Largest Number After Mutating Substring

Problem

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 <= 10^5
num consists of only digits 0-9.
change.length == 10
0 <= change[d] <= 9

Solution (Java)

class Solution {
    public String maximumNumber(String num, int[] change) {
        int n = num.length();
        char[] nums = num.toCharArray();
        char[] arr = new char[n];
        for (int i = 0; i < n; i++) {
            int val = nums[i] - '0';
            arr[i] = (char) (change[val] + '0');
        }
        boolean flag = false;
        for (int i = 0; i < n; i++) {
            if (nums[i] < arr[i]) {
                nums[i] = arr[i];
                flag = true;
            } else if (flag && nums[i] > arr[i]) {
                break;
            }
        }
        return String.valueOf(nums);
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).