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Maximum of minimums of every window size in a given array

Last Updated : 19 Feb, 2025
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Given an integer array arr[] of size n, the task is to find the maximum of the minimums for every window size in the given array, where the window size ranges from 1 to n.

Example:

Input: arr[] = [10, 20, 30]
Output: [30, 20, 10]
Explanation:
First element in output indicates maximum of minimums of all windows of size 1. Minimums of windows of size 1 are [10], [20], [30]. Maximum of these minimums are 30 and similarly other outputs can be computed

Input: arr[] = [10, 20, 30, 50, 10, 70, 30]
Output: [70, 30, 20, 10, 10, 10, 10]
Explanation: The first element in the output indicates the 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
The second element in the output indicates the 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
The third element in the output indicates the 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. 

Table of Content

  • [Naive Approach] - O(n^3) Time and O(1) Space
  • [Expected Approach] Using Stack - O(n) Time and O(n) Space

[Naive Approach] - O(n^3) Time and O(1) Space

The idea is to calculate the minimum of every window separately and print the maximum of each window size.

C++ 
// C++ program to find the maximum of the minimums  // for every window size in the array  #include <iostream> #include <vector> #include <climits> using namespace std;  vector<int> maxOfMins(vector<int>& arr) {     int n = arr.size();     vector<int> res(n, 0);      // 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 minVal = arr[i];             for (int j = 1; j < k; j++) {                 if (arr[i + j] < minVal)                     minVal = arr[i + j];             }              // Update maxOfMin if required             if (minVal > maxOfMin)                 maxOfMin = minVal;         }          // Store max of min for current window size         res[k - 1] = maxOfMin;     }      return res; }  int main() {     vector<int> arr = { 10, 20, 30, 50, 10, 70, 30 };     vector<int> res = maxOfMins(arr);     for (int x : res)         cout << x << " ";     return 0; }
C 
// C program to find the maximum of the minimums  // for every window size in the array  #include <stdio.h> #include <limits.h>  void maxOfMins(int arr[], int n, int res[]) {        // 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 minVal = arr[i];             for (int j = 1; j < k; j++) {                 if (arr[i + j] < minVal) {                     minVal = arr[i + j];                 }             }              // Update maxOfMin if required             if (minVal > maxOfMin) {                 maxOfMin = minVal;             }         }          // Store max of min for current window size         res[k - 1] = maxOfMin;     } }  int main() {     int arr[] = {10, 20, 30, 50, 10, 70, 30};     int n = sizeof(arr) / sizeof(arr[0]);     int res[n];          maxOfMins(arr, n, res);     for (int i = 0; i < n; i++) {         printf("%d ", res[i]);     }     return 0; }
Java 
// Java program to find the maximum of the minimums  // for every window size in the array  import java.util.*;  class GfG {     static ArrayList<Integer> maxOfMins(int[] arr) {         int n = arr.length;         ArrayList<Integer> res = new ArrayList<>(Collections.nCopies(n, 0));          // 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 minVal = arr[i];                 for (int j = 1; j < k; j++) {                     if (arr[i + j] < minVal)                         minVal = arr[i + j];                 }                  // Update maxOfMin if required                 if (minVal > maxOfMin)                     maxOfMin = minVal;             }              // Store max of min for current window size             res.set(k - 1, maxOfMin);         }          return res;     }      public static void main(String[] args) {         int[] arr = {10, 20, 30, 50, 10, 70, 30};         ArrayList<Integer> res = maxOfMins(arr);         for (int x : res)             System.out.print(x + " ");     } }
Python 
# Python program to find the maximum of the minimums  # for every window size in the array  def maxOfMins(arr):     n = len(arr)     res = [0] * 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 = float('-inf')          # Traverse through all windows of current size k         for i in range(n - k + 1):                        # Find minimum of current window             minVal = arr[i]             for j in range(1, k):                 minVal = min(minVal, arr[i + j])              # Update maxOfMin if required             maxOfMin = max(maxOfMin, minVal)          # Store max of min for current window size         res[k - 1] = maxOfMin      return res  if __name__ == "__main__":     arr = [10, 20, 30, 50, 10, 70, 30]     res = maxOfMins(arr)     print(' '.join(map(str, res)))
C# 
// C# program to find the maximum of the minimums  // for every window size in the array  using System; using System.Collections.Generic;  class GfG {     static List<int> maxOfMins(int[] arr) {         int n = arr.Length;         List<int> res = new List<int>(new 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 minVal = arr[i];                 for (int j = 1; j < k; j++) {                     if (arr[i + j] < minVal)                         minVal = arr[i + j];                 }                  // Update maxOfMin if required                 if (minVal > maxOfMin)                     maxOfMin = minVal;             }              // Store max of min for current window size             res[k - 1] = maxOfMin;         }          return res;     }      static void Main() {         int[] arr = {10, 20, 30, 50, 10, 70, 30};         List<int> res = maxOfMins(arr);         Console.WriteLine(string.Join(" ", res));     } }
JavaScript 
// JavaScript program to find the maximum of the minimums  // for every window size in the array  function maxOfMins(arr) {     const n = arr.length;     const res = new Array(n).fill(0);      // Consider all windows of different sizes starting from size 1     for (let k = 1; k <= n; k++) {              // Initialize max of min for current window size k         let maxOfMin = -Infinity;          // Traverse through all windows of current size k         for (let i = 0; i <= n - k; i++) {                      // Find minimum of current window             let minVal = arr[i];             for (let j = 1; j < k; j++) {                 minVal = Math.min(minVal, arr[i + j]);             }              // Update maxOfMin if required             maxOfMin = Math.max(maxOfMin, minVal);         }          // Store max of min for current window size         res[k - 1] = maxOfMin;     }      return res; }  // Driver program const arr = [10, 20, 30, 50, 10, 70, 30]; const res = maxOfMins(arr); console.log(res.join(" "));

