Introduction to Monotonic Stack – Data Structure and Algorithm Tutorials

Last Updated : 24 Jan, 2024

What is a Monotonic Stack?

A monotonic stack is a stack whose elements are monotonically increasing or decreasing. It contains all qualities that a typical stack has and its elements are all monotonic decreasing or increasing.

Below are the features of a monotonic stack:

• It is a range of queries in an array situation
• The minima/maxima elements
• When an element is popped from the monotonic stack, it will never be utilised again.

The monotonic stack problem is mainly the previous/next smaller/larger problem. It maintains monotonicity while popping elements when a new item is pushed into the stack.

Monotonic Stack

Let’s understand the term Monotonic Stacks by breaking it down.

Monotonic: It is a word for mathematics functions. A function y = f(x) is monotonically increasing or decreasing when it follows the below conditions:Â

• As x increases, y also increases always, then it’s a monotonically increasing function.Â
• As x increases, y decreases always, then it’s a monotonically decreasing function.

See the below examples:

• y = 2x +5, it’s a monotonically increasing function.
• y = -(2x), it’s a monotonically decreasing function. Â

Similarly, A stack is called a monotonic stack if all the elements starting from the bottom of the stack is either in increasing or in decreasing order.

Types of Monotonic Stack:

There are 2 types of monotonic stacks:

• Monotonic Increasing Stack
• Monotonic Decreasing Stack

Monotonic Increasing Stack:

It is a stack in which the elements are in increasing order from the bottom to the top of the stack.Â

Example: 1, 3, 10, 15, 17

How to achieve Monotonic Increasing Stack?

To create a Monotonic Increasing Stack, start with an empty stack, then, while iterating through elements in a sequence, keep removing elements from the stack as long as they are smaller than the current element, and push the current element onto the stack. This process ensures the stack maintains a strictly increasing order from bottom to top.

Steps to implement:

• As we need monotonically increasing stack, we should not have a smaller element at top of a bigger element.
• So Iterate the given list of elements one by one :
• Before pushing into the stack, POP all the elements till either of one condition fails:
• Stack is not empty
• Stack’s top is bigger than the element to be inserted.
• Then push the element into the stack.

See the illustration below to understand the idea:

Consider an array Arr[] = {1, 4, 5, 3, 12, 10}
For i = 0: stk = {1}
For i = 1: stk = {1, 4}
For i = 2: stk = {1, 4, 5}
For i = 3: stk = {1, 3} Â [pop 4 and 5 as 4 > 3 and 5 > 3]
For i = 4: stk = {1, 3, 12}
For i = 5: stk = {1, 3, 10} [pop 12 as 12 > 10]Â

Below is the code for the above approach:

C++

 // C++ code to implement the approach   #include using namespace std;   // Function to build Monotonic // increasing stack void increasingStack(int arr[], int N) {     // Initialise stack     stack stk;       for (int i = 0; i < N; i++) {           // Either stack is empty or         // all bigger nums are popped off         while (stk.size() > 0 && stk.top() > arr[i]) {             stk.pop();         }         stk.push(arr[i]);     }       int N2 = stk.size();     int ans[N2] = { 0 };     int j = N2 - 1;       // Empty Stack     while (!stk.empty()) {         ans[j] = stk.top();         stk.pop();         j--;     }       // Displaying the original array     cout << "The Array: ";     for (int i = 0; i < N; i++) {         cout << arr[i] << " ";     }     cout << endl;       // Displaying Monotonic increasing stack     cout << "The Stack: ";     for (int i = 0; i < N2; i++) {         cout << ans[i] << " ";     }     cout << endl; }   // Driver code int main() {     int arr[] = { 1, 4, 5, 3, 12, 10 };     int N = sizeof(arr) / sizeof(arr[0]);       // Function Call     increasingStack(arr, N);       return 0; } //Code done by Balakrishnan R (rbkraj000)

