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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 <= 105
  • num consists of only digits 0-9.
  • change.length == 10
  • 0 <= change[d] <= 9

Solution 1. Greedy

Intuition: Try to change the digit from left to right. We only greedily make the change if we can make it greater or the same.

Algorithm:

One caveat is that, we should skip the leading digits if making changes won't result in greater digit. So we can keep track of a boolean changed indicating whether we have made changes.

If we haven't made any changes:

  • If the change results in a greater digit, we do it.
  • Otherwise, we just skip the digit.

If we have made changes:

  • If the change results in a greater or equal digit, we do it.
  • Otherwise, we should stop the loop.
// OJ: https://leetcode.com/problems/largest-number-after-mutating-substring/
// Author: github.com/lzl124631x
// 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;
    }
};

Or

// OJ: https://leetcode.com/problems/largest-number-after-mutating-substring/
// Author: github.com/lzl124631x
// 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 (changed) {
                if (after >= before) c = '0' + after;
                else break;
            } else {
                if (after > before) {
                    c = '0' + after;
                    changed = true;
                } 
            }
        }
        return num;
    }
};