Problem 3

July 11, 2023

34 comments on “Problem 3”


  • varsha

    #include
    void main()
    {
    int n,i,key,count=0;
    scanf(“%d”,&n);
    int arr[n];
    for(i=0;i<n;i++)
    {
    scanf("%d",&arr[i]);
    }
    key=3;
    for(i=0;i<n;i++)
    {
    if(key==arr[i])
    {
    count++;
    }
    }
    if(count==0)
    {
    printf("-1");
    }
    else
    {
    printf("%d",count);
    }

    }


  • GAURAV

    #include

    using namespace std;
    int left(int arr[],int n){
    int l=0;
    int h=n-1;
    int mid;
    int val=2;
    while(lval)
    h=mid;
    else if(arr[mid]<val)
    l=mid;
    else if(arr[mid]==val&&arr[mid-1]<val){
    return mid;
    }
    else if(arr[mid]==val&&arr[mid-1]==val){
    h=mid;
    }

    }
    }
    int right(int arr[],int n){
    int l=0;
    int h=n-1;
    int mid;
    int val=2;
    while(lval)
    h=mid;
    else if(arr[mid]val){
    return mid;
    }
    else if(arr[mid]==val&&arr[mid+1]==val){
    l=mid;
    }

    }
    }

    int main()
    {
    int arr[]={ 1, 2, 2, 2, 2, 3, 4, 7, 8, 8 };
    int n=sizeof(arr)/sizeof(arr[0]);
    int l=left(arr,n);
    int r=right(arr,n);
    cout<<r-l+1;

    return 0;
    }


  • Ashish

    #include
    #include
    int main()
    {
    int arr[] = {1, 1, 2, 2, 2, 2, 3,},x;
    scanf(“%d”,&x);
    int i,count=0;
    for(i=0;i<(sizeof(arr)/sizeof(int));i++)
    {
    if(arr[i]==x)
    count++;
    }
    printf("%d",count);
    return 0;
    }


  • Vijaykumar

    a=[]
    for i in range(5):
    a.append(int(input()))
    b=int(input(“\nEnter x: “))
    c=a.count(b)
    print(c)


  • Kritika

    import math
    num=[1,2,3,5,5,5,6,7]

    def binarysearch(num,left,right,target):
    mid=math.floor((left+right)/2)
    while(lefttarget:
    right=mid-1
    else:
    left=mid+1
    return -1

    def occurance(num):

    mid=(0+len(num)-1)//2
    target=num[mid]
    firstpos=binarysearch(num,0,len(num)-1,target)
    if firstpos==-1:
    return -1
    else:
    startpos=firstpos
    endpos=firstpos
    temp1=0
    temp2=0

    while startpos!=-1:
    temp1=startpos
    startpos=binarysearch(num,0,startpos-1,target)
    startpos=temp1

    while endpos!=-1:
    temp2=endpos
    endpos=binarysearch(num,endpos+1,len(num)-1,target)
    endpos=temp2

    return [startpos,endpos]

    if __name__==”__main__”:
    x=occurance(num)
    print(x[1]-x[0]+1)


  • Ankita

    package PrepInsta;

    import java.util.Scanner;

    public class Frequency2 {

    public static int countfreq(int[] arr,int n,int num)
    {
    int count=0;
    for(int i=0;i<n;i++)
    {
    if(arr[i]==num)
    count++;
    }
    return count;
    }

    public static void main(String[] args)
    {
    Scanner sc=new Scanner(System.in);

    int[] arr= {2,5,6,2,4,5,6,1,3,6};

    System.out.println("Enter number:");

    int num=sc.nextInt();

    int count=countfreq(arr,arr.length,num);
    System.out.println(count);
    }
    }


  • Rohan

    #include
    int main()
    {
    int n, a[100], count=0,i, num;
    printf(“Enter a number\n”);
    scanf(“%d”,&num);
    printf(“enter the size of the array\n”);
    scanf(“%d”,&n);
    printf(“enter the array elements\n”);
    for(i=0; i<n;i++)
    {
    scanf("%d",&a[i]);
    }

    for(i=0; i<n;i++)
    {
    if (num==a[i])
    count++;
    }

    printf("%d ocuured %d times",num, count);
    return 0;

    }


  • amit

    #include

    int main()
    {
    int ch[]={1, 1, 2, 2, 2, 2, 3,};
    int input,size,count=0;
    scanf(“%d”,&input);
    size=sizeof(ch)/sizeof(int);
    for(int i=0;i<size;i++)
    {
    if(input==ch[i])
    {
    count++;
    }
    }
    if(count)
    printf("%d",count);
    else
    printf("%d does not occure in given array",input);
    return 0;
    }


  • Shubham

    #include
    int count_occurence(int arr[],int n,int m){
    int count = 0;
    for(int i=0;i<n;i++){
    if(arr[i] == m){
    count++;
    }
    }
    printf("%d",count);
    }
    int main()
    {
    int n,m;

    scanf("%d %d",&n,&m);
    int arr[n];
    for(int i=0;i<n;i++){
    scanf("%d",&arr[i]);
    }

    count_occurence(arr,n,m);

    return 0;
    }


  • Technophilic

    k=[1,2,2,3,4]
    n=5
    count=0
    if n in k:
    for i in range(len(k)):
    if k[i]==n:
    count+=1
    print(count)
    else:
    print(“doesn’t occur in arr”)


  • Nilesh

    def countocurrences(arr,n,x):
    count = 0
    for i in range(len(A)):
    if A[i] ==x:
    count+=1
    return count
    arr = [1,2,2,2,2,3,3,2,2,2,2]
    x = 2
    n = len(arr)
    print(countocurrences(arr,n,x))

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