Next Greater Element Using Stack in DSA C# - Build from Scratch

Next Greater Element Using Stack

📖 Scenario: Imagine you have a list of daily temperatures, and you want to find for each day the next day when the temperature will be higher. This helps you plan your activities better.

🎯 Goal: You will build a program that finds the next greater element for each number in a list using a stack. The next greater element for a number is the first number to the right that is bigger than it. If there is no bigger number, we use -1.

📋 What You'll Learn

Create an integer array called numbers with the exact values: 4, 5, 2, 25

Create an empty stack of integers called stack

Create an integer array called result of the same length as numbers

Use a for loop with variable i to iterate over numbers from right to left

Inside the loop, use a while loop to pop from stack while the top of stack is less than or equal to numbers[i]

If stack is empty after popping, set result[i] to -1, else set result[i] to the top element of stack

Push numbers[i] onto stack

Print the result array as space-separated values

💡 Why This Matters

🌍 Real World

Finding the next greater element is useful in stock price analysis, weather forecasting, and many other areas where you want to know the next time a value increases.

💼 Career

Understanding stacks and this pattern helps in technical interviews and real-world coding tasks involving arrays and efficient lookups.

1

Create the input array

Create an integer array called numbers with the exact values: 4, 5, 2, 25

DSA C#

# Create the integer array numbers with values 4, 5, 2, 25
# Your code here

Hint

Use int[] numbers = {4, 5, 2, 25}; to create the array.

2

Create the stack and result array

Create an empty stack of integers called stack and an integer array called result of the same length as numbers

DSA C#

int[] numbers = {4, 5, 2, 25};
// Create an empty stack of integers called stack
// Create an integer array called result with the same length as numbers
# Your code here

Hint

Use Stack<int> stack = new Stack<int>(); and int[] result = new int[numbers.Length];.

3

Implement the next greater element logic

Use a for loop with variable i to iterate over numbers from right to left. Inside the loop, use a while loop to pop from stack while the top of stack is less than or equal to numbers[i]. If stack is empty after popping, set result[i] to -1, else set result[i] to the top element of stack. Then push numbers[i] onto stack.

DSA C#

using System.Collections.Generic;

int[] numbers = {4, 5, 2, 25};
Stack stack = new Stack();
int[] result = new int[numbers.Length];

// Use a for loop from right to left over numbers
// Inside, use a while loop to pop from stack while top <= numbers[i]
// Set result[i] to -1 if stack empty else top of stack
// Push numbers[i] onto stack
# Your code here

Hint

Loop from the end of the array. Use stack.Peek() to check the top. Pop while top is smaller or equal. Set result accordingly. Push current number.

4

Print the result array

Print the result array as space-separated values using Console.WriteLine.

DSA C#

using System;
using System.Collections.Generic;

int[] numbers = {4, 5, 2, 25};
Stack stack = new Stack();
int[] result = new int[numbers.Length];

for (int i = numbers.Length - 1; i >= 0; i--)
{
    while (stack.Count > 0 && stack.Peek() <= numbers[i])
    {
        stack.Pop();
    }
    result[i] = stack.Count == 0 ? -1 : stack.Peek();
    stack.Push(numbers[i]);
}

// Print the result array as space-separated values
# Your code here

Hint

Use Console.WriteLine(string.Join(" ", result)); to print the array.

Practice

1. What does the Next Greater Element problem using a stack find for each element in an array?

easy

A. The first greater element to the right of each element

B. The first smaller element to the left of each element

C. The sum of all elements greater than the current element

D. The index of the maximum element in the array

Solution

1. Step 1: Understand the problem goal

   The Next Greater Element problem asks to find, for each element, the first element to its right that is greater than it.

2. Step 2: Identify the role of the stack

   A stack helps track elements for which we haven't found a greater element yet, allowing efficient lookup of the next greater element.

3. Final Answer:

   The first greater element to the right of each element -> Option A

4. Quick Check:

   Next Greater Element = first greater right element [OK]

Hint: Next Greater Element means first bigger number on right [OK]

Common Mistakes:

• Confusing next greater with next smaller element
• Thinking it finds the maximum element overall
• Assuming it finds elements to the left instead of right

2. Which of the following C# code snippets correctly initializes a stack to solve the Next Greater Element problem?

easy

A. int[] stack = new int[10];

B. Stack stack = new Stack();

C. List<int> stack = new List<int>();

D. Stack<int> stack = new Stack<int>();

Solution

1. Step 1: Identify correct stack initialization syntax in C#

   Using generic Stack<int> with new Stack<int>() is the modern and type-safe way to create a stack of integers.

