Min Max Riddle


Problem Statement :


Given an integer array of size n, find the maximum of the minimum(s) of every window size in the array. The window size varies from 1 to n.

For example, given arr = [6, 3 , 5 ,  1,  12 ] consider window sizes of 1 through 5. Windows of size  are . The maximum value of the minimum values of these windows is . Windows of size  are  and their minima are ( 3, 3, 1, 1 ). The maximum of these values is . Continue this process through window size  to finally consider the entire array. All of the answers are 12,  3 , 3 , 1 , 1 .

Function Description

Complete the riddle function in the editor below. It must return an array of integers representing the maximum minimum value for each window size from  to .

riddle has the following parameter(s):

arr: an array of integers


Input Format

The first line contains a single integer, n, the size of arr.
The second line contains n space-separated integers, each an arr[ i ].


Constraints

1   <=   n   <=  10^6
1   <=   arr[ i ]  <=  10^9


Output Format

Single line containing  space-separated integers denoting the output for each window size from 1 to n.


Solution :



title-img 




                        Solution in C++ :

In   C ++   :





#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n,i;
    cin>>n;

    int a[n];
    for(i=0;i<n;++i) {
        cin>>a[i];
    }

    stack<int> s;
    int left[n],right[n],ans[n+1],len;

    for(i=0;i<n;++i) {
        left[i]=-1,right[i]=n;
    }

    for(i=0;i<n;++i) {
        while(!s.empty() && a[s.top()] >= a[i]) {
            s.pop();
        }
        if(!s.empty()) {
            left[i]=s.top();
        }
        s.push(i);
    }

    while(!s.empty()) {
        s.pop();
    }

    for(i=n-1;i>=0;--i) {
        while(!s.empty() && a[s.top()] >= a[i]) {
            s.pop();
        }
        if(!s.empty()) {
            right[i]=s.top();
        }
        s.push(i);
    }

    memset(ans, 0, sizeof ans);
    for(i=0;i<n;++i) {
        len = right[i]-left[i]-1;
        ans[len]=max(ans[len], a[i]);
    }

    for(i=n-1;i>=1;--i) {
        ans[i]=max(ans[i], ans[i+1]);
    }

    for(i=1;i<=n;++i) {
        cout<<ans[i]<<" ";
    }
    cout<<endl;
    return 0;
}
                    


                        Solution in Java :

In   Java   :





static long[] riddle(long[] arr) {
        // complete the function
       int n=arr.length;
       Stack<Integer> st=new Stack<>();
       int[] left=new int[n+1];
       int[] right=new int[n+1];
       for(int i=0;i<n;i++){
           left[i]=-1;
           right[i]=n;
       }
       for(int i=0;i<n;i++){
           while(!st.isEmpty() && arr[st.peek()]>=arr[i])
               st.pop();
           
           if(!st.isEmpty())
               left[i]=st.peek();
           
           st.push(i);
       }
       while(!st.isEmpty()){
           st.pop();
       }

       for(int i=n-1;i>=0;i--){
           while(!st.isEmpty() && arr[st.peek()]>=arr[i])
               st.pop();
           
           if(!st.isEmpty())
               right[i]=st.peek();
           
           st.push(i);
       }
        long ans[] = new long[n+1]; 
        for (int i=0; i<=n; i++) {
            ans[i] = 0; 
        }
         for (int i=0; i<n; i++) 
        { 
            int len = right[i] - left[i] - 1; 
            ans[len] = Math.max(ans[len], arr[i]); 
        }
        for (int i=n-1; i>=1; i--) {
            ans[i] = Math.max(ans[i], ans[i+1]);  
        }
       long[] res=new long[n];
        for (int i=1; i<=n; i++) {
            res[i-1]=ans[i];
        }
        return res;
    }




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