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 :
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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