Problem Statement
We have an array of integers of size ‘n’, We need to print the maximum of minimums of every valid window size in the array. Where the window size ranges from 1 to n.
Example –
Input  arr[ ] = {15, 20, 8, 30, 12, 20} Output  30, 15, 12, 8, 8, 8
Explanation:
 The first element (30) of output is the maximum of minimums of all windows of size 1.
 Windows of size 1 are – {15}, {20}, {8}, {30}, {12}, {20} out of which maximum is 30.
 The second element in output represents the maximum of minimums of all windows of size 2.
 Windows of size 2 are – {15,20}, {20,8}, {8,30}, {30,12}, {12,20} Their minimum are 15, 8, 8, 12, 12 respectively and their maximum is 15.
 The third element in output represents the maximum of minimums of all windows of size 3. Minimums of windows of size 3 are {8}, {8}, {8}, and {12} and their maximum is 12.
 Using the same idea, other elements are calculated.
A simple approach (Brute Force)
The brute force approach is to generate and check for all windows of size ‘x’ where x ranges from 1 to n. Then, we need to find the minimum element of all windows for a particular value of x and storing the maximum of those minimums.
To generate all windows, we need to know about starting and endpoints of windows which can be done easily by two nested loops. And for checking for windows of all sizes from 1 to n, we need another nested loop.
Ultimately, we have to use 3 nested loops for window size, starting point, and endpoint.
The implementation of the approach discussed above is as follows:
C++ Programming:
#include <bits/stdc++.h> using namespace std; void printMaxofMinForEveryWindow(int arr[],int n){ // Checking for windows of all valid sizes i.e. from 1 to n for(int windowSize=1;windowSize<=n;windowSize++) { // Initializing maximum with minimum possible value int maximum=INT_MIN; // Traversing through all windows of current windowSize for(int i=0;i<=nwindowSize;i++) { // Checking for minimum element in current window int minimum=arr[i]; for(int j=1;j<windowSize;j++) { minimum=min(minimum,arr[i+j]); } // Updating value of max if the minimum of this window is greater than max. maximum=max(maximum,minimum); } // Print max of min for all window of current windowSize cout<<maximum<<" "; } } int main() { int arr[]={15, 20, 8, 30, 12, 20}; int n = sizeof(arr)/sizeof(arr[0]); printMaxofMinForEveryWindow(arr,n); cout<<endl; return 0; }
Output
Java Programming:
import java.util.*; public class Main { public static void printMaxofMinForEveryWindow(int arr[],int n){ // Checking for windows of all valid sizes i.e. from 1 to n for(int windowSize=1;windowSize<=n;windowSize++) { // Initializing max with minimum possible value int max=Integer.MIN_VALUE; // Traversing through all windows of current windowSize for(int i=0;i<=nwindowSize;i++) { // Checking for minimum element in current window int min=arr[i]; for(int j=1;j<windowSize;j++) { min=Math.min(min,arr[i+j]); } // Updating value of max if the minimum of this window is greater than max. max=Math.max(max,min); } // Print max of min for all window of current windowSize System.out.print(max+" "); } } public static void main (String[] args) { int arr[]={15, 20, 8, 30, 12, 20}; printMaxofMinForEveryWindow(arr,arr.length); System.out.println(); } }
Output
Python Programming:
INT_MIN = 9999999 def printMaxofMinForEveryWindow(arr, n): #Checking for windows of #all valid sizes i.e. from 1 to n for windowSize in range(1, n + 1): # Initializing maxOfMin # with very small value maxOfMin = INT_MIN; # Traversing through all # windows of current windowSize for i in range(n  windowSize + 1): # Checking for minimum element in current window min = arr[i] for j in range(windowSize): if (arr[i + j] < min): min = arr[i + j] # Updating value of max if minimum # of this window is greater than max. if (min > maxOfMin): maxOfMin = min # Print max of min for all window of current windowSize print(maxOfMin, end = " ") arr = [15, 20, 8, 30, 12, 20] n = len(arr) printMaxofMinForEveryWindow(arr, n)
Output:
30 15 12 8 8 8
Complexity Analysis
 Time Complexity: O(n^3). Since we have used 3 nested loops in the approach.
 Extra Space: O(1). Because we are using variables only for storing starting and points of the windows.
