LeetCode #1946 — MEDIUM

Largest Number After Mutating Substring

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

The Problem

Problem Statement

You are given a string num, which represents a large integer. You are also given a 0-indexed integer array change of length 10 that maps each digit 0-9 to another digit. More formally, digit d maps to digit change[d].

You may choose to mutate a single substring of num. To mutate a substring, replace each digit num[i] with the digit it maps to in change (i.e. replace num[i] with change[num[i]]).

Return a string representing the largest possible integer after mutating (or choosing not to) a single substring of num.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: num = "132", change = [9,8,5,0,3,6,4,2,6,8]
Output: "832"
Explanation: Replace the substring "1":
- 1 maps to change[1] = 8.
Thus, "132" becomes "832".
"832" is the largest number that can be created, so return it.

Example 2:

Input: num = "021", change = [9,4,3,5,7,2,1,9,0,6]
Output: "934"
Explanation: Replace the substring "021":
- 0 maps to change[0] = 9.
- 2 maps to change[2] = 3.
- 1 maps to change[1] = 4.
Thus, "021" becomes "934".
"934" is the largest number that can be created, so return it.

Example 3:

Input: num = "5", change = [1,4,7,5,3,2,5,6,9,4]
Output: "5"
Explanation: "5" is already the largest number that can be created, so return it.

Constraints:

1 <= num.length <= 10⁵
num consists of only digits 0-9.
change.length == 10
0 <= change[d] <= 9

Step 01

Brute Force Baseline

Problem summary: You are given a string num, which represents a large integer. You are also given a 0-indexed integer array change of length 10 that maps each digit 0-9 to another digit. More formally, digit d maps to digit change[d]. You may choose to mutate a single substring of num. To mutate a substring, replace each digit num[i] with the digit it maps to in change (i.e. replace num[i] with change[num[i]]). Return a string representing the largest possible integer after mutating (or choosing not to) a single substring of num. A substring is a contiguous sequence of characters within the string.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

"132"
[9,8,5,0,3,6,4,2,6,8]

Example 2

"021"
[9,4,3,5,7,2,1,9,0,6]

Example 3

"5"
[1,4,7,5,3,2,5,6,9,4]

Step 02

Core Insight

What unlocks the optimal approach

Should you change a digit if the new digit is smaller than the original?
If changing the first digit and the last digit both make the number bigger, but you can only change one of them; which one should you change?
Changing numbers closer to the front is always better

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 #1946: Largest Number After Mutating Substring
class Solution {
    public String maximumNumber(String num, int[] change) {
        char[] s = num.toCharArray();
        boolean changed = false;
        for (int i = 0; i < s.length; ++i) {
            char d = (char) (change[s[i] - '0'] + '0');
            if (changed && d < s[i]) {
                break;
            }
            if (d > s[i]) {
                changed = true;
                s[i] = d;
            }
        }
        return new String(s);
    }
}

// Accepted solution for LeetCode #1946: Largest Number After Mutating Substring
func maximumNumber(num string, change []int) string {
	s := []byte(num)
	changed := false
	for i, c := range num {
		d := byte('0' + change[c-'0'])
		if changed && d < s[i] {
			break
		}
		if d > s[i] {
			s[i] = d
			changed = true
		}
	}
	return string(s)
}

# Accepted solution for LeetCode #1946: Largest Number After Mutating Substring
class Solution:
    def maximumNumber(self, num: str, change: List[int]) -> str:
        s = list(num)
        changed = False
        for i, c in enumerate(s):
            d = str(change[int(c)])
            if changed and d < c:
                break
            if d > c:
                changed = True
                s[i] = d
        return "".join(s)

// Accepted solution for LeetCode #1946: Largest Number After Mutating Substring
impl Solution {
    pub fn maximum_number(num: String, change: Vec<i32>) -> String {
        let mut s: Vec<char> = num.chars().collect();
        let mut changed = false;
        for i in 0..s.len() {
            let d = (change[s[i] as usize - '0' as usize] + '0' as i32) as u8 as char;
            if changed && d < s[i] {
                break;
            }
            if d > s[i] {
                changed = true;
                s[i] = d;
            }
        }
        s.into_iter().collect()
    }
}

// Accepted solution for LeetCode #1946: Largest Number After Mutating Substring
function maximumNumber(num: string, change: number[]): string {
    const s = num.split('');
    let changed = false;
    for (let i = 0; i < s.length; ++i) {
        const d = change[+s[i]].toString();
        if (changed && d < s[i]) {
            break;
        }
        if (d > s[i]) {
            s[i] = d;
            changed = true;
        }
    }
    return s.join('');
}

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(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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.