Next Greater Element Using Stack in DSA Java - 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 for each item grows as the list gets bigger.

How does the number of steps change when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public 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;
}
    

This code finds the next greater number to the right for each element in the array using a stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop going through the array from right to left.
• How many times: The loop runs once for each element (n times).
• Stack operations: Each element is pushed once and popped at most once.

How Execution Grows With Input

As the input size grows, each element is processed a limited number of times due to stack push and pop.

Input Size (n)	Approx. Operations
10	About 20 stack operations (push/pop) plus 10 loop steps
100	About 200 stack operations plus 100 loop steps
1000	About 2000 stack operations plus 1000 loop steps

Pattern observation: The total 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 all next greater elements grows directly in proportion to the number of elements.

Common Mistake

[X] Wrong: "Because there is a nested while loop inside the for loop, the time complexity must be O(n²)."

[OK] Correct: Each element is pushed and popped at most once, so the total work inside the while loop across all iterations is limited to n, not n².

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 tasks.

Self-Check

"What if we changed the input array to be circular (the end connects to the start)? 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 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