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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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