | / _ \ | '__|| || __|| '_ \ | '_ ` _ \ / __| 
+/* | | | | \ V V / | __/| || (_| || (_) || | | || |_ | | | || | | | | |\__ \ 
+/* \_| |_/ \_/\_/ \___||_| \__, | \___/ |_| |_| \__||_| |_||_| |_| |_||___/ 
+/* __/ | 
+/* |___/ 
+/******************************************************************************************************/
+
+/***********************************
+/* Reverse An Array
+/***********************************/
+
+// Library imports
+#include <iostream>
+using namespace std;
+
+//Prototypes for the algorithms
+int unionAndIntersection(int* arr,int len, int* arr2,int len2);
+void someSortingAlgorithm(int*,int);
+int removeDuplicates(int*,int);
+
+int main(){
+// Lines to handle user input go here, be descriptive...
+// use cin, cout 
+int arr[] = {1, 3, 4, 5, 7};
+int arr2[] = {2, 3, 5, 6};
+
+int arrayLength = sizeof(arr)/sizeof(arr[0]);
+int arrayLength2 = sizeof(arr2)/sizeof(arr2[0]);
+// Call your algorithms here
+unionAndIntersection(arr, arrayLength, arr2, arrayLength2);
+
+// Union[] = {1, 2, 3, 4, 5, 6, 7}
+// Intersection[] = {3,5}
+
+return 0;
+}
+
+/**********************************************************
+/* <unionAndIntersection>
+/*
+/* @description : description here
+/* @param : Integer<variable> (Datatype<variable_name>)
+/* @return : datatype (int,float,double etc)
+/* @complexity : Big O notation. (O(n),O(n^2) etc)
+/* @explanation : Detailed description about what the algorithm does
+/* @author : <Your name> @<GitHub handle> (John Doe @johndoe)
+/**********************************************************/
+int unionAndIntersection(int* arr,int len, int* arr2,int len2){ //simple one first
+// Algorithm content goes here.
+// 
+int unionArray[len+len2];
+int counter = 0;
+
+// UNION START
+// *************************************************************
+for(int i=0;i<len;i++){
+unionArray[i] = arr[i];
+}
+
+for(int j=0; j<len2; j++){
+unionArray[j+len] = arr2[j];
+}
+
+cout<<endl;
+
+someSortingAlgorithm(unionArray, len+len2);
+
+int totalLen = removeDuplicates(unionArray,len+len2);
+
+cout<<"Union: ";
+for(int i=0;i<totalLen;i++){
+cout<<unionArray[i]<<" ";
+}
+cout<<endl;
+//UNION END
+//// *************************************************************
+
+// INTERSECTION START
+// *************************************************************
+cout<<"Intersection: ";
+for(int i=0;i<len;i++){
+for(int j=0;j<len2;j++){
+if(arr[i]==arr2[j]){
+cout<<arr[i]<<" ";
+}
+}
+}
+cout<<endl;
+return* arr;
+}
+
+
+void someSortingAlgorithm(int* arr,int len){
+
+//Bubble sort. Yeah, don't use this in real life situations
+for(int i=0;i<len-1;i++){
+for(int j=i;j<len;j++){
+if(arr[i]>arr[j]){
+int temp = arr[i];
+arr[i] = arr[j];
+arr[j] = temp;
+}
+
+}
+}
+}
+
+int removeDuplicates(int *arr,int len){
+int temp[len];
+int counter =0;
+
+for(int i=0;i<len;i++){
+if(arr[i]==arr[i+1]){
+continue;
+}
+else{
+temp[counter++]=arr[i];
+}
+}
+
+//Transferring values from temp array to original pointer
+for(int i=0;i<counter;i++){
+arr[i]=temp[i];
+}
+//Returning the original array.
+return counter;
+}

Arrays and Strings/README.md

@@ -65,22 +65,14 @@
 
   arr1[] = {1, 3, 4, 5, 7}
   arr2[] = {2, 3, 5, 6}
-  
+
   Union[] = {1, 2, 3, 4, 5, 6, 7}
   Intersection= {3,5}
 
 
 TODO :
 
 
-
-* Given two sorted arrays, find their union and intersection.
-
-For example, if the input arrays are: 
-arr1[] = {1, 3, 4, 5, 7}
-arr2[] = {2, 3, 5, 6}
-Then your program should print Union as {1, 2, 3, 4, 5, 6, 7} and Intersection as {3, 5}. 
-
 * Given an array A[], write a function that segregates even and odd numbers. The functions should put all even numbers first, and then odd numbers.
 
 Example