2. Step 2: Compare other options

   Stack stack = new Stack(); uses non-generic Stack (older style), C uses List which is not a stack, D uses array which is fixed size and not a stack.

3. Final Answer:

   Stack<int> stack = new Stack<int>(); -> Option D

4. Quick Check:

   Generic Stack<int> initialization is correct [OK]

Hint: Use generic Stack<int> for integer stacks in C# [OK]

Common Mistakes:

• Using non-generic Stack without type safety
• Using List instead of Stack for LIFO behavior
• Using fixed-size arrays instead of dynamic stack

3. What is the output of the Next Greater Element algorithm for the input array [4, 5, 2, 25]?

medium

A. [5, 25, 25, -1]

B. [5, -1, 25, -1]

C. [25, 25, 25, -1]

D. [-1, -1, -1, -1]

Solution

1. Step 1: Process each element to find next greater

   For 4, next greater is 5; for 5, next greater is 25; for 2, next greater is 25; for 25, no greater element, so -1.

2. Step 2: Form the output array

   Output array is [5, 25, 25, -1].

3. Final Answer:

   [5, 25, 25, -1] -> Option A

4. Quick Check:

   Next Greater Elements = [5, 25, 25, -1] [OK]

Hint: Check each element's right side for first bigger number [OK]

Common Mistakes:

• Assigning -1 too early for elements with greater elements later
• Confusing next greater with next smaller
• Misreading the input array order

4. Identify the error in the following C# code snippet for Next Greater Element:

int[] arr = {2, 7, 3, 5, 4, 6, 8};
Stack<int> stack = new Stack<int>();
int[] result = new int[arr.Length];
for (int i = 0; i < arr.Length; i++) {
  while (stack.Count > 0 && arr[stack.Peek()] < arr[i]) {
    result[stack.Pop()] = arr[i];
  }
  stack.Push(i);
}
while (stack.Count > 0) {
  result[stack.Pop()] = 0;
}

medium

A. Using arr[i] instead of arr[stack.Peek()] in while condition

B. Assigning 0 instead of -1 for elements with no greater element

C. Pushing values instead of indices onto the stack

D. Not initializing the result array

Solution

1. Step 1: Check how no greater elements are marked

   The code assigns 0 to elements with no next greater element, but conventionally -1 is used to indicate none.

2. Step 2: Verify other parts of the code

   Indices are pushed, arr[stack.Peek()] is correctly used, and result array is initialized, so no errors there.

3. Final Answer:

   Assigning 0 instead of -1 for elements with no greater element -> Option B

4. Quick Check:

   No greater element marked as -1, not 0 [OK]

Hint: Use -1 to mark no next greater element, not 0 [OK]

Common Mistakes:

• Using 0 instead of -1 for no greater element
• Pushing values instead of indices on stack
• Incorrect while loop condition

5. Given the array [13, 7, 6, 12], how would you modify the Next Greater Element algorithm to find the next greater element to the left instead of the right?

hard

A. Traverse the array from right to left and use a stack to track elements, popping greater ones.

B. Traverse the array from right to left and use a stack to track elements, popping smaller ones.

C. Traverse the array from left to right and use a stack to track elements, popping smaller ones.

D. Traverse the array from left to right and use a queue instead of a stack.

Solution

1. Step 1: Understand direction change for next greater to left

   To find next greater element on the left, we must scan from left to right, checking elements before current.

2. Step 2: Use stack to track candidates and pop smaller elements

   We push elements and pop smaller ones to find the nearest greater on the left efficiently.

3. Final Answer:

   Traverse the array from left to right and use a stack to track elements, popping smaller ones. -> Option C

4. Quick Check:

   Next greater left = left to right traversal with stack [OK]

Hint: For next greater left, scan left to right popping smaller elements [OK]

Common Mistakes:

• Scanning right to left for next greater left
• Using queue instead of stack
• Popping greater elements instead of smaller