MaximumValueAfterInsertion.cpp

// Source : https://leetcode.com/problems/maximum-value-after-insertion/
// Author : Hao Chen
// Date   : 2021-05-30

/***************************************************************************************************** 
 *
 * 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 '-'.
 ******************************************************************************************************/

class Solution {
public:
    string maxValue(string n, int x) {
        bool neg = false;
        if (n[0] == '-') neg = true;
        
        int i;
        
        for( i=neg?1:0; i<n.size(); i++){
            if (neg == true) {
                if ( n[i]-'0' > x) break;
            }else{ 
                if (n[i]-'0' < x) break; 
            }
        }
     
        n.insert(n.begin()+i, x+'0');
        return n;
    }
};