LeetCode #670 — MEDIUM

Maximum Swap

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

The Problem

Problem Statement

You are given an integer num. You can swap two digits at most once to get the maximum valued number.

Return the maximum valued number you can get.

Example 1:

Input: num = 2736
Output: 7236
Explanation: Swap the number 2 and the number 7.

Example 2:

Input: num = 9973
Output: 9973
Explanation: No swap.

Constraints:

0 <= num <= 10⁸

Step 01

Brute Force Baseline

Problem summary: You are given an integer num. You can swap two digits at most once to get the maximum valued number. Return the maximum valued number you can get.

Baseline thinking

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

Pattern signal: Math · Greedy

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 #670: Maximum Swap
class Solution {
    public int maximumSwap(int num) {
        char[] s = String.valueOf(num).toCharArray();
        int n = s.length;
        int[] d = new int[n];
        for (int i = 0; i < n; ++i) {
            d[i] = i;
        }
        for (int i = n - 2; i >= 0; --i) {
            if (s[i] <= s[d[i + 1]]) {
                d[i] = d[i + 1];
            }
        }
        for (int i = 0; i < n; ++i) {
            int j = d[i];
            if (s[i] < s[j]) {
                char t = s[i];
                s[i] = s[j];
                s[j] = t;
                break;
            }
        }
        return Integer.parseInt(String.valueOf(s));
    }
}

// Accepted solution for LeetCode #670: Maximum Swap
func maximumSwap(num int) int {
	s := []byte(strconv.Itoa(num))
	n := len(s)
	d := make([]int, n)
	for i := range d {
		d[i] = i
	}
	for i := n - 2; i >= 0; i-- {
		if s[i] <= s[d[i+1]] {
			d[i] = d[i+1]
		}
	}
	for i, j := range d {
		if s[i] < s[j] {
			s[i], s[j] = s[j], s[i]
			break
		}
	}
	ans, _ := strconv.Atoi(string(s))
	return ans
}

# Accepted solution for LeetCode #670: Maximum Swap
class Solution:
    def maximumSwap(self, num: int) -> int:
        s = list(str(num))
        n = len(s)
        d = list(range(n))
        for i in range(n - 2, -1, -1):
            if s[i] <= s[d[i + 1]]:
                d[i] = d[i + 1]
        for i, j in enumerate(d):
            if s[i] < s[j]:
                s[i], s[j] = s[j], s[i]
                break
        return int(''.join(s))

// Accepted solution for LeetCode #670: Maximum Swap
impl Solution {
    pub fn maximum_swap(mut num: i32) -> i32 {
        let mut list = {
            let mut res = Vec::new();
            while num != 0 {
                res.push(num % 10);
                num /= 10;
            }
            res
        };
        let n = list.len();
        let idx = {
            let mut i = 0;
            (0..n)
                .map(|j| {
                    if list[j] > list[i] {
                        i = j;
                    }
                    i
                })
                .collect::<Vec<usize>>()
        };
        for i in (0..n).rev() {
            if list[i] != list[idx[i]] {
                list.swap(i, idx[i]);
                break;
            }
        }
        let mut res = 0;
        for i in list.iter().rev() {
            res = res * 10 + i;
        }
        res
    }
}

// Accepted solution for LeetCode #670: Maximum Swap
function maximumSwap(num: number): number {
    const list = new Array();
    while (num !== 0) {
        list.push(num % 10);
        num = Math.floor(num / 10);
    }
    const n = list.length;
    const idx = new Array();
    for (let i = 0, j = 0; i < n; i++) {
        if (list[i] > list[j]) {
            j = i;
        }
        idx.push(j);
    }
    for (let i = n - 1; i >= 0; i--) {
        if (list[idx[i]] !== list[i]) {
            [list[idx[i]], list[i]] = [list[i], list[idx[i]]];
            break;
        }
    }
    let res = 0;
    for (let i = n - 1; i >= 0; i--) {
        res = res * 10 + list[i];
    }
    return res;
}

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(log M)

Space

O(log M)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.