Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1946.html
1946. Largest Number After Mutating Substring
Level
Medium
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 any 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) any 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 <= 10^5
num
consists of only digits0-9
.change.length == 10
0 <= change[d] <= 9
Solution
Use a greedy approach. Find the leftmost digit in num
such that mutating the digit will increase num
, and find the rightmost digit in num
such that mutating the digit will decrease num
. If such a leftmost digit does not exist, return num
without mutating. If such a rightmost digit does not exist, then mutate all digits from the determined leftmost digit to the end of num
. Finally, return the mutated num
.
-
class Solution { public String maximumNumber(String num, int[] change) { char[] array = num.toCharArray(); int length = array.length; int start = -1, end = length; for (int i = 0; i < length; i++) { int digit = array[i] - '0'; if (digit < change[digit]) { start = i; break; } } if (start < 0) return num; for (int i = start + 1; i < length; i++) { int digit = array[i] - '0'; if (digit > change[digit]) { end = i; break; } } for (int i = start; i < end; i++) { int digit = array[i] - '0'; char changed = (char) ('0' + change[digit]); array[i] = changed; } return new String(array); } } ############ class Solution { public String maximumNumber(String num, int[] change) { char[] s = num.toCharArray(); for (int i = 0; i < s.length; ++i) { if (change[s[i] - '0'] > s[i] - '0') { for (; i < s.length && s[i] - '0' <= change[s[i] - '0']; ++i) { s[i] = (char) (change[s[i] - '0'] + '0'); } break; } } return String.valueOf(s); } }
-
// OJ: https://leetcode.com/problems/largest-number-after-mutating-substring/ // Time: O(N) // Space: O(1) class Solution { public: string maximumNumber(string num, vector<int>& A) { bool changed = false; for (char &c : num) { int before = c - '0', after = A[before]; if (after < before) { if (changed) break; continue; } else if (after == before) { continue; } changed = true; c = '0' + after; } return num; } };
-
class Solution: def maximumNumber(self, num: str, change: List[int]) -> str: s = list(num) for i, c in enumerate(s): if change[int(c)] > int(c): while i < len(s) and int(s[i]) <= change[int(s[i])]: s[i] = str(change[int(s[i])]) i += 1 break return ''.join(s) ############ # 1946. Largest Number After Mutating Substring # https://leetcode.com/problems/largest-number-after-mutating-substring/ class Solution: def maximumNumber(self, nums: str, change: List[int]) -> str: nums = list(nums) done = False for i, num in enumerate(nums): pos = int(num) if change[pos] >= pos: nums[i] = str(change[pos]) if change[pos] > pos: done = True else: if done: break return "".join(nums)
-
func maximumNumber(num string, change []int) string { s := []byte(num) for i, c := range num { if change[c-'0'] > int(c-'0') { for ; i < len(s) && change[s[i]-'0'] >= int(s[i]-'0'); i++ { s[i] = byte(change[s[i]-'0']) + '0' } break } } return string(s) }