LeetCode #2259 — EASY

Remove Digit From Number to Maximize Result

Build confidence with an intuition-first walkthrough focused on greedy fundamentals.

The Problem

Problem Statement

You are given a string number representing a positive integer and a character digit.

Return the resulting string after removing exactly one occurrence of digit from number such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit occurs at least once in number.

Example 1:

Input: number = "123", digit = "3"
Output: "12"
Explanation: There is only one '3' in "123". After removing '3', the result is "12".

Example 2:

Input: number = "1231", digit = "1"
Output: "231"
Explanation: We can remove the first '1' to get "231" or remove the second '1' to get "123".
Since 231 > 123, we return "231".

Example 3:

Input: number = "551", digit = "5"
Output: "51"
Explanation: We can remove either the first or second '5' from "551".
Both result in the string "51".

Constraints:

2 <= number.length <= 100
number consists of digits from '1' to '9'.
digit is a digit from '1' to '9'.
digit occurs at least once in number.

Step 01

Brute Force Baseline

Problem summary: You are given a string number representing a positive integer and a character digit. Return the resulting string after removing exactly one occurrence of digit from number such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit occurs at least once in number.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"123"
"3"

Example 2

"1231"
"1"

Example 3

"551"
"5"

Core Insight

What unlocks the optimal approach

The maximum length of number is really small.
Iterate through the digits of number and every time we see digit, try removing it.
To remove a character at index i, concatenate the substring from index 0 to i - 1 and the substring from index i + 1 to number.length - 1.

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 #2259: Remove Digit From Number to Maximize Result
class Solution {
    public String removeDigit(String number, char digit) {
        String ans = "0";
        for (int i = 0, n = number.length(); i < n; ++i) {
            char d = number.charAt(i);
            if (d == digit) {
                String t = number.substring(0, i) + number.substring(i + 1);
                if (ans.compareTo(t) < 0) {
                    ans = t;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2259: Remove Digit From Number to Maximize Result
func removeDigit(number string, digit byte) string {
	ans := "0"
	for i, d := range number {
		if d == rune(digit) {
			t := number[:i] + number[i+1:]
			if strings.Compare(ans, t) < 0 {
				ans = t
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #2259: Remove Digit From Number to Maximize Result
class Solution:
    def removeDigit(self, number: str, digit: str) -> str:
        return max(
            number[:i] + number[i + 1 :] for i, d in enumerate(number) if d == digit
        )

// Accepted solution for LeetCode #2259: Remove Digit From Number to Maximize Result
/**
 * [2259] Remove Digit From Number to Maximize Result
 *
 * You are given a string number representing a positive integer and a character digit.
 * Return the resulting string after removing exactly one occurrence of digit from number such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit occurs at least once in number.
 *  
 * Example 1:
 *
 * Input: number = "123", digit = "3"
 * Output: "12"
 * Explanation: There is only one '3' in "123". After removing '3', the result is "12".
 *
 * Example 2:
 *
 * Input: number = "1231", digit = "1"
 * Output: "231"
 * Explanation: We can remove the first '1' to get "231" or remove the second '1' to get "123".
 * Since 231 > 123, we return "231".
 *
 * Example 3:
 *
 * Input: number = "551", digit = "5"
 * Output: "51"
 * Explanation: We can remove either the first or second '5' from "551".
 * Both result in the string "51".
 *
 *  
 * Constraints:
 *
 * 	2 <= number.length <= 100
 * 	number consists of digits from '1' to '9'.
 * 	digit is a digit from '1' to '9'.
 * 	digit occurs at least once in number.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/remove-digit-from-number-to-maximize-result/
// discuss: https://leetcode.com/problems/remove-digit-from-number-to-maximize-result/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn remove_digit(number: String, digit: char) -> String {
        let mut result = String::new();

        for (i, v) in number.chars().enumerate() {
            if v == digit {
                let mod_number = format!("{}{}", &number[..i], &number[i + 1..]);
                if mod_number > result {
                    result = mod_number;
                }
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2259_example_1() {
        let number = "123".to_string();
        let digit = '3';

        let result = "12".to_string();

        assert_eq!(Solution::remove_digit(number, digit), result);
    }

    #[test]
    fn test_2259_example_2() {
        let number = "1231".to_string();
        let digit = '1';

        let result = "231".to_string();

        assert_eq!(Solution::remove_digit(number, digit), result);
    }

    #[test]
    fn test_2259_example_3() {
        let number = "551".to_string();
        let digit = '5';

        let result = "51".to_string();

        assert_eq!(Solution::remove_digit(number, digit), result);
    }
}

// Accepted solution for LeetCode #2259: Remove Digit From Number to Maximize Result
function removeDigit(number: string, digit: string): string {
    const n = number.length;
    let last = -1;
    for (let i = 0; i < n; ++i) {
        if (number[i] === digit) {
            last = i;
            if (i + 1 < n && number[i] < number[i + 1]) {
                break;
            }
        }
    }
    return number.substring(0, last) + number.substring(last + 1);
}

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^2)

Space

O(n)

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.

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.