Java

 // Java code to implement the approach import java.io.*; import java.util.*;   class GFG {     // Function to build Monotonic   // increasing stack   static void increasingStack(int[] arr, int N)   {     // Initialise stack     Stack stk = new Stack<>();       for (int i = 0; i < N; i++) {         // Either stack is empty or       // all bigger nums are popped off       while (stk.size() > 0 && stk.peek() > arr[i]) {         stk.pop();       }       stk.push(arr[i]);     }       int N2 = stk.size();     int[] ans = new int[N2];     Arrays.fill(ans, 0);     int j = N2 - 1;       // Empty Stack     while (!stk.isEmpty()) {       ans[j] = stk.peek();       stk.pop();       j--;     }       // Displaying the original array     System.out.print("The Array: ");     for (int i = 0; i < N; i++) {       System.out.print(arr[i] + " ");     }     System.out.println();       // Displaying Monotonic increasing stack     System.out.print("The Stack: ");     for (int i = 0; i < N2; i++) {       System.out.print(ans[i] + " ");     }     System.out.println();   }     public static void main(String[] args)   {     int[] arr = { 1, 4, 5, 3, 12, 10 };     int N = arr.length;       // Function call     increasingStack(arr, N);   } }   // This code is contributed by lokeshmvs21.

Python3

 # Python code to implement the approach   # Function to build Monotonic # increasing stack def increasingStack(arr, N):     # Initialise stack     stk=[]           for i in range(N):         # Either stack is empty or         # all bigger nums are popped off         while(len(stk) > 0 and stk[len(stk) - 1] > arr[i]):             stk.pop()         stk.append(arr[i])               N2 = len(stk)     ans = [0]*N2     j = N2 - 1           # Empty Stack     while(len(stk) != 0):         ans[j] = stk[len(stk) - 1]         stk.pop()         j = j - 1           # Displaying the original array     print("The Array: ",end="")     for i in range(N):         print(arr[i],end=" ")     print()           # Displaying Monotonic increasing stack     print("The Stack: ",end="")     for i in range(N2):         print(ans[i],end=" ")     print()       # Driver code arr = [1, 4, 5, 3, 12, 10] N = len(arr)   # Function Call increasingStack(arr,N)   # This code is contributed by Pushpesh Raj.

C#

 // C# code to implement the approach   using System; using System.Collections.Generic;   public class GFG{     // Function to build Monotonic   // increasing stack   static void increasingStack(int[] arr, int N)   {     // Initialise stack     Stack stk = new Stack();       for (int i = 0; i < N; i++)     {       // Either stack is empty or       // all bigger nums are popped off       while (stk.Count > 0 && stk.Peek() > arr[i])       {         stk.Pop();       }       stk.Push(arr[i]);     }       int N2 = stk.Count;     int[] ans = new int[N2];     Array.Fill(ans, 0);     int j = N2 - 1;       // Empty Stack     while (stk.Count > 0)     {       ans[j] = stk.Peek();       stk.Pop();       j--;     }       // Displaying the original array     Console.Write("The Array: ");     for (int i = 0; i < N; i++)     {       Console.Write(arr[i] + " ");     }     Console.WriteLine();       // Displaying Monotonic increasing stack     Console.Write("The Stack: ");     for (int i = 0; i < N2; i++)     {       Console.Write(ans[i] + " ");     }     Console.WriteLine();   }     static public void Main (){       // Code     int[] arr = { 1, 4, 5, 3, 12, 10 };     int N = arr.Length;       // Function call     increasingStack(arr, N);   } }   // This code is contributed by lokesh.

Javascript

 // JavaScript code for the above approach      // Function to build Monotonic    // increasing stack    function increasingStack(a, N) {      // Initialise stack      let stk = [];        for (let i = 0; i < N; i++) {          // Either stack is empty or        // all bigger nums are popped off        while (stk.length > 0 && stk[stk.length - 1] > arr[i]) {          stk.pop();        }        stk.push(arr[i]);      }        let N2 = stk.length;      let ans = new Array(N2);      let j = N2 - 1;        // Empty Stack      while (stk.length != 0) {        ans[j] = stk[stk.length - 1];        stk.pop();        j--;      }        // Displaying the original array      console.log("The Array: ");      for (let i = 0; i < N; i++) {        console.log(arr[i] + " ");      }      console.log("
");        // Displaying Monotonic increasing stack      console.log("The Stack: ");      for (let i = 0; i < N2; i++) {        console.log(ans[i] + " ");      }      console.log("
");    }      // Driver code      let arr = [1, 4, 5, 3, 12, 10];    let N = arr.length;      // Function Call    increasingStack(arr, N);   // This code is contributed by Potta Lokesh

Output

The Array: 1 4 5 3 12 10
The Stack: 1 3 10

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

Monotonic Decreasing Stack:

A stack is monotonically decreasing if It’s elements are in decreasing order from the bottom to the top of the stack.Â

Example: 17, 14, 10, 5, 1

How to achieve Monotonic Decreasing Stack?

