Search (Easy Version) Editorial

editorial
1

Problem Link

Explanation:
We need to output if a given query/number is present in the given list of numbers.
a brute force solution is to do a liner search for each query Time Complexity: O(Q*N)
another solution is to sort the numbers and then for each query do a binary search Time Complexity : O(nlogn + Qlogn)
An easier solution for this and optimal solution is to use a count array which stores frequency of each element or rather use a bool count array to check if a element is there Time Complexity : O(n+q)

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[n];
    bool ck[1001];
    memset(ck,false,sizeof(ck));
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        ck[a[i]]=true;
    }
    int q;
    cin>>q;
    while(q--)
    {
        int x;
        cin>>x;
        if(ck[x])
        {
            cout<<"1\n";
        }
        else
        {
            cout<<"0\n";
        }
    }
}

Code In C:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    int a[n];
    int ck[1001]={0};
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        ck[a[i]]=1;
    }
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int x;
        scanf("%d",&x);
        if(ck[x])
        {
            printf("1\n");
        }
        else
        {
            printf("0\n");
        }
    }
}