Next Greater Element Using Stack in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a stack of integers.

DSA C++

std::stack<int> [1];

Drag options to blanks, or click blank then click option'

Astk

Bstack

Cs

Dst

Attempts:

3 left

💡 Hint

Common Mistakes

Using reserved keywords like 'stack' as variable name.

Using names that are too long or unclear.

The variable name 'stk' is commonly used for stack objects and matches the code style.

2fill in blank

medium

Complete the code to push the first element of the array into the stack.

DSA C++

stk.[1](arr[0]);

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cinsert

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' instead of 'push'.

Using 'insert' which is not a stack method.

The 'push' method adds an element to the top of the stack.

3fill in blank

hard

Fix the error in the while loop condition to check if the stack is not empty.

DSA C++

while (!stk.[1]()) {

Drag options to blanks, or click blank then click option'

Atop

Bsize

Cpop

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size()' which returns a number, not a boolean.

Using 'top()' which returns the top element, not a boolean.

The 'empty()' method returns true if the stack is empty, so '!empty()' checks if it is not empty.

4fill in blank

hard

Fill both blanks to complete the if condition that compares the current array element with the top of the stack.

DSA C++

if (arr[i] [1] stk.[2]()) {

Drag options to blanks, or click blank then click option'

A>

B<

Ctop

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>'.

Using 'empty()' instead of 'top()'.

We compare if the current element is greater than the top element of the stack using '>' and 'top()'.

5fill in blank

hard

Fill all three blanks to correctly assign the next greater element and update the stack.

DSA C++

result[stk.[1]()] = arr[i];
stk.[2]();
stk.[3](i);

Drag options to blanks, or click blank then click option'

Atop

Bpop

Cpush

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'empty()' instead of 'top()'.

Using 'push()' instead of 'pop()' to remove elements.

We use 'top()' to get the index, 'pop()' to remove it, and 'push()' to add the current index.

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