# Day One

Just do Linear Search

Given an array of nonnegative integers, design a linear algorithm and implement it using a program to find whether given key element is present in the array or not. Also, find total number of comparisons for each input case. (Time Complexity = O(n), where n is the size of input)

``````#include<iostream>using namespace std;
int main(){    int times ;    cin>>times;    while(times--){    int number;    cin>>number;    int arr[number];    for(int i=0;i<number;i++){        cin>>arr[i];    }    int numberToFind = 0;    cin>>numberToFind;    bool isFound = false;    for(int i=0;i<number;i++){        if(arr[i]==numberToFind){            cout<<"Present"<<" ";            cout<<i<<" ";            isFound = true;        }    }    if(!isFound){    cout<<"Not Present ";    cout<<number<<"\n";    }        }        return 0;}``````

Given an already sorted array of positive integers, design an algorithm and implement it using a program to find whether given key element is present in the array or not. Also, find total number of comparisons for each input case. (Time Complexity = O(nlogn), where n is the size of input).

``````#include<iostream>using namespace std;
int main(){    int times ;    cin>>times;    while(times--){    int number;    cin>>number;    int arr[number];    for(int i=0;i<number;i++){        cin>>arr[i];    }    int numberToFind = 0;    cin>>numberToFind;    bool isFound = false;    int left = 0;    int right = number - 1;    int count = 0;    while(arr[left] <= arr[right]){        int mid = (left + right)/2;        count++;                if(arr[mid]== numberToFind){         cout<<"Present:"<<count;             isFound = true;         break;        }else{            if(arr[mid]>numberToFind){                right = mid - 1;            }else{                left = mid + 1;             }        }    }    if(!isFound){    cout<<"Not Present ";    cout<<count<<"\n";    }        }        return 0;}``````