To create a Monotonic Decreasing Stack, begin with an empty stack, then, while iterating through elements in a sequence, continuously remove elements from the stack as long as they are smaller than or equal to the current element, and finally, push the current element onto the stack. This process ensures the stack maintains a monotonic decreasing order from bottom to top.

Steps to implement:

• As we need monotonically decreasing stack, we should not have a bigger element at top of a smaller element.
• So Iterate the elements of the list one by one:
• Before pushing into the stack, POP all the elements till either of one condition fails:
• Stack is not empty
• Stack’s top is smaller than the element to be Inserted.
• Then push the element into the stack.

See the below illustration for a better understanding:

Consider an array: arr[] = {15, 17, 12, 13, 14, 10}
For i = 0: stk = {15}
For i = 1: stk = {17} [pop 15 as 15 < 17]
For i = 2: stk = {17, 12}
For i = 3: stk = {17, 13} Â [pop 12 as 12 < 13]
For i = 4: stk = {17, 14} Â [pop 13 as 13 < 14]
For i = 5: stk = {17, 14, 10}

Below is the implementation of the above approach:

C++

 // C++ code to implement the approach   #include using namespace std;   // Function to find a Monotonic // decreasing stack void decreasingStack(int arr[], int N) {     // Initialising Stack     stack stk;       for (int i = 0; i < N; i++) {           // Either stack empty or         // all smaller nums are popped off         while (stk.size() > 0 && stk.top() < arr[i]) {             stk.pop();         }         stk.push(arr[i]);     }       int N2 = stk.size();     int ans[N2] = { 0 };     int j = N2 - 1;       // Empty stack     while (!stk.empty()) {           ans[j] = stk.top();         stk.pop();         j--;     }       // Displaying the original array     cout << "The Array: ";     for (int i = 0; i < N; i++) {           cout << arr[i] << " ";     }     cout << endl;       // Displaying Monotonic Decreasing Stack     cout << "The Stack: ";     for (int i = 0; i < N2; i++) {           cout << ans[i] << " ";     }     cout << endl; }   // Driver code int main() {     int arr[] = { 15, 17, 12, 13, 14, 10 };     int N = sizeof(arr) / sizeof(arr[0]);       // Function call     decreasingStack(arr, N);       return 0; } //Code done by Balakrishnan R (rbkraj000)

Java

 // Java code to implement the approach import java.util.*;   public class Main {       // Function to find a Monotonic     // decreasing stack     static void decreasingStack(int arr[], int N)     {         // Initialising Stack         Stack stk = new Stack();           for (int i = 0; i < N; i++) {               // Either stack empty or             // all smaller nums are popped off             while (stk.size() > 0 && stk.peek() < arr[i]) {                 stk.pop();             }             stk.push(arr[i]);         }           int N2 = stk.size();         int ans[] = new int[N2];         int j = N2 - 1;           // Empty stack         while (!stk.empty()) {               ans[j] = stk.peek();             stk.pop();             j--;         }           // Displaying the original array         System.out.print("The Array: ");         for (int i = 0; i < N; i++) {               System.out.print(arr[i] + " ");         }         System.out.println();           // Displaying Monotonic Decreasing Stack         System.out.print("The Stack: ");         for (int i = 0; i < N2; i++) {               System.out.print(ans[i] + " ");         }         System.out.println();     }       // Driver code     public static void main(String args[])     {         int arr[] = { 15, 17, 12, 13, 14, 10 };         int N = arr.length;           // Function call         decreasingStack(arr, N);     } }

Python3

 # Python code to implement the approach   # Function to find a Monotonic # decreasing stack def decreasingStack(arr, N):     stack = []     for i in range(N):                 # Either stack empty or         # all smaller nums are popped off         while len(stack)>0 and stack[-1] < arr[i]:             stack.pop()         stack.append(arr[i])               N2 = len(stack)     ans = [0]*N2     j = N2-1           # Empty Stack     while stack != []:         ans[j] = stack.pop()         j -= 1           # Displaying the original array     print('The array: ',end = ' ')     for i in range(N):         print(arr[i],end = ' ')     print()           # Displaying Monotonic Decreasing Stack     print('The array: ',end = ' ')     for i in range(N2):         print(ans[i],end = ' ')     print()           # Driver code arr = [15, 17, 12, 13, 14, 10] N = len(arr)   # Function call decreasingStack(arr, N)   # This code is contributed by hardikkhuswaha.

