LeetCode #1432 — MEDIUM

Max Difference You Can Get From Changing an Integer

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer num. You will apply the following steps to num two separate times:

Pick a digit x (0 <= x <= 9).
Pick another digit y (0 <= y <= 9). Note y can be equal to x.
Replace all the occurrences of x in the decimal representation of num by y.

Let a and b be the two results from applying the operation to num independently.

Return the max difference between a and b.

Note that neither a nor b may have any leading zeros, and must not be 0.

Example 1:

Input: num = 555
Output: 888
Explanation: The first time pick x = 5 and y = 9 and store the new integer in a.
The second time pick x = 5 and y = 1 and store the new integer in b.
We have now a = 999 and b = 111 and max difference = 888

Example 2:

Input: num = 9
Output: 8
Explanation: The first time pick x = 9 and y = 9 and store the new integer in a.
The second time pick x = 9 and y = 1 and store the new integer in b.
We have now a = 9 and b = 1 and max difference = 8

Constraints:

1 <= num <= 10⁸

Step 01

Brute Force Baseline

Problem summary: You are given an integer num. You will apply the following steps to num two separate times: Pick a digit x (0 <= x <= 9). Pick another digit y (0 <= y <= 9). Note y can be equal to x. Replace all the occurrences of x in the decimal representation of num by y. Let a and b be the two results from applying the operation to num independently. Return the max difference between a and b. Note that neither a nor b may have any leading zeros, and must not be 0.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

We need to get the max and min value after changing num and the answer is max - min.
Use brute force, try all possible changes and keep the minimum and maximum values.

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 #1432: Max Difference You Can Get From Changing an Integer
class Solution {
    public int maxDiff(int num) {
        String a = String.valueOf(num);
        String b = a;
        for (int i = 0; i < a.length(); ++i) {
            if (a.charAt(i) != '9') {
                a = a.replace(a.charAt(i), '9');
                break;
            }
        }
        if (b.charAt(0) != '1') {
            b = b.replace(b.charAt(0), '1');
        } else {
            for (int i = 1; i < b.length(); ++i) {
                if (b.charAt(i) != '0' && b.charAt(i) != '1') {
                    b = b.replace(b.charAt(i), '0');
                    break;
                }
            }
        }
        return Integer.parseInt(a) - Integer.parseInt(b);
    }
}

// Accepted solution for LeetCode #1432: Max Difference You Can Get From Changing an Integer
func maxDiff(num int) int {
	a, b := num, num
	s := strconv.Itoa(num)
	for i := range s {
		if s[i] != '9' {
			a, _ = strconv.Atoi(strings.ReplaceAll(s, string(s[i]), "9"))
			break
		}
	}
	if s[0] > '1' {
		b, _ = strconv.Atoi(strings.ReplaceAll(s, string(s[0]), "1"))
	} else {
		for i := 1; i < len(s); i++ {
			if s[i] != '0' && s[i] != '1' {
				b, _ = strconv.Atoi(strings.ReplaceAll(s, string(s[i]), "0"))
				break
			}
		}
	}
	return a - b
}

# Accepted solution for LeetCode #1432: Max Difference You Can Get From Changing an Integer
class Solution:
    def maxDiff(self, num: int) -> int:
        a, b = str(num), str(num)
        for c in a:
            if c != "9":
                a = a.replace(c, "9")
                break
        if b[0] != "1":
            b = b.replace(b[0], "1")
        else:
            for c in b[1:]:
                if c not in "01":
                    b = b.replace(c, "0")
                    break
        return int(a) - int(b)

// Accepted solution for LeetCode #1432: Max Difference You Can Get From Changing an Integer
impl Solution {
    pub fn max_diff(num: i32) -> i32 {
        let a = num.to_string();
        let mut a = a.clone();
        let mut b = a.clone();

        for c in a.chars() {
            if c != '9' {
                a = a.replace(c, "9");
                break;
            }
        }

        let chars: Vec<char> = b.chars().collect();
        if chars[0] != '1' {
            b = b.replace(chars[0], "1");
        } else {
            for &c in &chars[1..] {
                if c != '0' && c != '1' {
                    b = b.replace(c, "0");
                    break;
                }
            }
        }

        a.parse::<i32>().unwrap() - b.parse::<i32>().unwrap()
    }
}

// Accepted solution for LeetCode #1432: Max Difference You Can Get From Changing an Integer
function maxDiff(num: number): number {
    let a = num.toString();
    let b = a;
    for (let i = 0; i < a.length; ++i) {
        if (a[i] !== '9') {
            a = a.split(a[i]).join('9');
            break;
        }
    }
    if (b[0] !== '1') {
        b = b.split(b[0]).join('1');
    } else {
        for (let i = 1; i < b.length; ++i) {
            if (b[i] !== '0' && b[i] !== '1') {
                b = b.split(b[i]).join('0');
                break;
            }
        }
    }
    return +a - +b;
}

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 \textitnum)

Space

O(log \textitnum)

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.