LeetCode #556 — MEDIUM

Next Greater Element III

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1.

Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.

Example 1:

Input: n = 12
Output: 21

Example 2:

Input: n = 21
Output: -1

Constraints:

1 <= n <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1. Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Two Pointers

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #556: Next Greater Element III
class Solution {
    public int nextGreaterElement(int n) {
        char[] cs = String.valueOf(n).toCharArray();
        n = cs.length;
        int i = n - 2, j = n - 1;
        for (; i >= 0 && cs[i] >= cs[i + 1]; --i)
            ;
        if (i < 0) {
            return -1;
        }
        for (; cs[i] >= cs[j]; --j)
            ;
        swap(cs, i, j);
        reverse(cs, i + 1, n - 1);
        long ans = Long.parseLong(String.valueOf(cs));
        return ans > Integer.MAX_VALUE ? -1 : (int) ans;
    }

    private void swap(char[] cs, int i, int j) {
        char t = cs[i];
        cs[i] = cs[j];
        cs[j] = t;
    }

    private void reverse(char[] cs, int i, int j) {
        for (; i < j; ++i, --j) {
            swap(cs, i, j);
        }
    }
}

// Accepted solution for LeetCode #556: Next Greater Element III
func nextGreaterElement(n int) int {
	s := []byte(strconv.Itoa(n))
	n = len(s)
	i, j := n-2, n-1
	for ; i >= 0 && s[i] >= s[i+1]; i-- {
	}
	if i < 0 {
		return -1
	}
	for ; j >= 0 && s[i] >= s[j]; j-- {
	}
	s[i], s[j] = s[j], s[i]
	for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
		s[i], s[j] = s[j], s[i]
	}
	ans, _ := strconv.Atoi(string(s))
	if ans > math.MaxInt32 {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #556: Next Greater Element III
class Solution:
    def nextGreaterElement(self, n: int) -> int:
        cs = list(str(n))
        n = len(cs)
        i, j = n - 2, n - 1
        while i >= 0 and cs[i] >= cs[i + 1]:
            i -= 1
        if i < 0:
            return -1
        while cs[i] >= cs[j]:
            j -= 1
        cs[i], cs[j] = cs[j], cs[i]
        cs[i + 1 :] = cs[i + 1 :][::-1]
        ans = int(''.join(cs))
        return -1 if ans > 2**31 - 1 else ans

// Accepted solution for LeetCode #556: Next Greater Element III
struct Solution;

impl Solution {
    fn next_greater_element(n: i32) -> i32 {
        let mut s: Vec<char> = format!("{}", n).chars().collect();
        let n = s.len();
        let mut l = n;
        for i in (0..n - 1).rev() {
            if s[i] < s[i + 1] {
                l = i;
                break;
            }
        }
        if l == n {
            return -1;
        }
        let mut max = s[l + 1];
        let mut r = l + 1;
        for i in l + 2..n {
            if s[i] > s[l] && s[i] < max {
                max = s[i];
                r = i;
            }
        }
        s.swap(l, r);
        s[l + 1..].sort_unstable();
        s.iter().collect::<String>().parse::<i32>().unwrap_or(-1)
    }
}

#[test]
fn test() {
    let n = 12;
    let res = 21;
    assert_eq!(Solution::next_greater_element(n), res);
    let n = 21;
    let res = -1;
    assert_eq!(Solution::next_greater_element(n), res);
    let n = 1_999_999_999;
    let res = -1;
    assert_eq!(Solution::next_greater_element(n), res);
}

// Accepted solution for LeetCode #556: Next Greater Element III
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #556: Next Greater Element III
// class Solution {
//     public int nextGreaterElement(int n) {
//         char[] cs = String.valueOf(n).toCharArray();
//         n = cs.length;
//         int i = n - 2, j = n - 1;
//         for (; i >= 0 && cs[i] >= cs[i + 1]; --i)
//             ;
//         if (i < 0) {
//             return -1;
//         }
//         for (; cs[i] >= cs[j]; --j)
//             ;
//         swap(cs, i, j);
//         reverse(cs, i + 1, n - 1);
//         long ans = Long.parseLong(String.valueOf(cs));
//         return ans > Integer.MAX_VALUE ? -1 : (int) ans;
//     }
// 
//     private void swap(char[] cs, int i, int j) {
//         char t = cs[i];
//         cs[i] = cs[j];
//         cs[j] = t;
//     }
// 
//     private void reverse(char[] cs, int i, int j) {
//         for (; i < j; ++i, --j) {
//             swap(cs, i, j);
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.