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402. Remove K Digits

Description

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.

 

Constraints:

• 1 <= k <= num.length <= 105
• num consists of only digits.
• num does not have any leading zeros except for the zero itself.

Solutions

Solution 1: Greedy Algorithm

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public String removeKdigits(String num, int k) {
          StringBuilder stk = new StringBuilder();
          for (char c : num.toCharArray()) {
              while (k > 0 && stk.length() > 0 && stk.charAt(stk.length() - 1) > c) {
                  stk.deleteCharAt(stk.length() - 1);
                  --k;
              }
              stk.append(c);
          }
          for (; k > 0; --k) {
              stk.deleteCharAt(stk.length() - 1);
          }
          int i = 0;
          for (; i < stk.length() && stk.charAt(i) == '0'; ++i) {
          }
          String ans = stk.substring(i);
          return "".equals(ans) ? "0" : ans;
      }
  }
  

• class Solution {
  public:
      string removeKdigits(string num, int k) {
          string stk;
          for (char& c : num) {
              while (k && stk.size() && stk.back() > c) {
                  stk.pop_back();
                  --k;
              }
              stk += c;
          }
          while (k--) {
              stk.pop_back();
          }
          int i = 0;
          for (; i < stk.size() && stk[i] == '0'; ++i) {
          }
          string ans = stk.substr(i);
          return ans == "" ? "0" : ans;
      }
  };
  

• class Solution:
      def removeKdigits(self, num: str, k: int) -> str:
          stk = []
          remain = len(num) - k
          for c in num:
              while k and stk and stk[-1] > c:
                  stk.pop()
                  k -= 1
              stk.append(c)
          return ''.join(stk[:remain]).lstrip('0') or '0'
  
  

• func removeKdigits(num string, k int) string {
  	stk, remain := make([]byte, 0), len(num)-k
  	for i := 0; i < len(num); i++ {
  		n := len(stk)
  		for k > 0 && n > 0 && stk[n-1] > num[i] {
  			stk = stk[:n-1]
  			n, k = n-1, k-1
  		}
  		stk = append(stk, num[i])
  	}
  
  	for i := 0; i < len(stk) && i < remain; i++ {
  		if stk[i] != '0' {
  			return string(stk[i:remain])
  		}
  	}
  	return "0"
  }
  

• function removeKdigits(num: string, k: number): string {
      let nums = [...num];
      while (k > 0) {
          let idx = 0;
          while (idx < nums.length - 1 && nums[idx + 1] >= nums[idx]) {
              idx++;
          }
          nums.splice(idx, 1);
          k--;
      }
      return nums.join('').replace(/^0*/g, '') || '0';
  }
  
  

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