How do you solve Next Greater Element III on LeetCode?

Next Greater Element III (LeetCode #556) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to manipulate digits is crucial.

What is the brute force solution for Next Greater Element III?

The brute force approach for Next Greater Element III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the digits of the number and checking each one to find the smallest permutation that is greater than the original number. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Next Greater Element III?

Next Greater Element III uses the following concepts: Math, Two Pointers, String. The recommended patterns to study are: Two Pointers, Greedy Algorithms.

What are the interview tips for Next Greater Element III?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice writing clean and efficient code.

What are common mistakes when solving Next Greater Element III?

Forgetting to check for integer overflow. Not considering edge cases like the largest possible number.

#556

Next Greater Element III

Medium

MathTwo PointersStringTwo PointersGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a systematic method to find the next greater permutation of the digits without generating all permutations. This is efficient and avoids unnecessary computations.

⚙️

Algorithm

5 steps

1Step 1: Convert the number to a list of digits.
2Step 2: Find the first pair of digits from the right where the left digit is smaller than the right digit.
3Step 3: Find the smallest digit on the right side of this pair that is larger than the left digit.
4Step 4: Swap these two digits.
5Step 5: Reverse the digits to the right of the original position of the left digit to get the smallest order.

solution.py17 lines

1# Full working Python code
2
3def nextGreaterElement(n):
4    digits = list(str(n))
5    length = len(digits)
6    i = length - 2
7    while i >= 0 and digits[i] >= digits[i + 1]:
8        i -= 1
9    if i == -1:
10        return -1
11    j = length - 1
12    while digits[j] <= digits[i]:
13        j -= 1
14    digits[i], digits[j] = digits[j], digits[i]
15    digits[i + 1:] = reversed(digits[i + 1:])
16    result = int(''.join(digits))
17    return result if result < 2**31 else -1

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the digits to find the swap positions and then reverse the tail. The space complexity is O(1) since we only use a few variables for indexing.

1Understanding how to manipulate digits is crucial.
2Recognizing the need for efficient searching and swapping.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.