Output
70 30 20 10 10 10 10

[Expected Approach] Using Stack - O(n) Time and O(n) Space

The idea is to find the next smaller and previous smaller of each element and update the maximum of window with size as the difference in their indices. This problem is mainly a variation of Largest Area in a Histogram.

Below is the implementation of the above approach:

  • Initialize res[] to store maximum minimums and len[] to store window sizes. Use a stack s for efficient processing.
  • Traverse the array to determine window sizes. For each element, pop larger or equal elements from the stack, calculate their window sizes, and update len[]. Push the current index onto the stack.
  • Process remaining stack elements to determine their window sizes using the right boundary.
  • Populate res[] using len[] and arr[], storing the maximum minimum for each window size.
  • Update res[] to ensure values correctly reflect the maximum of the minimums.
  • Return res[], containing the maximum of minimums for every window size.
C++ 
// C++ program to find the maximum of the minimums  // for every window size in the array  #include <iostream> #include <stack> #include <vector> using namespace std;  vector<int> maxOfMins(vector<int>& arr) {     int n = arr.size();     vector<int> res(n, 0);      stack<int> s;      // Array to store the length of the window      // where each element is the minimum     vector<int> len(n, 0);      // Traverse through array to determine      // window sizes using a stack     for (int i = 0; i < n; i++) {                // Process elements to find next smaller          // element on the left         while (!s.empty() && arr[s.top()] >= arr[i]) {             int top = s.top();             s.pop();             int windowSize = s.empty() ? i : i - s.top() - 1;             len[top] = windowSize;         }         s.push(i);     }      // Process remaining elements in the stack     // for right boundaries     while (!s.empty()) {         int top = s.top();         s.pop();         int windowSize = s.empty() ? n : n - s.top() - 1;         len[top] = windowSize;     }      // Fill res[] based on len[] and arr[]     for (int i = 0; i < n; i++) {         int windowSize = len[i] - 1;  // 0-based indexing         res[windowSize] = max(res[windowSize], arr[i]);     }      // Fill remaining entries in res[] to ensure      // all are max of min     for (int i = n - 2; i >= 0; i--)         res[i] = max(res[i], res[i + 1]);      return res; }  int main() {     vector<int> arr = { 10, 20, 30, 50, 10, 70, 30 };     vector<int> res = maxOfMins(arr);     for (int x : res)         cout << x << " ";     return 0; }
Java 
// Java program to find the maximum of the minimums  // for every window size in the array  import java.util.*;  class GfG {     static ArrayList<Integer> maxOfMins(int[] arr) {         int n = arr.length;         ArrayList<Integer> res = new ArrayList<>(Collections.nCopies(n, 0));         Stack<Integer> s = new Stack<>();          // Array to store the length of the window          // where each element is the minimum         ArrayList<Integer> lenArr = new ArrayList<>(Collections.nCopies(n, 0));          // Traverse through array to determine          // window sizes using a stack         for (int i = 0; i < n; i++) {             // Process elements to find next smaller              // element on the left             while (!s.isEmpty() && arr[s.peek()] >= arr[i]) {                 int top = s.pop();                 int windowSize = s.isEmpty() ? i : i - s.peek() - 1;                 lenArr.set(top, windowSize);             }             s.push(i);         }          // Process remaining elements in the stack         // for right boundaries         while (!s.isEmpty()) {             int top = s.pop();             int windowSize = s.isEmpty() ? n : n - s.peek() - 1;             lenArr.set(top, windowSize);         }          // Fill ressult based on lenArr[] and arr[]         for (int i = 0; i < n; i++) {             int windowSize = lenArr.get(i) - 1;  // 0-based indexing             res.set(windowSize, Math.max(res.get(windowSize), arr[i]));         }          // Fill remaining entries in res[] to ensure          // all are max of min         for (int i = n - 2; i >= 0; i--)             res.set(i, Math.max(res.get(i), res.get(i + 1)));          return res;     }      public static void main(String[] args) {         int[] arr = {10, 20, 30, 50, 10, 70, 30};         ArrayList<Integer> res = maxOfMins(arr);         for (int x : res)             System.out.print(x + " ");     } }
Python 
