#include
void main()
{
int i,n,y[30],x,count=0;
printf(“Enter number of terms”);
scanf(“%d”,&n);
printf(“Enter the numbers”);
for(i=0;i<n;++i)
scanf("%d",&y[i]);
printf("Enter value of x \n");
scanf("%d", &x);
for(i=0;i<n;++i)
{
if(y[i]==x)
count++;
}
if(count==0)
{
printf("value doesnt exist in given array");
}
else
{
printf("%d",count);
}
}
import java.util.*;
class Main
{
public static void main (String[] args)
{
Scanner n=new Scanner(System.in);
int a=n.nextInt();
int x=n.nextInt();
int a1[]=new int[a];
int count=0;
System.out.println(“enter the elements”);
for(int i=0;i<a;i++)
{
a1[i]=n.nextInt();
}
for(int i=0;i<a;i++)
{
if(x==a1[i])
{
count=count+1;
#Simple python Code whose expected Time Complexity is O(log n)
arr=[int(x) for x in input().split(‘ ‘)]
n=int(input())
low=0
ct=0
high=len(arr)-1
for i in range(low,high+1,1):
mid=(low+high)//2
if arr[low]==n:
while True:
if arr[low]==n:
ct+=1
low=low+1
else:
break
break
elif arr[high]==n:
while True:
if arr[high]==n:
ct+=1
high-=1
else:
break
break
elif arr[mid]>n:
high=mid-1
elif arr[mid]<n:
low=mid+1
else:
p=mid-1
while True:
if arr[mid]==n:
ct+=1
mid+=1
else:
break
while True:
if arr[p]==n:
ct+=1
p-=1
else:
break
break
#include
void main()
{
int n;
printf(“enter the length: “);
scanf(“%d”, &n);
int arr[n];
for (int i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
}
int freq, count = 0;
printf("enter the freq: ");
scanf("%d",&freq);
for (int i = 0; i 0)
{
printf(“%d”, count);
}
else
{
printf(“-1 the number is not found!”);
}
}
// Online C compiler to run C program online
#include
int main() {
int x,count=0;
int arr[] = {1, 1, 2, 2, 2, 2, 3,};
int r=sizeof(arr)/sizeof(arr[0]);
scanf(“%d”,&x);
for(int i=0;i<r;i++){
if(arr[i]==x){
count++;
}
}
if(count==0){
printf("it do not appear");
}
else{
printf("appears %d times ",count);
}
return 0;
}
int freq(int arr[],int start,int end,int find){
if(start>end){
return -1;
}
int mid=(start+end)/2;
if(arr[mid]==find){
int count=1;
for(int i=mid-1;i>=start;i–){
if(arr[i]==find){
count++;
}
else{
break;
}
}
for(int i=mid+1;i<=end;i++){
if(arr[i]==find){
count++;
}
else{
break;
}
}
return count;
}
else if (arr[mid]<find)
{
return freq(arr,mid+1,end,find);
}
else{
return freq(arr,start,mid-1,find);
}
}
main(){
int arr[]={1,1,2,2,3,3,3,3,4,4,5,5,6,6,6};
int occ=freq(arr,0,14,1);
cout<<occ;
}
#include
int main()
{
int arr[100],n,x=2,count=0,i;
scanf(“%d”,&n);
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
for(i=0;i<n;i++)
{
if(arr[i]==x)
count++;
}
printf("%d",count);
return 0;
}
#include
void main()
{
int i,n,y[30],x,count=0;
printf(“Enter number of terms”);
scanf(“%d”,&n);
printf(“Enter the numbers”);
for(i=0;i<n;++i)
scanf("%d",&y[i]);
printf("Enter value of x \n");
scanf("%d", &x);
for(i=0;i<n;++i)
{
if(y[i]==x)
count++;
}
if(count==0)
{
printf("value doesnt exist in given array");
}
else
{
printf("%d",count);
}
}
def foo(m,k):
l=0
r=len(m)-1
count=0
while lk:
r=mid-1
elif(m[mid]<k):
l=mid+1
return count
m=[1,2,3,4,5,5,5,6]
print(foo(m,5))
// O(log(n)) Solution in C++
#include
using namespace std;
int main() {
int arr[] = {1, 1, 2, 2, 2, 2, 3};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 2, cnt = 0;
for(int i=0;ix)
break;
}
cout<<cnt<<endl;
return 0;
}
import java.util.*;
class Main
{
public static void main (String[] args)
{
Scanner n=new Scanner(System.in);
int a=n.nextInt();
int x=n.nextInt();
int a1[]=new int[a];
int count=0;
System.out.println(“enter the elements”);
for(int i=0;i<a;i++)
{
a1[i]=n.nextInt();
}
for(int i=0;i<a;i++)
{
if(x==a1[i])
{
count=count+1;
}
}
System.out.println("the elemnt");
System.out.println(count);
}
}
#Simple python Code whose expected Time Complexity is O(log n)
arr=[int(x) for x in input().split(‘ ‘)]
n=int(input())
low=0
ct=0
high=len(arr)-1
for i in range(low,high+1,1):
mid=(low+high)//2
if arr[low]==n:
while True:
if arr[low]==n:
ct+=1
low=low+1
else:
break
break
elif arr[high]==n:
while True:
if arr[high]==n:
ct+=1
high-=1
else:
break
break
elif arr[mid]>n:
high=mid-1
elif arr[mid]<n:
low=mid+1
else:
p=mid-1
while True:
if arr[mid]==n:
ct+=1
mid+=1
else:
break
while True:
if arr[p]==n:
ct+=1
p-=1
else:
break
break
print(ct)
#include
void main()
{
int n;
printf(“enter the length: “);
scanf(“%d”, &n);
int arr[n];
for (int i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
}
int freq, count = 0;
printf("enter the freq: ");
scanf("%d",&freq);
for (int i = 0; i 0)
{
printf(“%d”, count);
}
else
{
printf(“-1 the number is not found!”);
}
}
arr=[7,4,5,2]
for i in range(0,len(arr)):
for j in range(i):
if(arr[j]>arr[j+1]):
temp=arr[j]
arr[j]=arr[j+1]
arr[j+1]=temp
print(arr)
arr=[14,33,27,10,35,19,42,44]
for i in range(0,len(arr)):
for j in range(i):
if(arr[j]>arr[j+1]):
temp=arr[j]
arr[j]=arr[j+1]
arr[j+1]=temp
print(arr)
#include
void bubblesort(int [],int);
int main(){
int arr[] = {33,65,23,1,3};
int i;
bubblesort(arr,sizeof(arr)/sizeof(arr[0]));
for(i=0;i<sizeof(arr)/sizeof(int);i++){
printf("%d\t",arr[i]);
}
return 0;
}
void bubblesort(int arr[] , int n){
int i,j,temp;
for(i=1;i<=n-1;i++){
for(j=0;jarr[j+1]){
temp = arr[j];
arr[j]=arr[j+1];
arr[j+1]=temp;
}
}
}
}