# Find maximum of minimum for every window size in a given array

Given an integer array of size n, find the maximum of the minimum’s of every window size in the array. Note that window size varies from 1 to n.

Example:

```Input:  arr[] = {10, 20, 30, 50, 10, 70, 30}
Output:         70, 30, 20, 10, 10, 10, 10

First element in output indicates maximum of minimums of all
windows of size 1.
Minimums of windows of size 1 are {10}, {20}, {30}, {50}, {10},
{70} and {30}.  Maximum of these minimums is 70

Second element in output indicates maximum of minimums of all
windows of size 2.
Minimums of windows of size 2 are {10}, {20}, {30}, {10}, {10},
and {30}.  Maximum of these minimums is 30

Third element in output indicates maximum of minimums of all
windows of size 3.
Minimums of windows of size 3 are {10}, {20}, {10}, {10} and {10}.
Maximum of these minimums is 20

Similarly other elements of output are computed.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A Simple Solution is to go through all windows of every size, find maximum of all windows. Below is implementation of this idea.

## C++

```// A naive method to find maximum of minimum of all windows of
// different sizes
#include <iostream>
using namespace std;

void printMaxOfMin(int arr[], int n)
{
// Consider all windows of different sizes starting
// from size 1
for (int k=1; k<=n; k++)
{
// Initialize max of min for current window size k
int maxOfMin = arr[0];

// Traverse through all windows of current size k
for (int i = 0; i <= n-k; i++)
{
// Find minimum of current window
int min = arr[i];
for (int j = 1; j < k; j++)
{
if (arr[i+j] < min)
min = arr[i+j];
}

// Update maxOfMin if required
if (min > maxOfMin)
maxOfMin = min;
}

// Print max of min for current window size
cout << maxOfMin << " ";
}
}

// Driver program
int main()
{
int arr[] = {10, 20, 30, 50, 10, 70, 30};
int n = sizeof(arr)/sizeof(arr[0]);
printMaxOfMin(arr, n);
return 0;
}```

## Java

```// A naive method to find maximum of minimum of all windows of
// different sizes

class Test
{
static int arr[] = {10, 20, 30, 50, 10, 70, 30};

static void printMaxOfMin(int n)
{
// Consider all windows of different sizes starting
// from size 1
for (int k=1; k<=n; k++)
{
// Initialize max of min for current window size k
int maxOfMin = arr[0];

// Traverse through all windows of current size k
for (int i = 0; i <= n-k; i++)
{
// Find minimum of current window
int min = arr[i];
for (int j = 1; j < k; j++)
{
if (arr[i+j] < min)
min = arr[i+j];
}

// Update maxOfMin if required
if (min > maxOfMin)
maxOfMin = min;
}

// Print max of min for current window size
System.out.print(maxOfMin + " ");
}
}

// Driver method
public static void main(String[] args)
{
printMaxOfMin(arr.length);
}
}
```

Output:
`70 30 20 10 10 10 10`

Time complexity of above solution can be upper bounded by O(n3).

We can solve this problem in O(n) time using an Efficient Solution. The idea is to extra space. Below are detailed steps.
Step 1: Find indexes of next smaller and previous smaller for every element. Next smaller is the nearest smallest element on right side of arr[i]. Similarly, previous smaller element is the nearest smallest element on left side of arr[i].
If there is no smaller element on right side, then next smaller is n.If there is no smaller on left side, then previous smaller is -1.

For input {10, 20, 30, 50, 10, 70, 30}, array of indexes of next smaller is {7, 4, 4, 4, 7, 6, 7}.
For input {10, 20, 30, 50, 10, 70, 30}, array of indexes of previous smaller is {-1, 0, 1, 2, -1, 4, 4}

This step can be done in O(n) time using the approach discussed in next greater element.

Step 2: Once we have indexes of next and previous smaller, we know that arr[i] is a minimum of a window of length “right[i] – left[i] – 1”. Lengths of windows for which the elements are minimum are {7, 3, 2, 1, 7, 1, 2}. This array indicates, first element is minimum in window of size 7, second element is minimum in window of size 3, and so on.

