LeetCode 1946. Largest Number After Mutating Substring Solution in Java, C++, Python & More | Explanation + Code

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 num into a character array s and use a variable changed to record whether a change has already occurred, initially changed = false.

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

Finally, we convert the character array 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 num.

PythonJavaC++GoTypeScriptRustJavaScript

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)(code-box)

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); } }(code-box)

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; } };(code-box)

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) }(code-box)

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(''); }(code-box)

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() } }(code-box)

/** * @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(''); };(code-box)