# Python program to find the maximum of the minimums  # for every window size in the array  def maxOfMins(arr):     n = len(arr)     res = [0] * n     s = []      # Array to store the length of the window      # where each element is the minimum     lenArr = [0] * n      # Traverse through array to determine      # window sizes using a stack     for i in range(n):                # Process elements to find next smaller          # element on the left         while s and arr[s[-1]] >= arr[i]:             top = s.pop()             windowSize = i if not s else i - s[-1] - 1             lenArr[top] = windowSize         s.append(i)      # Process remaining elements in the stack     # for right boundaries     while s:         top = s.pop()         windowSize = n if not s else n - s[-1] - 1         lenArr[top] = windowSize      # Fill res[] based on len_arr[] and arr[]     for i in range(n):         windowSize = lenArr[i] - 1  # 0-based indexing         res[windowSize] = max(res[windowSize], arr[i])      # Fill remaining entries in res[] to ensure      # all are max of min     for i in range(n - 2, -1, -1):         res[i] = max(res[i], res[i + 1])      return res  if __name__ == '__main__':     arr = [10, 20, 30, 50, 10, 70, 30]     res = maxOfMins(arr)     print(' '.join(map(str, res)))
C# 
using System; using System.Collections.Generic;  class GfG {     static List<int> maxOfMins(int[] arr) {         int n = arr.Length;         List<int> res = new List<int>(new int[n]);         Stack<int> s = new Stack<int>();          // Array to store the length of the window where each element is the minimum         List<int> lenArr = new List<int>(new int[n]);           // Traverse through array to determine          // window sizes using a stack         for (int i = 0; i < n; i++) {                          // Process elements to find next smaller              // element on the left             while (s.Count > 0 && arr[s.Peek()] >= arr[i]) {                 int top = s.Pop();                 int windowSize = s.Count == 0 ? i : i - s.Peek() - 1;                 lenArr[top] = windowSize;             }             s.Push(i);         }          // Process remaining elements in the stack         // for right boundaries         while (s.Count > 0) {             int top = s.Pop();             int windowSize = s.Count == 0 ? n : n - s.Peek() - 1;             lenArr[top] = windowSize;         }          // Fill res[] based on lenArr[] and arr[]         for (int i = 0; i < n; i++) {             int windowSize = lenArr[i] - 1;  // 0-based indexing             if (windowSize >= 0 && windowSize < n)  // Prevent out-of-bounds access                 res[windowSize] = Math.Max(res[windowSize], arr[i]);         }          // Fill remaining entries in res[] to ensure          // all are max of min         for (int i = n - 2; i >= 0; i--) {             res[i] = Math.Max(res[i], res[i + 1]);         }          return res;     }      static void Main() {         int[] arr = {10, 20, 30, 50, 10, 70, 30};         List<int> res = maxOfMins(arr);         Console.WriteLine(string.Join(" ", res));     } }
JavaScript 
// JavaScript program to find the maximum of the minimums  // for every window size in the array  function maxOfMins(arr) {     let n = arr.length;     let res = new Array(n).fill(0);     let s = [];      // Array to store the length of the window      // where each element is the minimum     let len = new Array(n).fill(0);      // Traverse through array to determine      // window sizes using a stack     for (let i = 0; i < n; i++) {          // Process elements to find next smaller          // element on the left         while (s.length > 0 && arr[s[s.length - 1]] >= arr[i]) {             let top = s.pop();             let windowSize = s.length === 0 ? i : i - s[s.length - 1] - 1;             len[top] = windowSize;         }         s.push(i);     }      // Process remaining elements in the stack     // for right boundaries     while (s.length > 0) {         let top = s.pop();         let windowSize = s.length === 0 ? n : n - s[s.length - 1] - 1;         len[top] = windowSize;     }      // Fill res[] based on len[] and arr[]     for (let i = 0; i < n; i++) {         let windowSize = len[i] - 1; // 0-based indexing         res[windowSize] = Math.max(res[windowSize], arr[i]);     }      // Fill remaining entries in res[] to ensure      // all are max of min     for (let i = n - 2; i >= 0; i--) {         res[i] = Math.max(res[i], res[i + 1]);     }      return res; }  // Driver program const arr = [10, 20, 30, 50, 10, 70, 30]; const res = maxOfMins(arr); console.log(res.join(' '));

