LeetCode #1881 — MEDIUM

Maximum Value after Insertion

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

The Problem

Problem Statement

You are given a very large integer n, represented as a string, and an integer digit x. The digits in n and the digit x are in the inclusive range [1, 9], and n may represent a negative number.

You want to maximize n's numerical value by inserting x anywhere in the decimal representation of n. You cannot insert x to the left of the negative sign.

For example, if n = 73 and x = 6, it would be best to insert it between 7 and 3, making n = 763.
If n = -55 and x = 2, it would be best to insert it before the first 5, making n = -255.

Return a string representing the maximum value of n after the insertion.

Example 1:

Input: n = "99", x = 9
Output: "999"
Explanation: The result is the same regardless of where you insert 9.

Example 2:

Input: n = "-13", x = 2
Output: "-123"
Explanation: You can make n one of {-213, -123, -132}, and the largest of those three is -123.

Constraints:

1 <= n.length <= 10⁵
1 <= x <= 9
The digits in n are in the range [1, 9].
n is a valid representation of an integer.
In the case of a negative n, it will begin with '-'.

Step 01

Brute Force Baseline

Problem summary: You are given a very large integer n, represented as a string, and an integer digit x. The digits in n and the digit x are in the inclusive range [1, 9], and n may represent a negative number. You want to maximize n's numerical value by inserting x anywhere in the decimal representation of n. You cannot insert x to the left of the negative sign. For example, if n = 73 and x = 6, it would be best to insert it between 7 and 3, making n = 763. If n = -55 and x = 2, it would be best to insert it before the first 5, making n = -255. Return a string representing the maximum value of n after the insertion.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"99"
9

Example 2

"-13"
2

Step 02

Core Insight

What unlocks the optimal approach

Note that if the number is negative it's the same as positive but you look for the minimum instead.
In the case of maximum, if s[i] < x it's optimal that x is put before s[i].
In the case of minimum, if s[i] > x it's optimal that x is put before s[i].

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 #1881: Maximum Value after Insertion
class Solution {
    public String maxValue(String n, int x) {
        int i = 0;
        if (n.charAt(0) == '-') {
            ++i;
            while (i < n.length() && n.charAt(i) - '0' <= x) {
                ++i;
            }
        } else {
            while (i < n.length() && n.charAt(i) - '0' >= x) {
                ++i;
            }
        }
        return n.substring(0, i) + x + n.substring(i);
    }
}

// Accepted solution for LeetCode #1881: Maximum Value after Insertion
func maxValue(n string, x int) string {
	i := 0
	y := byte('0' + x)
	if n[0] == '-' {
		i++
		for i < len(n) && n[i] <= y {
			i++
		}
	} else {
		for i < len(n) && n[i] >= y {
			i++
		}
	}
	return n[:i] + string(y) + n[i:]
}

# Accepted solution for LeetCode #1881: Maximum Value after Insertion
class Solution:
    def maxValue(self, n: str, x: int) -> str:
        i = 0
        if n[0] == "-":
            i += 1
            while i < len(n) and int(n[i]) <= x:
                i += 1
        else:
            while i < len(n) and int(n[i]) >= x:
                i += 1
        return n[:i] + str(x) + n[i:]

// Accepted solution for LeetCode #1881: Maximum Value after Insertion
impl Solution {
    pub fn max_value(n: String, x: i32) -> String {
        let s = n.as_bytes();
        let mut i = 0;
        if n.starts_with('-') {
            i += 1;
            while i < s.len() && (s[i] - b'0') as i32 <= x {
                i += 1;
            }
        } else {
            while i < s.len() && (s[i] - b'0') as i32 >= x {
                i += 1;
            }
        }
        let mut ans = String::new();
        ans.push_str(&n[0..i]);
        ans.push_str(&x.to_string());
        ans.push_str(&n[i..]);
        ans
    }
}

// Accepted solution for LeetCode #1881: Maximum Value after Insertion
function maxValue(n: string, x: number): string {
    let i = 0;
    if (n[0] === '-') {
        i++;
        while (i < n.length && +n[i] <= x) {
            i++;
        }
    } else {
        while (i < n.length && +n[i] >= x) {
            i++;
        }
    }
    return n.slice(0, i) + x + n.slice(i);
}

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 log n)

Space

O(1)

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.