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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

• Java
• C++
• Python
• Go
• Javascript

• 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('');
  };
  
  

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