Efficient Approach
For each element, we can find what is the maximum size of the window in which that element is minimum and that too in linear time complexity. We use this idea to build our algorithm.
Algorithm
1. For each element find the index of the first smaller element on the right side (next smaller element) and on the left side (previous smaller element). If there is no element on the left side that is smaller than a[i], then the previous smaller element will be 1 and if there is no element on the right side that is smaller than a[i], then the next smaller element will be n. For input {15, 20, 8, 30, 12, 20}, the array of indexes of the previous smaller elements are {1, 0, 1, 2, 2, 4}. For the input {15, 20, 8, 30, 12, 20}, the array of indexes next smaller elements are {2, 2, 6, 4, 6, 6}.
2. Now that we have indexes of the next smaller and previous smaller elements, by observation, we can determine the length of the window in which a[i] is minimum and that would be “right[i] – left[i] – 1”. Using the same approach, we can find that the length of windows for which a[i] is minimum can be represented by the array {2, 1, 6, 1, 3, 1}. This means that the first element is minimum for a window size of 2, second for a window size of 1, and so on. Now we can create an auxiliary answer array of size n+1, ans[n+1] and fill the values according to values of left[] and right[].
for (int i=0; i < n; i++)
{
// length of the interval for which a[i] is minimum
int length = right[i] – left[i] – 1;
// arr[i] may be a possible answer for
// this length length interval, so we store
// maximum possible answer.
ans[length] = max(ans[length], arr[i]);
}
Using this we get the ans[] array as {0, 30, 15, 12, 0, 0, 8}. Here, ans[0] represents the maximum of the minimum of the window of size 0 which is not required.
3. We can see that some elements in the answer array are 0 which means that they are yet to be filled. In the above case, we do not know the maximum of minimums of windows of length 4, and 5. We can make two observations here –

 Answers for window of length i, i.e. ans[i] would always be greater or equal to the result for window of length i+1, i.e., ans[i+1].
 If ans[i] is not filled (means if it 0), it means there is no single element which is minimum of the window of length i and therefore either the element of length ans[i+1], or ans[i+2], and so on would be the answer to ans[i].
So we fill the remaining of the entries using the given for loop.
for (int i=n1; i>=1; i–)
ans[i] = max(ans[i+1], ans[i]);
Below is the implementation of the above algorithm.
C++ Programming:
#include <bits/stdc++.h>
using namespace std;
void printMaxofMinForEveryWindow(int arr[],int n){
// Creating left and right array of size n,
// And initializing with 1 and n respectively.
int left[n];
int right[n];
for(int i=0;i<n;i++)
{
left[i]=1;
right[i]=n;
}
// Fill the left array.
stack st;
for(int i=0;i<n;i++) { while(!st.empty()&&arr[st.top()]>=arr[i])
st.pop();
if(st.empty()) left[i]=1;
else left[i]=st.top();
st.push(i);
}
// Clear the array
while(!st.empty()) st.pop();
// Fill the right array.
for(int i=n1;i>1;i)
{
while(!st.empty()&&arr[st.top()]>=arr[i])
st.pop();
if(st.empty()) right[i]=n;
else right[i]=st.top();
st.push(i);
}
//Creating answer array of size n+1 for all windows
//ranging from 0 to n1 of which 0 size window is of no use
// and initializing with 0.
int ans[n+1];
for(int i=0;i<=n;i++)
ans[i]=0;
for(int i=0;i<n;i++) { // Finding length of window in which arr[i] is the minimum possible element. int length=right[i]left[i]1; // arr[i] is a possible answer for this length 'length' interval, //check if arr[i] is more than max for 'length' if(arr[i]>ans[length]) ans[length]=arr[i];
}
// Fill the values which are not
//filled yet by checking values on the right side of the array.
for(int i=n1;i>0;i)
{
ans[i]=max(ans[i],ans[i+1]);
}
//Printing max of min for all window sizes from 1 to n.
