Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1881.html

1881. Maximum Value after Insertion

Level

Medium

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 <= 10^5
• 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 '-'.

Solution

If n is positive, find the leftmost digit in n that is less than x and insert x to the left of the digit. If no such digit is found, insert x at the end of n.

If n is negative, find the leftmost digit in n that is greater than x and insert x to the left of the digit. If no such digit is found, insert x and the end of n.

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

• class Solution {
      public String maxValue(String n, int x) {
          char xChar = (char) (x + '0');
          if (n.charAt(0) != '-')
              return maxValuePositive(n, xChar);
          else
              return maxValueNegative(n, xChar);
      }
  
      public String maxValuePositive(String n, char x) {
          StringBuffer sb = new StringBuffer(n);
          int length = n.length();
          int insertIndex = length;
          for (int i = 0; i < length; i++) {
              if (n.charAt(i) < x) {
                  insertIndex = i;
                  break;
              }
          }
          sb.insert(insertIndex, x);
          return sb.toString();
      }
  
      public String maxValueNegative(String n, char x) {
          StringBuffer sb = new StringBuffer(n);
          int length = n.length();
          int insertIndex = length;
          for (int i = 1; i < length; i++) {
              if (n.charAt(i) > x) {
                  insertIndex = i;
                  break;
              }
          }
          sb.insert(insertIndex, x);
          return sb.toString();
      }
  }
  
  ############
  
  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);
      }
  }
  

• // OJ: https://leetcode.com/problems/maximum-value-after-insertion/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      string maxValue(string n, int x) {
          x += '0';
          int i = n[0] == '-';
          if (n[0] == '-') {
              for (; i < n.size() && n[i] <= x; ++i) ;
          } else {
              for (; i < n.size() && n[i] >= x; ++i) ;
          }
          n.insert(begin(n) + i, x);
          return n;
      }
  };
  

• 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)
  
  ############
  
  # 1881. Maximum Value after Insertion
  # https://leetcode.com/problems/maximum-value-after-insertion/
  
  class Solution:
      def maxValue(self, word: str, x: int) -> str:
          
          if word[0] == '-':
              pos = 0
              for i, c in enumerate(word):
                  if c == '-': continue
                      
                  if x < int(c):
                      return word[:i] + str(x) + word[i:]
              
              return word + str(x)
          else:
              pos = 0
              for i, c in enumerate(word):
                  
                  if x > int(c):
                       return word[:i] + str(x) + word[i:]
              
              return word + 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('');
  };
  
  

All Problems

All Solutions