# Java O(n*k) solution, 'n' is string length. how to make the code shorter?

• this solution works, but I don't like it since the lines are too long. How to make shorter ?
The thought:
Removing 'k' elements from the numeric string, then the digit on the maximum position must be among the first 'k+1' digits in the numeric string, then find the first smallest in the first 'k+1' digits, clear those before (index is smaller than) it, then mark the right element of the smallest one as the new start, removing "k - cleared"

``````public class Solution {
public String removeKdigits(String num, int k) {
if(num.length() <= k) {
return "0";
}
char[] chars = num.toCharArray();
char[] r = new char[chars.length - k];
int start = 0;
while(k > 0 && start < chars.length){
if(start + k == chars.length){
// remove all 'k' elements until the last
Arrays.fill(chars, start, chars.length, (char)0);
break;
}
// find the smallest element in the window
int windowSize = k + 1;
int idxSmallest = -1;
int smallest = Integer.MAX_VALUE;
for(int i = start; i < start + windowSize && i < chars.length; i++){
if(chars[i] < smallest){
smallest = chars[i];
idxSmallest = i;
}
}
// clear all those before the smallest element
Arrays.fill(chars, start, idxSmallest, (char)0);
k -= idxSmallest - start;
start = idxSmallest + 1;
}

StringBuffer sb = new StringBuffer();
boolean firstNonZeroFound = false;  // introducing this var to avoid heading '0' in result
for(int i = 0; i < chars.length; i++){
if(chars[i] != 0) {
if(!firstNonZeroFound && chars[i] != '0'){
firstNonZeroFound = true;
}

if(firstNonZeroFound){
sb.append(chars[i]);
}
}
}
String result = sb.toString();
return result.length() == 0 ? "0" : result;
}
}
``````

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