LeetCode 1881. Maximum Value after Insertion Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Greedy

If n is negative, we need to find the first position greater than x and insert x at that position. If n is positive, we need to find the first position less than x and insert x at that position.

The time complexity is O(m), where m is the length of n. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScript

class Solution: def maxValue(self, n: str, x: int) -> str: i = 0 if n[0] == "-": i += 1 while i < len(n) and int(n[i]) <= x: i += 1 else: while i < len(n) and int(n[i]) >= x: i += 1 return n[:i] + str(x) + n[i:](code-box)

class Solution { public String maxValue(String n, int x) { int i = 0; if (n.charAt(0) == '-') { ++i; while (i < n.length() && n.charAt(i) - '0' <= x) { ++i; } } else { while (i < n.length() && n.charAt(i) - '0' >= x) { ++i; } } return n.substring(0, i) + x + n.substring(i); } }(code-box)

class Solution { public: string maxValue(string n, int x) { int i = 0; if (n[0] == '-') { ++i; while (i < n.size() && n[i] - '0' <= x) { ++i; } } else { while (i < n.size() && n[i] - '0' >= x) { ++i; } } n.insert(i, 1, x + '0'); return n; } };(code-box)

func maxValue(n string, x int) string { i := 0 y := byte('0' + x) if n[0] == '-' { i++ for i < len(n) && n[i] <= y { i++ } } else { for i < len(n) && n[i] >= y { i++ } } return n[:i] + string(y) + n[i:] }(code-box)

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); }(code-box)

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 } }(code-box)

/** * @param {string} n * @param {number} x * @return {string} */ var maxValue = function (n, x) { 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); };(code-box)