Next Greater Element Using Stack in DSA C++ - Time & Space Complexity

Time Complexity: Next Greater Element Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the next greater element changes as the input size grows.

How does the number of steps grow when we check each element against others using a stack?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <iostream>
#include <stack>
#include <vector>

std::vector<int> nextGreaterElement(const std::vector<int>& nums) {
    std::stack<int> s;
    std::vector<int> result(nums.size(), -1);
    for (int i = 0; i < (int)nums.size(); ++i) {
        while (!s.empty() && nums[i] > nums[s.top()]) {
            result[s.top()] = nums[i];
            s.pop();
        }
        s.push(i);
    }
    return result;
}
    

This code finds the next greater element for each number in the list using a stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop through the array with a nested while loop that pops from the stack.
• How many times: Each element is pushed once and popped at most once from the stack.

How Execution Grows With Input

As the input size grows, each element is processed a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 operations (push and pop combined)
100	About 200 operations
1000	About 2000 operations

Pattern observation: The operations grow roughly in a straight line with input size, not faster.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the next greater element grows directly in proportion to the number of elements.

Common Mistake

[X] Wrong: "The nested while loop makes this code O(n²) because it looks like two loops inside each other."

[OK] Correct: Each element is pushed and popped only once, so the total work inside the while loop across all iterations is at most n, keeping it linear.

Interview Connect

Understanding this pattern shows you how to use stacks to solve problems efficiently, a skill that helps in many coding challenges and real projects.

Self-Check

"What if we changed the input to a linked list instead of an array? How would the time complexity change?"

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 last element in the array

Solution

1. Step 1: Understand the problem goal

   The Next Greater Element problem aims to find, for each element, the first element to its right that is larger 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 bigger right element [OK]

Hint: Look right for the first bigger number using a stack [OK]

Common Mistakes:

• Confusing next greater with next smaller element
• Thinking it finds elements to the left
• Assuming it sums elements instead of finding one

2. Which of the following C++ code snippets correctly declares a stack to be used for Next Greater Element?

easy

A. stack<int> s = new stack<int>();

B. stack<int> s();

C. std::stack s<int>;

D. std::stack<int> s;

Solution

1. Step 1: Recall correct stack declaration syntax in C++

   The standard way to declare a stack of integers is std::stack<int> s; which creates an empty stack named s.

2. Step 2: Identify incorrect syntax in other options

   stack<int> s = new stack<int>(); uses 'new' which is not needed and incorrect here. std::stack s<int>; has wrong template syntax. stack<int> s(); declares a function, not a stack variable.

3. Final Answer:

   std::stack<int> s; -> Option D

4. Quick Check:

   std::stack<int> s; is correct stack declaration [OK]

Hint: Use std::stack<int> s; to declare stack of integers [OK]

Common Mistakes:

• Using 'new' keyword incorrectly
• Wrong template syntax like s<int>
• Declaring stack as a function

3. What is the output of the following C++ code for Next Greater Element?

std::vector<int> arr = {4, 5, 2, 25};
std::vector<int> nge(arr.size(), -1);
std::stack<int> s;
for (int i = 0; i < arr.size(); i++) {
  while (!s.empty() && arr[i] > arr[s.top()]) {
    nge[s.top()] = arr[i];
    s.pop();
  }
  s.push(i);
}
for (int x : nge) std::cout << x << " ";

medium

A. -1 -1 -1 -1

B. 25 25 -1 -1

C. 5 25 25 -1

D. 5 2 25 -1

Solution

1. Step 1: Trace the loop and stack operations

   Start with empty stack and nge = [-1, -1, -1, -1]. - i=0: push 0 (value 4) - i=1: 5 > 4, so nge[0]=5, pop 0, push 1 - i=2: 2 < 5, push 2 - i=3: 25 > 2, nge[2]=25, pop 2; 25 > 5, nge[1]=25, pop 1; push 3

2. Step 2: Final nge array after loop

   nge = [5, 25, 25, -1]. Print outputs with spaces: "5 25 25 -1 "

3. Final Answer:

   5 25 25 -1 -> Option C

4. Quick Check:

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

Hint: Push indices, update nge when current > top element [OK]

Common Mistakes:

• Printing values instead of indices
• Not popping stack after updating nge
• Confusing greater with smaller elements

4. Identify the bug in this Next Greater Element code snippet:

std::vector<int> arr = {2, 7, 3, 5, 4, 6, 8};
std::vector<int> nge(arr.size(), -1);
std::stack<int> s;
for (int i = arr.size() - 1; i >= 0; i--) {
  while (!s.empty() && arr[i] > s.top()) {
    s.pop();
  }
  if (!s.empty()) nge[i] = s.top();
  s.push(arr[i]);
}

medium

A. Stack should store indices, not values

B. Loop direction should be from 0 to arr.size() - 1

C. nge vector should be initialized with zeros

D. Comparing arr[i] with s.top() instead of arr[s.top()]

Solution

1. Step 1: Analyze stack usage in code

   The stack stores values (arr[i]) but comparisons and nge assignments expect indices to access arr elements correctly.

2. Step 2: Explain why storing indices is needed

   Storing indices allows referencing original array elements for comparison and updating nge. Storing values breaks logic when multiple same values exist.

3. Final Answer:

   Stack should store indices, not values -> Option A

4. Quick Check:

   Stack stores indices for correct comparisons [OK]

Hint: Stack must hold indices to access original array [OK]

Common Mistakes:

• Storing values instead of indices in stack
• Looping in wrong direction without adjusting logic
• Initializing nge with wrong default values

5. Given an array {13, 7, 6, 12}, how would you modify the Next Greater Element algorithm using a stack to find the Next Greater Element to the LEFT instead of the right?

hard

A. Traverse array from right to left, use stack to track elements, pop smaller elements, assign top as NGE to left.

B. Traverse array from left to right, use stack to track elements, pop smaller elements, assign top as NGE to left.

C. Traverse array from left to right, use stack to track elements, pop greater elements, assign top as NGE to left.

D. Traverse array from right to left, use stack to track elements, pop greater elements, assign top as NGE to left.

Solution

1. Step 1: Understand direction change for Next Greater to LEFT

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

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

   While scanning, pop elements smaller than current from stack. The top of stack after popping is the next greater element to the left.

3. Final Answer:

   Traverse array from left to right, use stack to track elements, pop smaller elements, assign top as NGE to left. -> Option B

4. Quick Check:

   Left NGE needs left-to-right scan and pop smaller elements [OK]

Hint: Scan left to right, pop smaller, top is left NGE [OK]

Common Mistakes:

• Scanning right to left for left NGE
• Popping greater elements instead of smaller
• Confusing next greater with next smaller element