Create an auxiliary array ans[n+1] to store the result. Values in ans[] can be filled by iterating through right[] and left[]

```    for (int i=0; i < n; i++)
{
// length of the interval
int len = right[i] - left[i] - 1;

// a[i] is the possible answer for
// this length len interval
ans[len] = max(ans[len], arr[i]);
}```

We get the ans[] array as {0, 70, 30, 20, 0, 0, 0, 10}. Note that ans[0] or answer for length 0 is useless.

Step 3:Some entries in ans[] are 0 and yet to be filled. For example, we know maximum of minimum for lengths 1, 2, 3 and 7 are 70, 30, 20 and 10 respectively, but we don’t know the same for lengths 4, 5 and 6.
Below are few important observations to fill remaining entries
a) Result for length i, i.e. ans[i] would always be greater or same as result for length i+1, i.e., ans[i+1].
b) If ans[i] is not filled it means there is no direct element which is minimum of length i and therefore either the element of length ans[i+1], or ans[i+2], and so on is same as ans[i]
So we fill rest of the entries using below loop.

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

Below is implementation of above algorithm.

## C++

```// An efficient C++ program to find maximum of all minimums of
// windows of different sizes
#include <iostream>
#include<stack>
using namespace std;

void printMaxOfMin(int arr[], int n)
{
stack<int> s; // Used to find previous and next smaller

// Arrays to store previous and next smaller
int left[n+1];
int right[n+1];

// Initialize elements of left[] and right[]
for (int i=0; i<n; i++)
{
left[i] = -1;
right[i] = n;
}

// Fill elements of left[] using logic discussed on
// http://www.geeksforgeeks.org/next-greater-element/
for (int i=0; i<n; i++)
{
while (!s.empty() && arr[s.top()] >= arr[i])
s.pop();

if (!s.empty())
left[i] = s.top();

s.push(i);
}

// Empty the stack as stack is going to be used for right[]
while (!s.empty())
s.pop();

// Fill elements of right[] using same logic
for (int i = n-1 ; i>=0 ; i-- )
{
while (!s.empty() && arr[s.top()] >= arr[i])
s.pop();

if(!s.empty())
right[i] = s.top();

s.push(i);
}

// Create and initialize answer array
int ans[n+1];
for (int i=0; i<=n; i++)
ans[i] = 0;

// Fill answer array by comparing minimums of all
// lengths computed using left[] and right[]
for (int i=0; i<n; i++)
{
// length of the interval
int len = right[i] - left[i] - 1;

// arr[i] is a possible answer for this length
// 'len' interval, check if arr[i] is more than
// max for 'len'
ans[len] = max(ans[len], arr[i]);
}

// Some entries in ans[] may not be filled yet. Fill
// them by taking values from right side of ans[]
for (int i=n-1; i>=1; i--)
ans[i] = max(ans[i], ans[i+1]);

// Print the result
for (int i=1; i<=n; i++)
cout << ans[i] << " ";
}

// Driver program
int main()
{
int arr[] = {10, 20, 30, 50, 10, 70, 30};
int n = sizeof(arr)/sizeof(arr[0]);
printMaxOfMin(arr, n);
return 0;
}
```

## Java

```// An efficient Java program to find maximum of all minimums of
// windows of different size

import java.util.Stack;

class Test
{
static int arr[] = {10, 20, 30, 50, 10, 70, 30};

static void printMaxOfMin(int n)
{
// Used to find previous and next smaller
Stack<Integer> s = new Stack<>();

// Arrays to store previous and next smaller
int left[] = new int[n+1];
int right[]  = new int[n+1];

// Initialize elements of left[] and right[]
for (int i=0; i<n; i++)
{
left[i] = -1;
right[i] = n;
}

// Fill elements of left[] using logic discussed on
// http://www.geeksforgeeks.org/next-greater-element/
for (int i=0; i<n; i++)
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();

if (!s.empty())
left[i] = s.peek();

s.push(i);
}

// Empty the stack as stack is going to be used for right[]
while (!s.empty())
s.pop();

// Fill elements of right[] using same logic
for (int i = n-1 ; i>=0 ; i-- )
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();

if(!s.empty())
right[i] = s.peek();

s.push(i);
}

// Create and initialize answer array
int ans[] = new int[n+1];
for (int i=0; i<=n; i++)
ans[i] = 0;

// Fill answer array by comparing minimums of all
// lengths computed using left[] and right[]
for (int i=0; i<n; i++)
{
// length of the interval
int len = right[i] - left[i] - 1;

// arr[i] is a possible answer for this length
// 'len' interval, check if arr[i] is more than
// max for 'len'
ans[len] = Math.max(ans[len], arr[i]);
}

// Some entries in ans[] may not be filled yet. Fill
// them by taking values from right side of ans[]
for (int i=n-1; i>=1; i--)
ans[i] = Math.max(ans[i], ans[i+1]);

// Print the result
for (int i=1; i<=n; i++)
System.out.print(ans[i] + " ");
}

// Driver method
public static void main(String[] args)
{
printMaxOfMin(arr.length);
}
}
```

Output:
`70 30 20 10 10 10 10`

Time Complexity: O(n)
Auxiliary Space: O(n)

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