# Remove K Digits LeetCode Solution

Here, We see Remove K Digits LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.

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.

## Remove K Digits Leetcode SolutionC++

``````class Solution {
public:
string removeKdigits(string num, int k) {
string ans="";
for(char &c:num)
{
while(ans.size() && ans.back()>c &&k)
{
ans.pop_back();
k--;
}
if(ans.size()||c!='0')ans.push_back(c);
}
while(ans.size()&&k--)
{
ans.pop_back();
}
return (ans=="")?"0":ans;
}
};```Code language: PHP (php)```

## Remove K Digits Leetcode SolutionJava

``````class Solution {
public String removeKdigits(String num, int k) {
int len = num.length();
if(k==len)
return "0";
Stack<Character> stack = new Stack<>();
int i =0;
while(i<num.length()){
while(k>0 && !stack.isEmpty() && stack.peek()>num.charAt(i)){
stack.pop();
k--;
}
stack.push(num.charAt(i));
i++;
}
while(k>0){
stack.pop();
k--;
}
StringBuilder sb = new StringBuilder();
while(!stack.isEmpty())
sb.append(stack.pop());
sb.reverse();
while(sb.length()>1 && sb.charAt(0)=='0')
sb.deleteCharAt(0);
return sb.toString();
}
}```Code language: JavaScript (javascript)```

## Remove K Digits SolutionJavaScript

``````var removeKdigits = function(num, k) {
const stack = [];
let removed = 0;
for(let n of num) {
while(stack.length && n < stack[stack.length-1] && removed < k) {
stack.pop();
removed += 1;
}
stack.push(n);
}
while(removed < k) {
stack.pop();
removed += 1;
}
while(stack.length && stack[0] === '0') {
stack.shift();
}
return stack.length ? stack.join('') : '0';
};```Code language: JavaScript (javascript)```

## Remove K Digits SolutionPython

``````class Solution(object):
def removeKdigits(self, num, k):
stack = []
for n in num:
while( stack and int(stack[-1]) > int(n) and k):
stack.pop()
k -= 1
stack.append(str(n))
while(k):
stack.pop()
k -= 1
i = 0
while( i <len(stack) and stack[i] == "0" ):
i += 1
return ''.join(stack[i:]) if (len(stack[i:]) > 0) else "0"  ```Code language: HTML, XML (xml)```
Scroll to Top