README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1946.Largest%20Number%20After%20Mutating%20Substring/README_EN.md	1445	Weekly Contest 251 Q2	Greedy Array String

1946. Largest Number After Mutating Substring

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Description

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

Solutions

Solution 1: Greedy

According to the problem description, we can start from the highest digit of the string and greedily perform continuous replacement operations until we encounter a digit smaller than the current digit.

First, we convert the string $\textit{num}$ into a character array $\textit{s}$ and use a variable $\textit{changed}$ to record whether a change has already occurred, initially $\textit{changed} = \text{false}$.

Then we traverse the character array $\textit{s}$. For each character $\textit{c}$, we convert it to a number $\textit{d} = \text{change}[\text{int}(\textit{c})]$. If a change has already occurred and $\textit{d} < \textit{c}$, it means we cannot continue changing, so we exit the loop immediately. Otherwise, if $\textit{d} > \textit{c}$, it means we can replace $\textit{c}$ with $\textit{d}$. At this point, we set $\textit{changed} = \text{true}$ and replace $\textit{s}[i]$ with $\textit{d}$.

Finally, we convert the character array $\textit{s}$ back to a string and return it.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{num}$.

Python3

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)

Java

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);
    }
}

C++

class Solution {
public:
    string maximumNumber(string num, vector<int>& change) {
        int n = num.size();
        bool changed = false;
        for (int i = 0; i < n; ++i) {
            char d = '0' + change[num[i] - '0'];
            if (changed && d < num[i]) {
                break;
            }
            if (d > num[i]) {
                changed = true;
                num[i] = d;
            }
        }
        return num;
    }
};

Go

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)
}

TypeScript

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('');
}

Rust

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()
    }
}

JavaScript

/**
 * @param {string} num
 * @param {number[]} change
 * @return {string}
 */
var maximumNumber = function (num, change) {
    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('');
};