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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

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 \(O(1)\).

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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:]

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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);
    }
}

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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;
    }
};

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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:]
}

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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);
}

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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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/**
 * @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);
};

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