Formatted question description: https://leetcode.ca/all/1881.html

1881. Maximum Value after Insertion

Level

Medium

Description

You are given a very large integer n, represented as a string, and an integer digit x. The digits in n and the digit x are in the inclusive range [1, 9], and n may represent a negative number.

You want to** maximize n’s numerical value** by inserting x anywhere in the decimal representation of n. You cannot insert x to the left of the negative sign.

  • For example, if n = 73 and x = 6, it would be best to insert it between 7 and 3, making n = 763.
  • If n = -55 and x = 2, it would be best to insert it before the first 5, making n = -255.

Return a string representing the maximum value of n after the insertion.

Example 1:

Input: n = “99”, x = 9

Output: “999”

Explanation: The result is the same regardless of where you insert 9.

Example 2:

Input: n = “-13”, x = 2

Output: “-123”

Explanation: You can make n one of {-213, -123, -132}, and the largest of those three is -123.

Constraints:

  • 1 <= n.length <= 10^5
  • 1 <= x <= 9
  • The digits in n are in the range [1, 9].
  • n is a valid representation of an integer.
  • In the case of a negative n, it will begin with '-'.

Solution

If n is positive, find the leftmost digit in n that is less than x and insert x to the left of the digit. If no such digit is found, insert x at the end of n.

If n is negative, find the leftmost digit in n that is greater than x and insert x to the left of the digit. If no such digit is found, insert x and the end of n.

  • class Solution {
        public String maxValue(String n, int x) {
            char xChar = (char) (x + '0');
            if (n.charAt(0) != '-')
                return maxValuePositive(n, xChar);
            else
                return maxValueNegative(n, xChar);
        }
    
        public String maxValuePositive(String n, char x) {
            StringBuffer sb = new StringBuffer(n);
            int length = n.length();
            int insertIndex = length;
            for (int i = 0; i < length; i++) {
                if (n.charAt(i) < x) {
                    insertIndex = i;
                    break;
                }
            }
            sb.insert(insertIndex, x);
            return sb.toString();
        }
    
        public String maxValueNegative(String n, char x) {
            StringBuffer sb = new StringBuffer(n);
            int length = n.length();
            int insertIndex = length;
            for (int i = 1; i < length; i++) {
                if (n.charAt(i) > x) {
                    insertIndex = i;
                    break;
                }
            }
            sb.insert(insertIndex, x);
            return sb.toString();
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-value-after-insertion/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        string maxValue(string n, int x) {
            x += '0';
            int i = n[0] == '-';
            if (n[0] == '-') {
                for (; i < n.size() && n[i] <= x; ++i) ;
            } else {
                for (; i < n.size() && n[i] >= x; ++i) ;
            }
            n.insert(begin(n) + i, x);
            return n;
        }
    };
    
  • # 1881. Maximum Value after Insertion
    # https://leetcode.com/problems/maximum-value-after-insertion/
    
    class Solution:
        def maxValue(self, word: str, x: int) -> str:
            
            if word[0] == '-':
                pos = 0
                for i, c in enumerate(word):
                    if c == '-': continue
                        
                    if x < int(c):
                        return word[:i] + str(x) + word[i:]
                
                return word + str(x)
            else:
                pos = 0
                for i, c in enumerate(word):
                    
                    if x > int(c):
                         return word[:i] + str(x) + word[i:]
                
                return word + str(x)
                
            
    
    

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