Output
70 30 20 10 10 10 10



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    Write a program to reverse a stack using recursion, without using any loop.Example: Input: elements present in stack from top to bottom 4 3 2 1Output: 1 2 3 4Input: elements present in stack from top to bottom 1 2 3Output: 3 2 1The idea of the solution is to hold all values in Function Call Stack un
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    Reverse individual words
    Given string str, we need to print the reverse of individual words.Examples: Input: Hello WorldOutput: olleH dlroWExplanation: Each word in "Hello World" is reversed individually, preserving the original order, resulting in "olleH dlroW".Input: Geeks for GeeksOutput: skeeG rof skeeG[Expected Approac
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    Reverse a String using Stack
    Given a string str, the task is to reverse it using stack. Example:Input: s = "GeeksQuiz"Output: ziuQskeeGInput: s = "abc"Output: cbaAlso read: Reverse a String – Complete Tutorial.As we all know, stacks work on the principle of first in, last out. After popping all the elements and placing them bac
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    Reversing a Queue
    You are given a queue Q, and your task is to reverse the elements of the queue. You are only allowed to use the following standard queue operations:enqueue(x): Add an item x to the rear of the queue.dequeue(): Remove an item from the front of the queue.empty(): Check if the queue is empty or not.Exa
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    Intermediate problems on Stack

