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);
}
}