for(int i=1;i<=n;i++)
cout<<ans[i]<<" ";
}
int main() {
int arr[]={15, 20, 8, 30, 12, 20};
int n = sizeof(arr)/sizeof(arr[0]);
printMaxofMinForEveryWindow(arr,n);
cout<<endl;
return 0;
}
Output:
Java Programming:
import java.util.*; public class Main { public static int[] prevSmaller(int[] a) { // Finding the index of the first smaller element on the left side, using Stacks. int n=a.length; int ans[]=new int[n]; Stack st=new Stack<>(); for(int i=0;i<n;i++) { while(!st.isEmpty()&&a[st.peek()]>=a[i]) st.pop(); if(st.isEmpty()) ans[i]=1; else ans[i]=st.peek(); st.push(i); } return ans; } public static int[] nextSmaller(int[] a) { // Finding index of first smaller element on right side, using Stacks. int n=a.length; int ans[]=new int[n]; Stack st=new Stack<>(); for(int i=n1;i>1;i) { while(!st.isEmpty()&&a[st.peek()]>=a[i]) st.pop(); if(st.isEmpty()) ans[i]=n; else ans[i]=st.peek(); st.push(i); } return ans; } public static void printMaxofMinForEveryWindow(int arr[],int n){ //Finding index of first smaller element on left side. int left[]=prevSmaller(arr); //Finding index of first smaller element on left side. int right[]=nextSmaller(arr); //Creating answer array of size n+1 for all windows //ranging from 0 to n1 of which 0 size window is of no use. int ans[]=new int[n+1]; for(int i=0;i<n;i++) { // Finding length of window in which //arr[i] is the minimum possible element. int length=right[i]left[i]1; // arr[i] is a possible answer for this length 'length' //interval, check if arr[i] is more than max for 'length' if(arr[i]>ans[length]) ans[length]=arr[i]; } // Fill the values which are not filled //yet by checking values on the right side of the array. for(int i=n1;i>0;i) { ans[i]=Math.max(ans[i],ans[i+1]); } //Printing max of min for all window sizes from 1 to n. for(int i=1;i<=n;i++) System.out.print(ans[i]+" "); } public static void main (String[] args) { int arr[]={15, 20, 8, 30, 12, 20}; printMaxofMinForEveryWindow(arr,arr.length); System.out.println(); } }
Output:
Python Programming:
def printMaxofMinForEveryWindow(arr, n): s = [] # Used to find previous # and next smaller # Create left and right array # to store previous and next smaller # element, initialize with 1 and n respectively. left = [1] * (n + 1) right = [n] * (n + 1) # Fill elements of left[] for i in range(n): while (len(s) != 0 and arr[s[1]] >= arr[i]): s.pop() if (len(s) != 0): left[i] = s[1] s.append(i) # Clear the stack # to use it again for # filling the right array. while (len(s) != 0): s.pop() # Fill elements of right[] using same approach for i in range(n  1, 1, 1): while (len(s) != 0 and arr[s[1]] >= arr[i]): s.pop() if(len(s) != 0): right[i] = s[1] s.append(i) # Create and initialize answer array # of size n+1. ans = [0] * (n + 1) for i in range(n + 1): ans[i] = 0 # Fill answer array by comparing minimums # of all. Lengths computed using left[] # and right[] for i in range(n): # Finding length of window in which # arr[i] is the minimum possible element. Length = right[i]  left[i]  1 # arr[i] is a possible answer for this # length 'Length' interval,check if arr[i] # is more than max for 'Length' ans[Length] = max(ans[Length], arr[i]) # Fill the values which are not # filled yet by checking values # on the right side of the array. for i in range(n  1, 0, 1): ans[i] = max(ans[i], ans[i + 1]) # Print the result for i in range(1, n + 1): print(ans[i], end = " ") # Driver Code if __name__ == '__main__': arr = [15, 20, 8, 30, 12, 20] n = len(arr) printMaxofMinForEveryWindow(arr, n) print()
Output :
30 15 12 8 8 8
Complexity Analysis
 Time Complexity: O(n). Finding the next smaller element and the previous smaller element takes linear time.
 Extra Space: O(n). We have used stack for calculating next and previous smaller and array to store the intermediate results.
Wrapping Up!
In this article, we have learned how to calculate the maximum of minimums for every possible window of valid size. We discussed two approaches in which we can solve this problem. This article also contains wellexplained codes of both the approaches in the three most popular languages which are c++, Java, and python along with their respective outputs attached to the article for a better understanding of a wide range of our readers.
We sincerely hope that this article has walked you through some deep and important concepts of Arrays and Stacks and you have learned how we should approach such kinds of problems. We hope that you will now be able to solve this problem as well as its other variations seamlessly and efficiently.
Happy Coding!