1881. Maximum Value after Insertion

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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 $O(1)$.

Python3

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:]

Java

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);
    }
}

C++

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;
    }
};

Go

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:]
}

TypeScript

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);
}

Rust

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
    }
}

JavaScript

/**
 * @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);
};