Why Next Greater Element Using Stack in DSA Rust?

The Big Idea

What if you could find the next bigger number for every item in a list without checking each one again and again?

The Scenario

Imagine you have a list of daily temperatures, and you want to know for each day when the next warmer day will be. Doing this by checking every following day manually is like flipping through a calendar one day at a time, which takes forever.

The Problem

Manually comparing each day with all the days after it means a lot of repeated work. It's slow and tiring, especially if the list is long. You might miss some days or get confused, making mistakes easy.

The Solution

Using a stack helps you keep track of days you haven't found a warmer day for yet. You only look ahead once, and the stack helps you remember which days to check next. This way, you quickly find the next warmer day for each day without repeating work.

Before vs After

✗ Before

for i in 0..arr.len() {
    let mut next_greater = -1;
    for j in i+1..arr.len() {
        if arr[j] > arr[i] {
            next_greater = arr[j];
            break;
        }
    }
    println!("{} -> {}", arr[i], next_greater);
}

✓ After

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

What It Enables

This method lets you find the next bigger number for every item in a list quickly and efficiently, even for very long lists.

Real Life Example

Stock traders use this to find the next day when a stock price will be higher, helping them decide the best time to sell.

Key Takeaways

Manual checking is slow and error-prone for finding next greater elements.

Using a stack remembers which elements need a next greater element.

This approach speeds up the process and reduces mistakes.

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