Next Greater Element Using Stack in DSA Rust - 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 Rust code that finds the next greater element for each item in the array?

DSA Rust

fn next_greater_element(nums: Vec<i32>) -> Vec<i32> {
    let mut stack = Vec::new();
    let mut result = vec![-1; nums.len()];
    for i in 0..nums.len() {
        while let Some(&top) = stack.last() {
            if nums[i] > nums[top] {
                result[top] = nums[i];
                stack.pop();
            } else {
                break;
            }
        }
        stack.push(i);
    }
    result
}

fn main() {
    let nums = vec![4, 5, 2, 25];
    let res = next_greater_element(nums);
    println!("{:?}", res);
}

A [5, 25, 25, -1]

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

C [-1, -1, -1, -1]

D [5, -1, 25, -1]

💡 Hint

Think about how the stack keeps indexes of elements waiting for a greater element.

The code uses a stack to track indexes of elements for which the next greater element is not found yet. When a bigger element is found, it updates the result for those indexes. For the input [4, 5, 2, 25], the next greater elements are [5, 25, 25, -1].

❓ Predict Output

intermediate

Next Greater Element output for descending array

What is the output of the next greater element function for the input array [9, 7, 5, 3]?

DSA Rust

fn next_greater_element(nums: Vec<i32>) -> Vec<i32> {
    let mut stack = Vec::new();
    let mut result = vec![-1; nums.len()];
    for i in 0..nums.len() {
        while let Some(&top) = stack.last() {
            if nums[i] > nums[top] {
                result[top] = nums[i];
                stack.pop();
            } else {
                break;
            }
        }
        stack.push(i);
    }
    result
}

fn main() {
    let nums = vec![9, 7, 5, 3];
    let res = next_greater_element(nums);
    println!("{:?}", res);
}

A [9, 9, 9, -1]

B [7, 5, 3, -1]

C [-1, -1, -1, -1]

D [-1, 9, 7, 5]

💡 Hint

In a strictly descending array, no element has a greater element to its right.

Since the array is strictly descending, no element has a next greater element to its right. So all values remain -1.

🔧 Debug

advanced

Identify the error in the Next Greater Element implementation

What error will the following Rust code produce when run?

DSA Rust

fn next_greater_element(nums: Vec<i32>) -> Vec<i32> {
    let mut stack = Vec::new();
    let mut result = vec![-1; nums.len()];
    for i in 0..nums.len() {
        while let Some(top) = stack.last() {
            if nums[i] > nums[*top] {
                result[*top] = nums[i];
                stack.pop();
            } else {
                break;
            }
        }
        stack.push(i);
    }
    result
}

fn main() {
    let nums = vec![2, 1, 3];
    let res = next_greater_element(nums);
    println!("{:?}", res);
}

A Output: [3, 3, -1]

B Compilation error: cannot move out of borrowed content

C Output: [-1, 3, -1]

D Runtime panic: index out of bounds

💡 Hint

Check how the stack's last element is accessed and used.

The code tries to move the value out of a reference when using while let Some(top) = stack.last(). This causes a compilation error because stack.last() returns a reference and the code tries to move it.

❓ Predict Output

advanced

Next Greater Element output for circular array

What is the output of the next greater element function modified for a circular array for input [1, 2, 1]?

DSA Rust

fn next_greater_element_circular(nums: Vec<i32>) -> Vec<i32> {
    let n = nums.len();
    let mut result = vec![-1; n];
    let mut stack = Vec::new();
    for i in 0..(2 * n) {
        let idx = i % n;
        while let Some(&top) = stack.last() {
            if nums[idx] > nums[top] {
                result[top] = nums[idx];
                stack.pop();
            } else {
                break;
            }
        }
        if i < n {
            stack.push(idx);
        }
    }
    result
}

fn main() {
    let nums = vec![1, 2, 1];
    let res = next_greater_element_circular(nums);
    println!("{:?}", res);
}

A [2, -1, 2]

B [2, 1, -1]

C [-1, -1, -1]

D [1, 2, 1]

💡 Hint

In a circular array, elements after the last wrap around to the start.

The circular logic allows the last element to find the next greater element at the start of the array. For [1, 2, 1], the next greater elements are [2, -1, 2].

🧠 Conceptual

expert

Time complexity of Next Greater Element using stack

What is the time complexity of the Next Greater Element algorithm using a stack for an input array of size n?

A O(n^2) because each element is compared with all others

B O(n log n) because of sorting the array first

C O(log n) due to binary search inside the stack

D O(n) because each element is pushed and popped at most once

💡 Hint

Consider how many times each element is pushed and popped from the stack.

Each element is pushed onto the stack once and popped at most once, so the total operations are proportional to n, making the time complexity O(n).

Practice

1. What is the main purpose of using a stack in the Next Greater Element problem?

easy

A. To keep track of elements for which we haven't found the next greater element yet

B. To sort the array elements in ascending order

C. To store the final output of next greater elements

D. To reverse the array elements before processing

Solution

1. Step 1: Understand the role of stack in the problem

   The stack stores elements that are waiting to find their next greater element as we scan the array.

