Next Greater Element Using Stack in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push the current element onto the stack.

DSA Java

stack.[1](arr[i]);

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

Apush

Bpop

Cpeek

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push

Using peek which only views the top element

We use push to add the current element to the stack.

2fill in blank

medium

Complete the code to check if the stack is empty before accessing its top element.

DSA Java

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

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

Apeek

Bpush

Cpop

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() as a condition

Using pop() which removes elements

The empty() method checks if the stack has no elements.

3fill in blank

hard

Fix the error in the condition to compare the top of the stack with the current element.

DSA Java

while (!stack.empty() && stack.[1]() <= arr[i]) {

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

Apeek

Bpop

Cpush

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which removes elements

Using empty() which returns a boolean

Use peek() to look at the top element without removing it.

4fill in blank

hard

Fill both blanks to assign the next greater element or -1 if none exists.

DSA Java

result[i] = !stack.[1]() ? stack.[2]() : -1;

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

Apeek

Bpop

Cempty

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() instead of peek()

Trying to check emptiness with peek()

Use empty() to check if the stack is empty and peek() to get the top element if it is not.

5fill in blank

hard

Fill all three blanks to complete the loop that processes the array from right to left.

DSA Java

for (int i = arr.length - 1; i [1] 0; i[2]) {
    while (!stack.empty() && stack.peek() <= arr[i]) {
        stack.[3]();
    }
    // rest of the code
}

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

Apop

B>=

C--

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using ++ instead of -- in the loop

Using > instead of >= in the condition

Using push() instead of pop() inside the while

The loop runs while i >= 0, decrements i--, and inside the loop, pop() removes smaller elements.

Practice

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

easy

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

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

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

D. The index of the maximum element in the array

Solution

1. Step 1: Understand the problem definition

   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

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

3. Final Answer:

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

4. Quick Check:

   Next Greater Element = first greater right element [OK]

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

Common Mistakes:

• Confusing next greater with next smaller element
• Searching to the left instead of right
• Thinking it finds max element overall

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

easy

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

B. Stack<Integer> stack = new Stack<>();

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

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

Solution

1. Step 1: Check Java generic syntax for Stack

   Java uses generics with angle brackets and wrapper classes, so Stack<Integer> stack = new Stack<>(); is correct.

2. Step 2: Identify errors in other options

   Stack stack = new Stack<int>(); misses generic type parameter syntax; Stack<int> stack = new Stack<>(); uses primitive int which is invalid; Stack<> stack = new Stack<Integer>(); has incorrect generic syntax.

3. Final Answer:

   Stack<Integer> stack = new Stack<>(); -> Option B

4. Quick Check:

   Use wrapper class Integer with generics [OK]

Hint: Use Stack<Integer> stack = new Stack<>(); for integers [OK]

Common Mistakes:

• Using primitive int instead of Integer in generics
• Missing generic type parameter <>
• Incorrect generic syntax placement

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

int[] arr = {4, 5, 2, 25};
int[] nge = nextGreaterElement(arr);
System.out.println(Arrays.toString(nge));

public static int[] nextGreaterElement(int[] nums) {
    int n = nums.length;
    int[] result = new int[n];
    Stack<Integer> stack = new Stack<>();
    for (int i = n - 1; i >= 0; i--) {
        while (!stack.isEmpty() && stack.peek() <= nums[i]) {
            stack.pop();
        }
        result[i] = stack.isEmpty() ? -1 : stack.peek();
        stack.push(nums[i]);
    }
    return result;
}

medium

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

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

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

D. [-1, 25, 25, -1]

Solution

1. Step 1: Trace from right to left

   Start with i=3 (25): stack empty, result[3] = -1, push 25.
   i=2 (2): stack top=25 > 2, result[2]=25, push 2.
   i=1 (5): pop 2 (<=5), stack top=25 > 5, result[1]=25, push 5.
   i=0 (4): stack top=5 > 4, result[0]=5, push 4.

2. Step 2: Collect results

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

3. Final Answer:

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

4. Quick Check:

   Right side bigger elements found correctly [OK]

Hint: Traverse array backward, use stack to track next greater [OK]

Common Mistakes:

• Traversing left to right instead of right to left
• Not popping smaller or equal elements from stack
• Assigning wrong default -1 positions

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

public static int[] nextGreaterElement(int[] nums) {
    int n = nums.length;
    int[] result = new int[n];
    Stack<Integer> stack = new Stack<>();
    for (int i = 0; i < n; i++) {
        while (!stack.isEmpty() && stack.peek() <= nums[i]) {
            stack.pop();
        }
        result[i] = stack.isEmpty() ? -1 : stack.peek();
        stack.push(nums[i]);
    }
    return result;
}

medium

A. Loop should run from right to left (i = n-1 to 0)

B. Stack should store indices, not values

C. Result array size is incorrect

D. Missing initialization of stack

Solution

1. Step 1: Analyze loop direction

   The code loops from left to right, but Next Greater Element using stack requires right to left traversal to find next greater on the right.

2. Step 2: Understand impact of loop direction

   Looping left to right causes incorrect results because stack does not have future elements yet.

3. Final Answer:

   Loop should run from right to left (i = n-1 to 0) -> Option A

4. Quick Check:

   Traverse backward for correct next greater [OK]

Hint: Traverse array backward to find next greater correctly [OK]

Common Mistakes:

• Looping forward instead of backward
• Using stack for values instead of indices (not always wrong but different approach)
• Not popping smaller or equal elements

5. Given the array {13, 7, 6, 12}, what is the Next Greater Element array using a stack? Also, explain why the last element's next greater element is -1.

hard

A. [ 7, 12, 13, -1 ]

B. [ -1, 13, 12, -1 ]

C. [ -1, 12, -1, -1 ]

D. [ -1, 12, 12, -1 ]

Solution

1. Step 1: Find next greater elements from right to left

   Start with 12 (last element): no elements to right, so result is -1.
   Next 6: next greater is 12.
   Next 7: next greater is 12.
   Next 13: no greater element to right, so -1.

2. Step 2: Explain last element's result

   The last element has no elements to its right, so no next greater element exists, hence -1.

3. Final Answer:

   [ -1, 12, 12, -1 ] -> Option D

4. Quick Check:

   Last element no right neighbor = -1 [OK]

Hint: Last element always -1, no elements to right [OK]

Common Mistakes:

• Assigning next greater for last element incorrectly
• Mixing up indices and values
• Not traversing array backward