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

1881. Maximum Value after Insertion

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)