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// Source : https://leetcode.com/problems/remove-k-digits/
// Author : Hao Chen
// Date : 2016-11-11
/***************************************************************************************
*
* Given a non-negative integer num represented as a string, remove k digits from the
* number so that the new number is the smallest possible.
*
* Note:
*
* The length of num is less than 10002 and will be ≥ k.
* The given num does not contain any leading zero.
*
* Example 1:
*
* Input: num = "1432219", k = 3
* Output: "1219"
* Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which
* is the smallest.
*
* Example 2:
*
* Input: num = "10200", k = 1
* Output: "200"
* Explanation: Remove the leading 1 and the number is 200. Note that the output must
* not contain leading zeroes.
*
* Example 3:
*
* Input: num = "10", k = 2
* Output: "0"
* Explanation: Remove all the digits from the number and it is left with nothing which
* is 0.
***************************************************************************************/
class Solution {
public:
string removeKdigits_pick(string& num, int k) {
int len = num.size();
string result;
int idx = 0;
for (int i=0; i < len - k; i++) {
int min_idx = idx;
for (int j=min_idx; j<=i+k; j++) {
if (num[min_idx] > num[j]) min_idx = j;
}
//don't put zero at the beginning
if ( !(result.empty() && num[min_idx]=='0') ){
result.push_back(num[min_idx]);
}
//select the number started from next one, to make the order correctness.
idx = min_idx + 1;
}
if (result.empty()) result = "0";
return result;
}
string removeKdigits_remove(string& num, int k) {
if ( num.size() <= k ) return "0";
int left_len = num.size() - k;
int idx = 0;
for (int i=0; i<k ;i++){
int len = num.size();
for (int j=0; j<num.size()-1; j++) {
//if the current is bigger than next one, then revmoe the current one.
//In other word, we always pick the smaller one number.
if ( num[j] > num[j+1] ) {
num.erase(j, 1);
idx = j;
break;
}
}
}
//remove all of ZEROs at the beginning.
for (int i=0; i<= num.size(); i++) {
if (num[i] != '0' || i == num.size()) {
num.erase(0, i);
break;
}
}
// if the digits in the array are sorted,
// then, we need remove the digits at the ends.
if (num.size() > left_len ) {
num.erase(num.begin() + left_len, num.end());
}
if (num.empty()) num = "0";
return num;
}
string removeKdigits(string num, int k) {
srand(time(0));
if (rand() % 2 ) {
return removeKdigits_pick(num, k);
} else {
return removeKdigits_remove(num, k);
}
}
};