Discover how a simple stack can turn a slow, confusing search into a fast, clear solution!
Why Next Greater Element Using Stack in DSA Javascript?
Imagine you have a list of daily temperatures, and you want to find out for each day when the next warmer day will be. Doing this by checking each day against all future days manually is like looking through a messy pile of papers one by one.
Checking every element against all others takes a lot of time and effort. It's slow and easy to make mistakes, especially with long lists. You might miss some comparisons or repeat work unnecessarily.
Using a stack helps you keep track of elements you haven't found a greater next element for yet. You only compare what's needed, making the process fast and neat, like organizing papers in a stack and quickly finding what you need.
for (let i = 0; i < arr.length; i++) { for (let j = i + 1; j < arr.length; j++) { if (arr[j] > arr[i]) { result[i] = arr[j]; break; } } }
const stack = []; for (let i = 0; i < arr.length; i++) { while (stack.length && arr[i] > arr[stack[stack.length - 1]]) { result[stack.pop()] = arr[i]; } stack.push(i); }
This method lets you quickly find the next greater element for each item in a list, enabling efficient solutions to many real-world problems involving sequences.
Stock market analysis: For each day's stock price, find the next day when the price is higher to decide the best time to sell.
Manual checking is slow and error-prone for large lists.
Stack helps track elements needing a next greater element efficiently.
Next Greater Element using stack speeds up comparisons and reduces mistakes.
Practice
Solution
Step 1: Understand the problem goal
The Next Greater Element problem asks to find, for each element, the first element to its right that is greater than it.Step 2: Recognize the role of the stack
A stack helps track elements waiting for their next greater element to appear on the right.Final Answer:
The first greater element to the right of each element -> Option DQuick Check:
Next Greater Element = first bigger right element [OK]
- Confusing next greater with next smaller element
- Looking to the left instead of right
- Thinking it sums elements instead of finding one
Solution
Step 1: Identify stack data structure in JavaScript
Stacks are commonly implemented using arrays in JavaScript, so initializing with [] is correct.Step 2: Check other options
{ } is an object, null is no value, 0 is a number, none suitable for stack operations.Final Answer:
const stack = []; -> Option CQuick Check:
Stack = array [] for push/pop [OK]
- Using object {} instead of array
- Assigning null or number instead of array
- Not initializing stack before use
[4, 5, 2, 25]?Solution
Step 1: Find next greater for each element
4's next greater is 5, 5's next greater is 25, 2's next greater is 25, 25 has none so -1.Step 2: Write output array
The output array is [5, 25, 25, -1].Final Answer:
[5, 25, 25, -1] -> Option BQuick Check:
Next Greater Element for [4,5,2,25] = [5,25,25,-1] [OK]
- Assigning smaller or equal numbers as next greater
- Using -1 for elements that do have greater elements
- Mixing order of output array
function nextGreater(arr) {
const stack = [];
const result = new Array(arr.length).fill(-1);
for (let i = 0; i <= arr.length; i++) {
while (stack.length && arr[i] > arr[stack[stack.length - 1]]) {
const idx = stack.pop();
result[idx] = arr[i];
}
stack.push(i);
}
return result;
}Solution
Step 1: Check loop boundary
The loop runs from i = 0 to i <= arr.length, which causes arr[i] to be undefined at i = arr.length, causing errors.Step 2: Correct loop condition
Change to i < arr.length to avoid out-of-bound access and fix logic.Final Answer:
Loop condition should be i < arr.length, not i <= arr.length -> Option AQuick Check:
Loop index must be within array length [OK]
- Using <= instead of < in loop condition
- Pushing values instead of indices to stack
- Filling result with wrong default values
[13, 7, 6, 12]. Using the Next Greater Element approach with a stack, what is the final output array?Solution
Step 1: Find next greater for each element
13 has no greater element to the right, so -1. 7's next greater is 12. 6's next greater is 12. 12 has none, so -1.Step 2: Check if any element has a greater element before it
13 is greater than 7 and 6, so their next greater is 12, but 7's next greater is 12, not 13. 13 itself has no greater element.Step 3: Confirm output array
The output is [-1, 12, 12, -1].Final Answer:
[-1, 12, 12, -1] -> Option AQuick Check:
Next Greater Element for [13,7,6,12] = [-1,12,12,-1] [OK]
- Assigning previous greater 13 instead of next greater 12 for 7
- Confusing next greater with previous greater
- Mixing output order
Ready to prove it?
Pass the test and this pattern moves unknown
