Next Greater Element Using Stack in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Next Greater Element (NGE) for an element in an array?

The Next Greater Element for an element is the first element to the right of it in the array that is greater than the element itself. If no such element exists, the NGE is -1.

intermediate

Why do we use a stack to solve the Next Greater Element problem efficiently?

A stack helps keep track of elements for which we haven't found the next greater element yet. It allows us to process elements in a way that avoids repeated scanning, reducing time complexity from O(n²) to O(n).

intermediate

In the Next Greater Element algorithm using a stack, what do we do when the current element is greater than the element at the top of the stack?

We pop elements from the stack until the stack becomes empty or the top element is greater than or equal to the current element. For each popped element, the current element is its Next Greater Element.

beginner

What is the time complexity of the Next Greater Element algorithm using a stack?

The time complexity is O(n) because each element is pushed and popped at most once from the stack.

beginner

What will be the Next Greater Element array for [4, 5, 2, 25]?

The Next Greater Element array is [5, 25, 25, -1]. Explanation: For 4, next greater is 5; for 5, next greater is 25; for 2, next greater is 25; for 25, no greater element to the right, so -1.

What does the stack store in the Next Greater Element algorithm?

A Only the largest element found so far

B Indices or values of elements waiting for their next greater element

C All elements in sorted order

D Elements that have already found their next greater element

The stack stores elements (or their indices) for which the next greater element has not been found yet.

If the current element is smaller than the top of the stack, what should you do?

A Push the current element onto the stack

B Pop the top element from the stack

C Ignore the current element

D Replace the top element with the current element

If the current element is smaller, it might have a next greater element later, so we push it onto the stack.

What is the Next Greater Element for the last element in the array?

A The largest element in the array

B The next element to the right

C The first element in the array

D -1 (no greater element to the right)

The last element has no elements to its right, so its Next Greater Element is -1.

What is the main advantage of using a stack for the Next Greater Element problem?

A It reduces time complexity to O(n)

B It finds the smallest element quickly

C It uses less memory than arrays

D It sorts the array automatically

Using a stack allows us to find the Next Greater Element for all elements in linear time O(n).

Which of these arrays has the Next Greater Element array as [-1, -1, -1]?

A [1, 2, 3]

B [2, 3, 1]

C [3, 2, 1]

D [1, 1, 1]

In [3, 2, 1], no element has a greater element to its right, so all NGEs are -1.

Explain step-by-step how to find the Next Greater Element for each element in an array using a stack.

Think about how the stack helps track elements waiting for their next greater element.

Describe why the Next Greater Element algorithm using a stack runs in O(n) time.

Consider how many times each element interacts with the stack.

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