2. Step 2: Clarify what the stack does not do

   The stack does not sort or store final output; it only helps track elements pending next greater element discovery.

3. Final Answer:

   To keep track of elements for which we haven't found the next greater element yet -> Option A

4. Quick Check:

   Stack tracks pending elements = A [OK]

Hint: Stack holds candidates waiting for next greater element [OK]

Common Mistakes:

• Thinking stack sorts the array
• Confusing stack with output storage
• Assuming stack reverses the array

2. Which of the following Rust code snippets correctly initializes an empty stack to use for Next Greater Element?

easy

A. let mut stack = Vec::empty();

B. let mut stack: Vec<i32> = Vec::new();

C. let stack = new Vec<i32>();

D. let stack = Vec<i32>;

Solution

1. Step 1: Identify correct Rust syntax for empty vector

   In Rust, an empty vector is created with Vec::new() and mutable if we want to push elements.

2. Step 2: Check each option

   let mut stack: Vec<i32> = Vec::new(); uses correct syntax with mut and Vec::new(). Options A, B, and C are invalid Rust syntax.

3. Final Answer:

   let mut stack: Vec<i32> = Vec::new(); -> Option B

4. Quick Check:

   Vec::new() with mut = D [OK]

Hint: Use let mut stack: Vec<i32> = Vec::new(); to create empty stack [OK]

Common Mistakes:

• Omitting mut when pushing elements
• Using incorrect Vec initialization syntax
• Confusing Vec::new() with other methods

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

let nums = vec![4, 5, 2, 25];
let mut stack: Vec<usize> = Vec::new();
let mut result = vec![-1; nums.len()];

for i in 0..nums.len() {
    while !stack.is_empty() && nums[*stack.last().unwrap()] < nums[i] {
        let idx = stack.pop().unwrap();
        result[idx] = nums[i];
    }
    stack.push(i);
}

println!("{:?}", result);

medium

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

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

2. Step 2: Finalize result array

   Stack has index 3 left, result[3] remains -1. Final result is [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: Pop smaller indices until current is smaller, assign next greater [OK]

Common Mistakes:

• Not updating result when popping from stack
• Pushing indices instead of values
• Confusing indices and values in comparisons

4. Identify the bug in this Rust code snippet for Next Greater Element and select the fix:

let nums = vec![2, 7, 3, 5, 4, 6, 8];
let mut stack: Vec<i32> = Vec::new();
let mut result = vec![-1; nums.len()];

for i in 0..nums.len() {
    while !stack.is_empty() && nums[stack.last().unwrap()] < nums[i] {
        let idx = stack.pop().unwrap();
        result[idx] = nums[i];
    }
    stack.push(i as i32);
}

println!("{:?}", result);

medium

A. Change stack type to Vec<usize> and remove i32 casts

B. Change result vector to Vec<i32> with zeros instead of -1

C. Replace while loop with if statement

D. Push nums[i] instead of index i into stack

Solution

1. Step 1: Identify type mismatch in stack usage

   The stack stores indices to access nums. Indices are usize, but stack is Vec<i32> and i is cast to i32. This causes type mismatch and indexing errors.

2. Step 2: Fix by using correct type for stack

   Change stack to Vec<usize> and push i directly without casting. This matches indexing type and avoids errors.

3. Final Answer:

   Change stack type to Vec<usize> and remove i32 casts -> Option A

4. Quick Check:

   Stack indices must be usize = A [OK]

Hint: Use usize for indices in stack to match array indexing [OK]

Common Mistakes:

• Using i32 for indices causing type errors
• Pushing values instead of indices
• Replacing while with if losing multiple pops

5. Given the array [13, 7, 6, 12], apply the Next Greater Element using stack approach and provide the final output array.

hard

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

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

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

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

Solution

1. Step 1: Initialize stack and result

   Start with empty stack and result = [-1, -1, -1, -1]. Process elements left to right.

2. Step 2: Process each element and update result

   - i=0 (13): stack empty, push 0 (stack=[0]) - i=1 (7): 13 > 7, push 1 (stack=[0,1]) - i=2 (6): 7 > 6, push 2 (stack=[0,1,2]) - i=3 (12): 6 < 12, pop 2, result[2]=12 7 < 12, pop 1, result[1]=12 13 > 12, push 3 (stack=[0,3])

3. Step 3: Finalize result

   Stack has indices 0 and 3 left, their result remains -1. Final result: [-1, 12, 12, -1]

4. Final Answer:

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

5. Quick Check:

   Next greater elements = [-1, 12, 12, -1] [OK]

Hint: Pop smaller elements until current is smaller, assign next greater [OK]

Common Mistakes:

• Assigning wrong next greater for first element
• Not popping all smaller elements before pushing
• Confusing values and indices in stack