C#

 // C# code using System; using System.Collections.Generic;   public class GFG {     // Function to find a Monotonic   // decreasing stack   static void decreasingStack(int[] arr, int N)   {     // Initialising Stack     Stack stk = new Stack();       for (int i = 0; i < N; i++) {         // Either stack empty or       // all smaller nums are popped off       while (stk.Count > 0 && stk.Peek() < arr[i]) {         stk.Pop();       }       stk.Push(arr[i]);     }       int N2 = stk.Count;     int[] ans = new int[N2];     int j = N2 - 1;       // Empty stack     while (stk.Count > 0) {         ans[j] = stk.Peek();       stk.Pop();       j--;     }       // Displaying the original array     Console.Write("The Array: ");     for (int i = 0; i < N; i++) {         Console.Write(arr[i] + " ");     }     Console.WriteLine();       // Displaying Monotonic Decreasing Stack     Console.Write("The Stack: ");     for (int i = 0; i < N2; i++) {         Console.Write(ans[i] + " ");     }     Console.WriteLine();   }     // Driver code   public static void Main(string[] args)   {     int[] arr = { 15, 17, 12, 13, 14, 10 };     int N = arr.Length;       // Function call     decreasingStack(arr, N);   } }   // This code is contributed by ishankhandelwals.

Javascript

 // // Js code to implement the approach // // Function to find a Monotonic // // decreasing stack function decreasingStack(arr, N) {       // initialising Stack     let stk = [];     for (let i = 0; i < N; i++)     {               // Either stack empty or         // all smaller nums are popped off         while (stk.length > 0 && stk[0] < arr[i]) {             stk.shift();         }         stk.unshift(arr[i]);     }     let N2 = stk.length;     let ans = Array(N2).fill(0);     let j = N2 - 1;     // Empty stack     while (stk.length != 0) {         ans[j] = stk[0];         stk.shift();         j--;     }           // Displaying the original array     console.log("The Array: ");     for (let i = 0; i < N; i++) {         console.log(arr[i]);     }           // Displaying Monotonic Decreasing Stack     console.log("The Stack: ");     for (let i = 0; i < N2; i++) {         console.log(ans[i]);     } }   // Driver code let arr = [15, 17, 12, 13, 14, 10]; let N = arr.length;   // Function call decreasingStack(arr, N);   // This code is contributed by ishankhandelwals.

Output

The Array: 15 17 12 13 14 10
The Stack: 17 14 10

Time Complexity: O(N)
Auxiliary Space: O(N)Â

Note that this implementation assumes that all elements in the input array are distinct. If there are duplicates, we need to modify the implementation to handle them correctly.

Applications of Monotonic Stack :

• Monotonic stack is generally used to deal with a typical problem like Next Greater Element. NGE (Find the first value on the right that is greater than the element.
• Also can be used for its varieties.
• Next Smaller Element
• Previous Greater Element
• Previous Smaller Element
• Also, we use it to get the greatest or smallest array or string by the given conditions (remaining size k/ no duplicate).
• To understand the optimization power of monotonic stacks, let’s take this example problem: Minimum Cost Tree From Leaf Values. This problem can be solved in 3 different algorithm ways, out of which the monotonic stack is the most optimized approach.
• Dynamic Programming Algorithmic Approach: O(N^3) Time O(N^2) Space
• Greedy Algorithmic Approach: O(N^2) Time O(1) Space
• Monotonic Stack Algorithmic Approach: O(N) Time O(N) Space

Follow the link to know How to Identify and Solve the Monotonic Stack Problems

• We can use the extra space of a monotonic stack to reduce the time complexity.
• We can get the nearest smaller or greater element depending on the monotonic stack type, by just retrieving the stack’s top element, which is just an O(1) operation.
• The monotonic stack helps us maintain maximum and minimum elements in the range and keeps the order of elements in the range. Therefore, we donâ€™t need to compare elements one by one again to get minima and maxima in the range. Meanwhile, because it keeps the elementâ€™s order, we only need to update the stack based on the newest added element.