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.

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

C++

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// A naive method to find maximum of minimum of all windows of
// different sizes
#include <iostream>
#include <climits>
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 = INT_MIN;
  
        // 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;
}

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Java

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// 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 = Integer.MIN_VALUE;
       
            // 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);
    }
}

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Python3

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# A naive method to find maximum of 
# minimum of all windows of different sizes 
INT_MIN = -1000000
def printMaxOfMin(arr, n): 
      
    # Consider all windows of different 
    # sizes starting from size 1 
    for k in range(1, n + 1): 
          
        # Initialize max of min for
        # current window size k 
        maxOfMin = INT_MIN; 
  
        # Traverse through all windows 
        # of current size k 
        for i in range(n - k + 1): 
              
            # Find minimum of current window 
            min = arr[i] 
            for j in range(k): 
                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 
        print(maxOfMin, end = " ")
  
# Driver Code
arr = [10, 20, 30, 50, 10, 70, 30
n = len(arr)
printMaxOfMin(arr, n)
  
# This code is contributed by sahilshelangia

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C#

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// C# program using Naive approach to find 
// maximum of minimum of all windows of
// different sizes
using System;
  
class GFG{
      
    static int []arr = {10, 20, 30, 50, 10, 70, 30};
      
    // Function to print maximum of minimum
    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 = int.MinValue;
      
            // 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
            Console.Write(maxOfMin + " ");
        }
    }
      
    // Driver Code
    public static void Main() 
    {
        printMaxOfMin(arr.Length);
    }
}
  
// This code is contributed by Sam007.

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PHP

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<?php
// PHP program to find maximum of 
// minimum of all windows of
// different sizes
  
// Method to find maximum of 
// minimum of all windows of
// different sizes
function printMaxOfMin($arr, $n)
{
      
    // Consider all windows of 
    // different sizes starting
    // from size 1
    for($k = 1; $k <= $n; $k++)
    {
          
        // Initialize max of min for
        // current window size k
        $maxOfMin = PHP_INT_MIN;
  
        // Traverse through all windows
        // of current size k
        for ($i = 0; $i <= $n-$k; $i++)
        {
              
            // Find minimum of current window
            $min = $arr[$i];
            for ($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
        echo $maxOfMin , " ";
    }
}
  
    // Driver Code
    $arr= array(10, 20, 30, 50, 10, 70, 30);
    $n = sizeof($arr);
    printMaxOfMin($arr, $n);
  
// This code is contributed by nitin mittal.
?>

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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 use 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++

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// 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
    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;
}

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Java

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// 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
        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);
    }
}

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Python3

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# An efficient Python3 program to find
# maximum of all minimums of windows of 
# different sizes
  
def printMaxOfMin(arr, n):
      
    s = [] # Used to find previous 
           # and next smaller 
  
    # Arrays to store previous and next 
    # smaller. Initialize elements of 
    # left[] and right[]
    left = [-1] * (n + 1
    right = [n] * (n + 1
  
    # Fill elements of left[] using logic discussed on 
    # https:#www.geeksforgeeks.org/next-greater-element
    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)
  
    # Empty the stack as stack is going 
    # to be used for right[] 
    while (len(s) != 0):
        s.pop()
  
    # Fill elements of right[] using same logic
    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 
    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):
          
        # Length of the interval 
        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 i in range(n - 1, 0, -1):
        ans[i] = max(ans[i], ans[i + 1]) 
  
    # Prthe result
    for i in range(1, n + 1):
        print(ans[i], end = " ")
  
# Driver Code
if __name__ == '__main__':
  
    arr = [10, 20, 30, 50, 10, 70, 30
    n = len(arr) 
    printMaxOfMin(arr, n)
  
# This code is contributed by PranchalK

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C#

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// An efficient C# program to find maximum 
// of all minimums of windows of different size 
using System;
using System.Collections.Generic;
  
class GFG
{
public static int[] arr = new int[] {10, 20, 30, 50,
                                     10, 70, 30};
  
public static void printMaxOfMin(int n)
{
    // Used to find previous and next smaller 
    Stack<int> s = new Stack<int>();
  
    // 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 
    for (int i = 0; i < n; i++)
    {
        while (s.Count > 0 && 
               arr[s.Peek()] >= arr[i])
        {
            s.Pop();
        }
  
        if (s.Count > 0)
        {
            left[i] = s.Peek();
        }
  
        s.Push(i);
    }
  
    // Empty the stack as stack is going 
    // to be used for right[] 
    while (s.Count > 0)
    {
        s.Pop();
    }
  
    // Fill elements of right[] using
    // same logic 
    for (int i = n - 1 ; i >= 0 ; i--)
    {
        while (s.Count > 0 && 
               arr[s.Peek()] >= arr[i])
        {
            s.Pop();
        }
  
        if (s.Count > 0)
        {
            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 x
        // 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++)
    {
        Console.Write(ans[i] + " ");
    }
}
  
// Driver Code 
public static void Main(string[] args)
{
    printMaxOfMin(arr.Length);
}
}
  
// This code is contributed by Shrikant13

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Output:

70 30 20 10 10 10 10

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

This article is contributed by Ekta Goel and Ayush Govil. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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