We ca use HashSet data structure to remove duplicats from an array i.pdf

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We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n) Solution We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n).

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Question 1 answer is A Global trade allows wealthy countries to use .pdf

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Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthened, there must be a terminal solid solution that has a decreasing solid solubility as the temperature decreases. The Al-Cu (Duralumin is an aluminum alloy of 2XXX group) phase diagram shown in Figure 1 shows this type of decrease along the solvus between the and + regions. Consider a 96wt%Al – 4wt%Cu alloy which is chosen since there is a large degrease in the solid solubility of solid solution in decreasing the temperature from 550°C to 75°C. Solution Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthene.

Precipitation hardening, or age hardening, provides one of the most .pdf

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Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate cell –to- cell communication through sensory perception obtained from “eyespot” finally neighboring cells join through “cytoplasmic strands” that promote cell –to-cell communication. Recent research has illustrated that cellular adhesion molecules (Algal-CAMs) are found in multicellular Volvox. However, the bulk transport has not observed in the Volvox, a primitive feature still yet to develop in them. Brown algae are the algae, which possess secondary chloroplasts as their functional units to perform their photosynthesis and these algae are commonly originated from the endosymbiotic red algae. Examples are diatoms. Phylogenetic relationship with land plants: Red algae do not possess property of photosynthesis & as they do not contain green pigments Ques-3: Outer coverings: Sphingolipids Pellicle Phospholipids & glycolipids Solution Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate.

Protists are beneficial to the ecosystem in generating oxygen as the.pdf

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Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2 Solution Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2.

Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf

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P1.javaimport java.util.Scanner;keyboard inputting package pub.pdf

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//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"<.

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Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies. Solution Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies..

Network Protocol1) A Protocol is used for the communication betwe.pdf

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