Array Questions(Program)

Find minimum and maximum element in an array

Given an array A of size N of integers. Your task is to find the minimum and maximum elements in the array.

pair<long long, long long> getMinMax(long long arr[], int n) {
    pair <long, long> p1;
    p1.first = p1.second = arr[0];

    for(int i=0;i<n;i++)
    {
        if(p1.first > arr[i])
            p1.first = arr[i];
        if(p1.second < arr[i])
            p1.second = arr[i];
    }
    return p1;
}

pair<long long, long long> getMinMax(long long a[], int n) {
    pair<long long, long long> p;
    long long max = a[0];
    long long min = a[0];
    for(int i=0;i<n;i++)
    {
        if(min > a[i])
            min = a[i];
        if(max < a[i])
            max = a[i];
    }
    p.first = min;
    p.second = max;
    return p;
}

Largest Element in Array

Given an array A[] of size n. The task is to find the largest element in it.

class Solution
{
public:
    int largest(vector<int> &arr, int n)
    {
        int max= arr[0];
        for(int i=0;i<n;i++)
        {
            if(max < arr[i])
                max = arr[i];
        }
        return max;
    }
};

Immediate Smaller Element

Given an integer array Arr of size N. For each element in the array, check whether the right adjacent element (on the next immediate position) of the array is smaller. If next element is smaller, update the current index to that element. If not, then -1.

class Solution{
public:    
    void immediateSmaller(vector<int>&arr, int n) {
        //  code here
        for(int i=0;i<n-1;i++)
        {
            if(arr[i] > arr[i+1])
                arr[i]= arr[i+1];
            else
                arr[i] =-1;
        }
        arr[n-1]= -1;
    }

};

First element to occur k times

Given an array of N integers. Find the first element that occurs at least K number of times.

class Solution{
    public:
    int firstElementKTime(int a[], int n, int k)
    {

        int max=a[0];
        for(int i=0;i<n;i++)
        {
            if(max < a[i])
                max = a[i];
        }    
        int freq[max+1]={0};
        for(int i=0;i<n;i++)
        {
            int index = a[i];
            freq[index] +=1;
            if (freq[index]>=k){
                return a[i];
            }
        }
        return -1;
    }
};

Implement stack using array

Write a program to implement a Stack using Array. Your task is to use the class as shown in the comments in the code editor and complete the functions push() and pop() to implement a stack.

//Function to push an integer into the stack.
void MyStack :: push(int x)
{
    // Your Code
    if(top == 999)
        return;
    else
    {
        arr[++top] = x;
    }
}

//Function to remove an item from top of the stack.
int MyStack :: pop()
{
    // Your Code 
    if(top == -1)
        return -1;
    else
        return arr[top--];
}

Implement Queue using array

Implement a Queue using an Array. Queries in the Queue are of the following type:
(i) 1 x (a query of this type means pushing 'x' into the queue)
(ii) 2 (a query of this type means to pop element from queue and print the poped element)

//Function to push an element x in a queue.
void MyQueue :: push(int x)
{
        // Your Code
        // if(rear == 100005)
        //     cout<<-1;
        // if(front==0)
        //     front = 1;
        arr[rear++]=x;
}

//Function to pop an element from queue and return that element.
int MyQueue :: pop()
{
        // Your Code 
        if(front == rear)
            return -1;
        else
        {
            int elem = arr[front];
            front++;
            if(front > rear)
                front=rear=0;
            return elem;
        }
}

Searching a number

Given an array Arr of N elements and a integer K. Your task is to return the position of first occurence of K in the given array.

class Solution{
public:
    int search(int arr[], int n, int k) {
        // code here
        for(int i=0;i<n;i++)
        {
            if(arr[i] == k)
                return i+1;
        }
        return -1;
    }
};

Rotating an Array

Given an array of size N. The task is to rotate array by D elements where D ≤ N.

//User function template for C++

class Solution{   
public:
    void leftRotate(int arr[], int n, int d) {
        // code here
        int arr2[n], i, j;
        for(i=d,j=0;i<n;i++)
            arr2[j++]= arr[i];
        for(i=0;i<d;i++)
            arr2[j++] = arr[i];
        for(i=0;i<n;i++)
            arr[i]= arr2[i];

    }
};

Binary Array Sorting

Given a binary array A[] of size N. The task is to arrange the array in increasing order.

       int c0=0, c1=0;
       for(int i=0;i<N;i++){
            if(arr[i] == 0)
                c0++;
            else if(arr[i]==1)
                c1++;
       }
       for(int i=0;i<c0;i++)
            arr[i] =0;
       for(int i=c0;i<c0+c1;i++)
            arr[i] =1;

Third largest element

Given an array of distinct elements. Find the third largest element in it.

class Solution{
  public:
    int thirdLargest(int a[], int n)
    {
         //Your code here
         if(n<=2)
            return -1;

        //  int max = INT_MIN;
        //  for(int i=0;i<n;i++)
        //  {
        //      if (max < a[i])
        //         max = a[i];
        //  }
        //  int smax = INT_MIN;
        //  for(int i=0;i<n;i++)
        //  {
        //      if (smax < a[i] && a[i] < max )
        //         smax = a[i];
        //  }
        //  int tmax = INT_MIN;
        //  for(int i=0;i<n;i++)
        //  {
        //      if (tmax < a[i] && a[i] < smax )
        //         tmax = a[i];
        //  }       
        //  return tmax;


        // sort(a, a+n);

        int max=INT_MIN, smax= INT_MIN, tmax = INT_MIN ;

        for(int i=0;i<n;i++)
        {
            if(a[i] > max)
            {
                tmax = smax;
                smax = max;
                max = a[i];
            }
            else if(a[i] >smax)
            {
                tmax = smax;
                smax = a[i];
            }
            else if(a[i]> tmax)
            {
                tmax = a[i];
            }

        }
        return tmax;
    }

};