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    Design a stack with the following operations. push(Stack s, x): Adds an item x to stack s pop(Stack s): Removes the top item from stack s merge(Stack s1, Stack s2): Merge contents of s2 into s1. Time Complexity of all above operations should be O(1). If we use array implementation of the stack, then
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    The Stock Span Problem
    The stock span problem is a financial problem where we have a series of daily price quotes for a stock denoted by an array arr[] and the task is to calculate the span of the stock's price for all days. The span of the stock's price on ith day represents the maximum number of consecutive days leading
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    Next Greater Element (NGE) for every element in given Array
    Given an array arr[] of integers, the task is to find the Next Greater Element for each element of the array in order of their appearance in the array. Note: The Next Greater Element for an element x is the first greater element on the right side of x in the array. Elements for which no greater elem
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    Next Greater Frequency Element
    Given an array, for each element find the value of the nearest element to the right which is having a frequency greater than that of the current element. If there does not exist an answer for a position, then make the value '-1'.Examples: Input: arr[] = [2, 1, 1, 3, 2, 1]Output: [1, -1, -1, 2, 1, -1
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    Maximum product of indexes of next greater on left and right
    Given an array arr[1..n], for each element at position i (1 <= i <= n), define the following:left(i) is the closest index j such that j < i and arr[j] > arr[i]. If no such j exists, then left(i) = 0.right(i) is the closest index k such that k > i and arr[k] > arr[i]. If no such k e
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    Iterative Tower of Hanoi
    The Tower of Hanoi is a mathematical puzzle with three poles and stacked disks of different sizes. The goal is to move all disks from the source pole to the destination pole using an auxiliary pole, following two rules:Only one disk can be moved at a time.A larger disk cannot be placed on a smaller
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    Sort a stack using a temporary stack
    Given a stack of integers, sort it in ascending order using another temporary stack.Examples: Input: [34, 3, 31, 98, 92, 23]Output: [3, 23, 31, 34, 92, 98]Explanation: After Sorting the given array it would be look like as [3, 23, 31, 34, 92, 98]Input: [3, 5, 1, 4, 2, 8]Output: [1, 2, 3, 4, 5, 8] Ap
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    Reverse a stack without using extra space in O(n)
    Reverse a Stack without using recursion and extra space. Even the functional Stack is not allowed. Examples: Input : 1->2->3->4 Output : 4->3->2->1 Input : 6->5->4 Output : 4->5->6 We have discussed a way of reversing a stack in the below post.Reverse a Stack using Recu
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    Delete middle element of a stack
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    Check if a queue can be sorted into another queue using a stack
    Given a Queue consisting of first n natural numbers (in random order). The task is to check whether the given Queue elements can be arranged in increasing order in another Queue using a stack. The operation allowed are: Push and pop elements from the stack Pop (Or Dequeue) from the given Queue. Push
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    Check if an array is stack sortable
    Given an array arr[] of n distinct elements, where each element is between 1 and n (inclusive), determine if it is stack-sortable.Note: An array a[] is considered stack-sortable if it can be rearranged into a sorted array b[] using a temporary stack stk with the following operations:Remove the first
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    Largest Rectangular Area in a Histogram
    Given a histogram represented by an array arr[], where each element of the array denotes the height of the bars in the histogram. All bars have the same width of 1 unit.Task is to find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of c
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    Maximum of minimums of every window size in a given array
    Given an integer array arr[] of size n, the task is to find the maximum of the minimums for every window size in the given array, where the window size ranges from 1 to n.Example:Input: arr[] = [10, 20, 30]Output: [30, 20, 10]Explanation: First element in output indicates maximum of minimums of all
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    Find index of closing bracket for a given opening bracket in an expression
    Given a string with brackets. If the start index of the open bracket is given, find the index of the closing bracket. Examples: Input : string = [ABC[23]][89] index = 0 Output : 8 The opening bracket at index 0 corresponds to closing bracket at index 8.Recommended PracticeClosing bracket indexTry It
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    Maximum difference between nearest left and right smaller elements
    Given an array of integers, the task is to find the maximum absolute difference between the nearest left and the right smaller element of every element in the array. Note: If there is no smaller element on right side or left side of any element then we take zero as the smaller element. For example f
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    Delete consecutive same words in a sequence
    Given an array of n strings arr[]. The task is to determine the number of words remaining after pairwise destruction. If two consecutive words in the array are identical, they cancel each other out. This process continues until no more eliminations are possible. Examples: Input: arr[] = ["gfg", "for
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    Check mirror in n-ary tree
    Given two n-ary trees, determine whether they are mirror images of each other. Each tree is described by e edges, where e denotes the number of edges in both trees. Two arrays A[]and B[] are provided, where each array contains 2*e space-separated values representing the edges of both trees. Each edg
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    Reverse a number using stack
    Given a number , write a program to reverse this number using stack.Examples: Input : 365Output : 563Input : 6899Output : 9986We have already discussed the simple method to reverse a number in this post. In this post we will discuss about how to reverse a number using stack.The idea to do this is to
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    Reversing the first K elements of a Queue
    Given an integer k and a queue of integers, The task is to reverse the order of the first k elements of the queue, leaving the other elements in the same relative order.Only following standard operations are allowed on the queue. enqueue(x): Add an item x to rear of queuedequeue(): Remove an item fr
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