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1881. Maximum Value after Insertion

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 <= 105
  • 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 '-'.

Solutions

  • class Solution { public String maxValue(String n, int x) { int i = 0; if (n.charAt(0) != '-') { for (; i < n.length() && n.charAt(i) - '0' >= x; ++i) ; } else { for (i = 1; i < n.length() && n.charAt(i) - '0' <= x; ++i) ; } return n.substring(0, i) + x + n.substring(i); } } 
  • class Solution { public: string maxValue(string n, int x) { int i = 0; if (n[0] != '-') for (; i < n.size() && n[i] - '0' >= x; ++i) ; else for (i = 1; i < n.size() && n[i] - '0' <= x; ++i) ; return n.substr(0, i) + to_string(x) + n.substr(i); } }; 
  • class Solution: def maxValue(self, n: str, x: int) -> str: if n[0] != '-': for i, c in enumerate(n): if int(c) < x: return n[:i] + str(x) + n[i:] return n + str(x) else: for i, c in enumerate(n[1:]): if int(c) > x: return n[: i + 1] + str(x) + n[i + 1 :] return n + str(x) 
  • func maxValue(n string, x int) string { i := 0 y := byte('0' + x) if n[0] != '-' { for ; i < len(n) && n[i] >= y; i++ { } } else { for i = 1; i < len(n) && n[i] <= y; i++ { } } return n[:i] + string(y) + n[i:] } 
  • /** * @param {string} n * @param {number} x * @return {string} */ var maxValue = function (n, x) { let nums = [...n]; let sign = 1, i = 0; if (nums[0] == '-') { sign = -1; i++; } while (i < n.length && (nums[i] - x) * sign >= 0) { i++; } nums.splice(i, 0, x); return nums.join(''); }; 
  • function maxValue(n: string, x: number): string { let i = 0; if (n[0] === '-') { i++; while (i < n.length && +n[i] <= x) { i++; } } else { while (i < n.length && +n[i] >= x) { i++; } } return n.slice(0, i) + x + n.slice(i); } 
  • impl Solution { pub fn max_value(n: String, x: i32) -> String { let s = n.as_bytes(); let mut i = 0; if n.starts_with('-') { i += 1; while i < s.len() && (s[i] - b'0') as i32 <= x { i += 1; } } else { while i < s.len() && (s[i] - b'0') as i32 >= x { i += 1; } } let mut ans = String::new(); ans.push_str(&n[0..i]); ans.push_str(&x.to_string()); ans.push_str(&n[i..]); ans } }

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