Next Greater Element Using Stack in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Next Greater Element for a given array

What is the output of the following C++ code that finds the next greater element for each element in the array?

DSA C++

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

int main() {
    std::vector<int> nums = {4, 5, 2, 25};
    std::vector<int> res = nextGreaterElement(nums);
    for (int x : res) {
        std::cout << x << " ";
    }
    return 0;
}

A 5 -1 25 -1

B 5 25 -1 -1

C 25 25 25 -1

D 5 25 25 -1

💡 Hint

Think about how the stack helps find the next greater element by popping smaller or equal elements.

The code iterates from right to left, using a stack to keep track of next greater elements. For 4, next greater is 5; for 5, next greater is 25; for 2, next greater is 25; for 25, no greater element exists, so -1.

❓ Predict Output

intermediate

Next Greater Element output for circular array

What is the output of the following code that finds the next greater element in a circular array?

DSA C++

std::vector<int> nextGreaterCircular(std::vector<int>& nums) {
    int n = nums.size();
    std::vector<int> result(n, -1);
    std::stack<int> s;
    for (int i = 2 * n - 1; i >= 0; --i) {
        while (!s.empty() && s.top() <= nums[i % n]) {
            s.pop();
        }
        if (i < n && !s.empty()) {
            result[i] = s.top();
        }
        s.push(nums[i % n]);
    }
    return result;
}

int main() {
    std::vector<int> nums = {1, 2, 1};
    std::vector<int> res = nextGreaterCircular(nums);
    for (int x : res) {
        std::cout << x << " ";
    }
    return 0;
}

A 2 -1 2

B 2 2 -1

C -1 2 2

D 2 2 2

💡 Hint

Remember the array is circular, so elements after the end wrap around to the start.

For the circular array [1,2,1], the next greater element for 1 is 2, for 2 there is no greater element so -1, and for the last 1 the next greater is 2 again.

🔧 Debug

advanced

Identify the error in Next Greater Element implementation

What error will the following code produce when run?

DSA C++

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

int main() {
    std::vector<int> nums = {2, 1, 2, 4, 3};
    std::vector<int> res = nextGreaterElement(nums);
    for (int x : res) {
        std::cout << x << " ";
    }
    return 0;
}

A Outputs: 4 2 4 -1 -1

B Outputs: 4 4 4 -1 -1

C Runtime error: stack underflow

D Outputs: 4 4 -1 -1 -1

💡 Hint

Check how the stack stores indices and how the result is assigned.

The code correctly finds the next greater element for each index by storing indices in the stack. For input {2,1,2,4,3}, the output is 4 2 4 -1 -1.

🧠 Conceptual

advanced

Stack usage in Next Greater Element algorithm

Why does the Next Greater Element algorithm use a stack to solve the problem efficiently?

A Because a stack helps keep track of elements for which the next greater element is not yet found, allowing O(n) time complexity.

B Because a stack sorts the array automatically to find the next greater element.

C Because a stack stores all elements in sorted order to find the maximum element quickly.

D Because a stack allows random access to elements to compare all pairs efficiently.

💡 Hint

Think about how the algorithm processes elements and remembers which ones need a next greater element.

The stack stores elements (or their indices) waiting for their next greater element. When a greater element is found, it resolves all smaller elements on the stack. This avoids nested loops and achieves O(n) time.

❓ Predict Output

expert

Output of Next Greater Element with duplicate values and large input

What is the output of the following code for the input array with duplicates?

DSA C++

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

int main() {
    std::vector<int> nums = {2, 2, 1, 3, 2, 3, 1};
    std::vector<int> res = nextGreaterElement(nums);
    for (int x : res) {
        std::cout << x << " ";
    }
    return 0;
}

A 3 3 3 -1 -1 -1 -1

B 3 3 3 3 3 -1 -1

C 3 3 3 -1 3 -1 -1

D 3 3 1 -1 3 -1 -1

💡 Hint

Duplicates do not affect the logic; the stack pops elements less than or equal to current.

For each element, the next greater element is the first element to the right that is strictly greater. For the input, the output is 3 3 3 -1 3 -1 -1.

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